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Samacheer, Kalvi, 10th, sslc, Science, Solutions, Chapter 6, Nuclear Physics, tamilnadu board, Physics,

Question 1.
Identify A, B, C, and D from the following nuclear reactions.

Solution:


Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 2

A is alpha particle, B is neutron, C is proton and D is electron.

Question 2.
A radon specimen emits radiation of 3.7 × 103 GBq per second. Convert this disintegration in terms of a curie, (one curie = 3.7 × 1010 disintegration per second)
Solution:
1 Bq = one disintegration per second
one curie = 3.7 × 1010 Bq

Question 3.
92U235 experiences one α – decay and one β – decay. Find the number of neutrons in the final daughter nucleus that is formed.
Solution:
Let X and Y be the resulting nucleus after the emission of the alpha and beta particles respectively.

Number of neutrons = Mass number – Atomic number = 231 – 91 = 140.

Question 4.
Calculate, the amount of energy released when a radioactive substance undergoes fusion and results in a mass defect of 2 kg.
Solution:
Mass defect in the reaction (m) = 2 kg
Velocity of light (c) = 3 × 108 ms-1
By Einstein’s equation,
Energy released E = mc2
= 2 × (3 × 108)2
= 1.8 × 1017 J.

Samacheer Kalvi 10th Science Nuclear Physics Textual Evaluation

I. Choose the correct answer

Question 1.
Man – made radioactivity is also known as _____.
(a) Induced radioactivity
(b) Spontaneous radioactivity
(c) Artificial radioactivity
(d) (a) & (c).
Answer:
(d) (a) & (c).

Question 2.
Unit of radioactivity is:
(a) roentgen
(b) curie
(c) becquerel
(d) all the above
Answer:
(d) all the above

Question 3.
Artificial radioactivity was discovered by _____.
(a) Becquerel
(b) Irene Curie
(c) Roentgen
(d) Neils Bohr.
Answer:
(b) Irene Curie

Question 4.
In which of the following, no change in mass number of the daughter nuclei takes place:
(i) a decay;
(ii) P decay
(iii) y decay
(iv) neutron decay
(a) (i) is correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Answer:
(b) (ii) and (iii) are correct

Question 5.
_____ isotope is used for the treatment of cancer.
(a) Radio Iodine
(b) Radio Cobalt
(c) Radio Carbon
(d) Radio Nickel.
Answer:
(b) Radio Cobalt

Question 6.
Gamma radiations are dangerous because:
(a) it affects eyes and bones
(b) it affects tissues
(c) it produces genetic disorder
(d) it produces an enormous amount of heat
Answer:
(c) it produces genetic disorder

Question 7.
_____ aprons are used to protect us from gamma radiations.
(a) Lead oxide
(b) Iron
(c) Lead
(d) Aluminium.
Answer:
(c) Lead

Question 8.
Which of the following statements is / are correct?
(i) α particles are photons
(ii) Penetrating power of γ radiation is very low
(iii) Ionization power is maximum for α rays
(iv) Penetrating power of γ radiation is very high
(a) (i) & (ii) are correct
(b) (ii) & (iii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct.
Answer:
(d) (iii) & (iv) are correct.

Question 9.
Proton-Proton chain reaction is an example of:
(a) Nuclear fission
(b) α – decay
(c) Nuclear fusion
(d) β – decay
Answer:
(c) Nuclear fusion

Question 10.
In the nuclear reaction X126⟶α decay zYA, the value of A & Z.
(a) 8, 6
(b) 8, 4
(c) 4, 8
(d) cannot be determined with the given data.
Answer:
(c) 4, 8

Question 11.
Kamini reactor is located at _____.
(a) Kalpakkam
(b) Koodankulam
(c) Mumbai
(d) Rajasthan.
Answer:
(a) Kalpakkam

Question 12.
Which of the following is/are correct?
(i) Chain reaction takes place in a nuclear reactor and an atomic bomb.
(ii) The chain reaction in a nuclear reactor is controlled.
(iii) The chain reaction in a nuclear reactor is not controlled.
(iv) No chain reaction takes place in an atom bomb.
(a) (i) only correct
(b) (i) & (ii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct
Answer:
(b) (i) & (ii) are correct

II. Fill in the blanks

Question 1.
One roentgen is equal to ______ disintegrations per second?
Answer:
3.7 × 1010.

Question 2.
Positron is an _____.
Answer:
antiparticle of electron.

Question 3.
Anaemia can be cured by _____ isotope.
Answer:
Radio iron (Fe59).

Question 4.
Abbreviation of ICRP _____.
Answer:
International Commission on Radiological Protection.

Question 5.
_____ is used to measure the exposure rate of radiation in humans.
Answer:
Roentgen.

Question 6.
_____ has the greatest penetration power.
Answer:
Gamma ray.

Question 7.
zYA→Z+1YA+X; Then X is _____.
Answer:
−1e0 (β decay).

Question 8.
zXA→YAZ This reaction is possible in _____ decay.
Answer:
Gamma (γ).

Question 9.
The average energy released in each fusion reaction is about _____ J.
Answer:
3.84 × 10-12.

Question 10.
Nuclear fusion is possible only at an extremely high temperature of the order of _____ K.
Answer:
107 to 109.

Question 11.
The radioisotope of _____ helps to increase the productivity of crops.
Answer:
phosphorous (P – 32).

Question 12.
If radiation exposure is 100 R, it may cause _____.
Answer:
fatal disease.

III. State whether the following statements are true or false: If false, correct the statement

Question 1.
Plutonium -239 is a fissionable material.
Answer:
True.

Question 2.
Elements having an atomic number greater than 83 can undergo nuclear fusion.
Answer:
False.
Correct Statement: Elements having an atomic number greater than 83 can undergo nuclear fusion.

Question 3.
Nuclear fusion is more dangerous than nuclear fission.
Answer:
False.
Correct Statement: Nuclear fission is more dangerous than nuclear fusion. Because the average energy released in fission (3.2 × 10-11 J) process is more than the average energy released in fusion (3.84 × 10-12 J).

Question 4.
Natural uranium U-238 is the core fuel used in a nuclear reactor.
Answer:
False.
Correct Statement: U-238 is not a fissile material but are abundant in nature. But in a reactor, this can be converted into a fissile material Pu239 and U233. Only fissile materials are used in the fuel of a nuclear reactor.

Question 5.
If a moderator is not present, then a nuclear reactor will behave like an atom bomb.
Answer:
True.

Question 6.
During one nuclear fission on an average, 2 to 3 neutrons are produced.
Answer:
True.

Question 7.
Einstein’s theory of mass-energy equivalence is used in nuclear fission and fusion.
Answer:
True.

IV. Match the following

Question 1.

1. BARC(a) Kalpakkam
2. India’s first atomic power station(b) Apsara
3. IGCAR(c) Mumbai
4. The first nuclear reactor in India(d) Tarapur

Answer:
1. (c) Mumbai
2. (d) Tarapur
3. (a) Kalpakkam
4. (b) Apsara

Question 2.

1. Fuel(a) lead
2. Moderator(b) heavy water
3. Coolant(c) Graphite
4. Shield(d) Uranium

Answer:
1. (d) uranium
2. (c) Graphite
3. (b) heavy water
4. (a) lead

Question 3.

1. Soddy Fagan(a) Natural radioactivity
2. Irene Curie(b) Displacement law
3. Henry Becquerel(c) Mass energy equivalence
4. Albert Einstein(d) Artificial Radioactivity

Answer:
1. (b) Displacement law
2. (d) Artificial Radioactivity
3. (a) Natural radioactivity
4. (c) Mass energy equivalence

Question 4.

1. Uncontrolled fission Reaction(a) Hydrogen Bomb
2. Fertile material(b) Nuclear Reactor
3. Controlled fission Reaction(c) Breeder reactor
4. Fusion reaction(d) Atom bomb

Answer:
1. (d) Atom bomb
2. (c) Breeder reactor
3. (b) Nuclear Reactor
4. (a) Hydrogen Bomb

Question 5.

1. Co – 60(a) Age of fossil
2. I – 13(b) Function of Heart
3. Na – 24(c) Leukaemia
4. C – 14(d) Thyroid disease

Answer:
1. (c) Leukemia
2. (d) Thyroid disease
3. (b) Function of Heart
4. (a) Age of fossil

V. Arrange the following in the correct sequence

Question 1.
Arrange in descending order, on the basis of their penetration power.

  1. Alpha rays
  2. Beta rays
  3. Gamma rays
  4. Cosmic rays.

Answer:

  1. Gamma rays
  2. Beta rays
  3. Alpha rays
  4. Cosmic rays.

Question 2.
Arrange the following in the chronological order of discovery.

  1. A nuclear reactor
  2. Radioactivity
  3. Artificial radioactivity
  4. Discovery of radium.

Answer:

  1. Radioactivity (1896)
  2. Discovery of radium (1898)
  3. Artificial radioactivity (1934)
  4. Nuclear reactor (1942).

VI. Use the analogy to fill in the blank

Question 1.
Spontaneous process : Natural Radioactivity, Induced process: _____.
Answer:
Artificial radioactivity
(or)
Man – made activity.

Question 2.
Nuclear Fusion : Extreme temperature, Nuclear Fission: _____.
Answer:
Room temperature.

Question 3.
Increasing crops : Radio phosphorous, Effective functioning of heart: _____.
Answer:
Radio sodium (Na24).

Question 4.
Deflected by electric field : α ray, Null Deflection: _____.
Answer:
γ ray (Gamma – ray).

VII. Numerical Problems

Question 1.
88Ra226 experiences three α-decay. Find the number of neutrons in the daughter element.
Solution:
88Ra226 consider as a parent element that is 88X226 and their daughter element is zYA
According to α decay process,
88X26⟶3α decay 82214+3α decay
During the 3α decay, the atomic number decreases by 6 and mass number decreases by 12.
So the number of neutrons in the daughter element
N = A – Z
N = 214 – 88 = 126
Number of neutrons in the daughter element N = 126.

Question 2.
A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 1010 Bq).
Solution:
Cobalt specimen emits induced radiation = 75.6 millicurie per second
(1 curie = 3.7 × 1010 Bq)
So 75.6 millicurie = 75.6 × 103 × 1 curie
= 75.6 × 10-3 × 3.7 × 1010 Bq
= 279.72 × 107
= 2.7972 × 109 Bq
75.6 millicurie per second is equivalent to 2.7972 × 109 Bq.

VIII. Assertion and Reason Type Questions

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: A neutron impinging on U235, splits it to produce Barium and Krypton.
Reason: U-235 is a fissile material.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion: In a β – decay, the neutron number decreases by one.
Reason: In β – decay atomic number increases by one.
Answer:
(d) The assertion is false, but the reason is true.
Explanation: In β – decay there is no change in the mass number of the daughter nucleus but the atomic number increases by one.

Question 3.
Assertion: Extreme temperature is necessary to execute nuclear fusion.
Reason: In nuclear fusion, the nuclei of the reactants combine releasing high energy.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 4.
Assertion: Control rods are known as ‘Neutron seeking rods’
Reason: Control rods are used to perform a sustained nuclear fission reaction.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Control rods are used to control the number of neutrons in order to have a sustained the chain reaction. They absorb the neutrons, (they seeking the neutrons)

IX. Answer in one or two words (VSA)

Question 1.
Who discovered natural radioactivity?
Answer:
Henri Becquerel was discovered natural radioactivity.

Question 2.
Which radioactive material is present in the ore of pitchblende?
Answer:
Uranium

Question 3.
Write any two elements which are used for inducing radioactivity?
Answer:

  1. Boron and Aluminium.
  2. Alpha particle and neutron.

Question 4.
Write the name of the electromagnetic radiation which is emitted during a natural radioactivity.
Answer:
Gamma rays

Question 5.
If A is a radioactive element which emits an α-particle and produces 104Rf259. Write the atomic number and mass number of the element A.
Answer:
In α decay
zXAα decay 263×z−2YA−4+2He4(α decay )106X263⟶α decay 104Rf259+2He4
In element A having atomic number is 106 and mass number is 263.

Question 6.
What is the average energy released from a single fission process?
Answer:
The average energy released from a single fission process is about 3.2 × 10-11 J.

Question 7.
Which hazardous radiation is the cause for the genetic disorders (or) effect?
Answer:
Radioactive radiations

Question 8.
What is the amount of radiation that may cause the death of a person when exposed to it?
Answer:
When the body is exposed to about 600 R, it leads to death.

Question 9.
When and where was the first nuclear reactor built?
Answer:
The first nuclear reactor was built in 1942 in Chicago, USA.

Question 10.
Give the SI unit of radioactivity.
Answer:
Becquerel

Question 11.
Which material protects us from radiation?
Answer:
Lead coated aprons and lead gloves should be used while working with the hazardous area. These materials are used to protects us from radiation.

X. Answer the following questions in a few sentences.

Question 1.
Write any three features of natural and artificial radioactivity.
Answer:

Natural radioactivityArtificial radioactivity
1. Emission of radiation due to the self-disintegration of a nucleus.1. Emission of radiation due to the disintegration of a nucleus through the induced process.
2. Alpha, Beta and Gamma radiations are emitted.2. Mostly elementary particles such as neutron, positron, etc. are emitted.
3. It is a spontaneous process.3. It is an induced process.

Question 2.
Define critical mass.
Answer:
The minimum mass of fissile material necessary to sustain the chain reaction is called ‘critical mass (mc). It depends on the nature, density and the size of the fissile material.

Question 3.
Define One roentgen.
Answer:
One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 4.
State Soddy and Fagan’s displacement law.
Answer:
During a radioactive disintegration, the nucleus which undergoes disintegration is called a parent nucleus and that which remains after the disintegration is called the daughter nucleus.

Question 5.
Give the function of control rods in a nuclear reactor.
Answer:
Control rods are used to control the number of neutrons in order to have sustained chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.

Question 6.
In Japan, some of the newborn children are having congenital diseases. Why?
Answer:
During the Second World War American, a bomber dropped the nuclear weapons over the Japanese cities of Hiroshima and Nagasaki. In the explosion of the atomic bomb to release the high energy dangerous radiation. In the explosion period, Japanese peoples are affected by radiation. This is the reason in Japan, some of the newborn children are having congenital diseases.

Question 7.
Mr Ramu is working as an X – ray technician in a hospital. But, he does not Wear the lead aprons. What suggestion will you give to Mr Ramu?
Answer:
X – rays have a destructive effect on living tissue. When the human body is exposed to X – rays, it causes redness of the skin, sores and serious injuries to the tissues and glands. They destroy the white corpuscles of the blood. If you don’t wear the lead aprons these kinds of diseases formed in your body. In my suggestion, you must wear lead aprons.

Question 8.
What is stellar energy?
Answer:
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as stellar energy.

Question 9.
Give any two uses of radioisotopes in the field of agriculture?
Answer:

  • The radioisotope of phosphorus (P – 32) helps to increase the productivity of crops.
  • The radiations from the radioisotopes can be used to kill the insects and parasites and prevent the wastage of agricultural products.

XI. Answer the following questions in detail.

Question 1.
Explain the process of controlled and uncontrolled chain reactions.
Answer:
(a) Controlled chain reaction

  • In the controlled chain reaction, the number of neutrons released is maintained to be one. This is achieved by absorbing the extra neutrons with a neutron absorber leaving only one neutron to produce further fission.
  • Thus, the reaction is sustained in a controlled manner. The energy released due to a controlled chain reaction can be utilized for constructive purposes.
  • The controlled chain reaction is used in a nuclear reactor to produce energy in a sustained and controlled manner.

(b) Uncontrolled chain reaction:

Question 2.
Compare the properties of Alpha, Beta and Gamma radiations.
Answer:

Propertiesα raysβ raysγ rays
What are they?Helium nucleus (2He4)consisting of two protons and two neutrons.They are electrons (−1e0), basic elementary particle in all atoms.They are electromagnetic waves consisting of photons.
ChargePositively charged particles. Charge of each alpha particle = +2eNegatively charged particles. Charge of each beta particle = -eNeutral particles. Charge of each gamma particle = zero
Ionising Power100 time greater than β rays and 10,000 times greater than γ raysComparatively lowVery less ionization power
Penetrating powerLow penetrating power (even stopped by a thick paper)Penetrating power is greater than that of α rays. They can penetrate through a thin metal foil.They have a very high penetrating power greater than that of β rays. They can penetrate through thick metal blocks.
Effect of an electric and magnetic fieldDeflected by both the fields. (in accordance with Fleming’s left-hand rule)Deflected by both the fields, but the direction of deflection is opposite to that for alpha rays. (in accordance with Fleming’s left-hand rule)They are not deflected by both the fields.
SpeedTheir speed ranges from 1/10 to 1/20 times the speed of light.Their speed can go up to 9/10 times the speed of light.They travel with the speed of light.

Question 3.
What is a nuclear reactor? Explain its essential parts with their functions.
Answer:
Nuclear reactor: A Nuclear reactor is a device in which the nuclear fission reaction takes place in a self – sustained and controlled manner to produce electricity.

Components of a Nuclear Reactor:
The essential components of a nuclear reactor are

  • Fuel: A fissile material is used as the fuel. The commonly used fuel material is uranium.
  • Moderator: A moderator is used to slow down the high energy neutrons to provide slow neutrons. Graphite and heavy water are commonly used moderators.
  • Control rod: Control rods are used to control the number of neutrons in order to have a sustained a chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.
  • Coolant: A coolant is used to remove the heat produced in the reactor core, to produce steam. This steam is used to run a turbine in order to produce electricity. Water, air and helium are some of the coolants.
  • Protection wall: A thick concrete lead wall is built around the nuclear reactor in order to prevent the harmful radiations from escaping into the environment.

XII. HOT Questions

Question 1.
Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
Answer:
Mass number A = 232
Atomic number Z = 90
Daughter element:
Mass number A = 208
Atomic number Z = 82
Difference in mass number = 232 – 208 = 24
Difference in atomic number
= 90 – 82 = 8
Atomic number of α = 2
Atomic number of β = -1
Mass number of α = 4
Mass number of β = 0
Difference in mass number in transformations
= 24
Number of a decays = 244 = 6
Difference in atomic number = 8
ΔZ = 6α + 4β
= 6(2) + 4(-1)
= 12 – 4
= 8
∴ Number of β decays = 4
∴ Number of α decays = 6
∴ Number of β decays = 4

Question 2.
‘X – rays should not be taken often’. Give the reason.
Answer:

  • Radiation does involve in X – rays tests and isotope scans (in nuclear medicine) are too low to cause immediate hazardous effects.
  • If should be taken often, X – ray radiation from medical examinations though slightly increases one’s risk for cancer which can occur year or decades after X-ray exposure.

Question 3.
Cell phone towers should be placed far away from the residential area. why?
Answer:

  1. Living near a cell phone tower is not healthy. There is multiple health risks associated with living near a cell phone tower.
  2. Cell phone towers communicate by use pulsed microwave signals (radiofrequency radiation) with each other.
  3. That is the reason cell phone towers should be placed far away from the residential area.

Samacheer Kalvi 10th Science Nuclear Physics Additional Questions

I. Choose the best Answer.

Question 1.
Radium was discovered by _____.
(a) Marie curie
(b) Irene curie
(c) Henri Becquerel
(d) F. Joliot.
Answer:
(a) Marie Curie

Question 2.
How many radioactive substances discovered so far?
(a) 83
(b) 92
(c) 43
(d) 29
Answer:
(d) 29

Question 3.
The SI unit of Radioactivity is _____.
(a) Curie
(b) Rutherford
(c) Becquerel
(d) Roentgen (R).
Answer:
(c) Becquerel

Question 4.
Radioactivity is _____.
(a) increases with increase in temperature
(b) increases with increase in pressure
(c) depends on the number of electrons
(d) purely a nuclear phenomenon.
Answer:
(d) purely a nuclear phenomenon

Question 5.
Which of the following processes is a spontaneous process?
(a) Artifical radioactivity
(b) Natural radioactivity
(c) Photoelectric effect
(d) Collisions
Answer:
(b) Natural radioactivity

Question 6.
The charge of the β rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
Answer:
(c) -e

Question 7.
The charge of the γ rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
Answer:
(b) 0

Question 8.
The atomic number of the elements that exhibit artifical radioactivity is:
(a) more than 82
(b) more than 83
(c) less than 83
(d) less than 82
Answer:
(c) less than 83

Question 9.
Arrange α, β, γ rays in the increasing order of their ionizing power.
(a) α, β, γ
(b) β, α, γ
(c) γ, β, α
(d) γ, α, β.
Answer:
(c) γ, β, α

Question 10.
Which produces a charge of 2.58 × 10-4Coulomb in 1 Kg of air?
(a) Curie
(b) Becquerel
(c) Rutherford
(d) Roentgen
Answer:
(d) Roentgen

Question 11.
Ionising power of the γ rays _____.
(a) Comparatively very high ionization power
(b) 100 times greater than the α rays
(c) 100 times greater than the β rays
(d) Comparatively very less ionization power.
Answer:
(d) Comparatively very less ionization power.

Question 12.
Ionization power maximum for _____.
(a) neutrons
(b) α particles
(c) γ rays
(d) β particles.
Answer:
(b) α particles

Question 13.
Charge of gamma particle is:
(a) +2e
(b) -e
(c) Zero
(d) +1e
Answer:
(c) Zero

Question 14.
Which has low penetrating power?
(a) α rays
(b) γ rays
(c) β rays
(d) X rays.
Answer:
(a) α rays

Question 15.
In β – decay _____.
(a) atomic number decreases by one
(b) the mass number decreases by one
(c) proton number remains the same
(d) neutron number decreases by one.
Answer:
(d) neutron number decreases by one

Question 16.
In which decay the energy level of the nucleus changes:
(a) α – decay
(b) β – decay
(c) γ – decay
(d) neutron decay
Answer:
(c) γ – decay

Question 17.
In γ – decay _____.
(a) atomic number decreases by one
(b) there is no change in atomic and mass number
(c) energy only changes in the decay process
(d) both (b) and (c).
Answer:
(d) both (b) and (c).

Question 18.
The unit of decay constant is _____.
(a) no unit
(b) second
(c) second-1
(d) curie.
Answer:
(c) second-1

Question 19.
The range of temperature required for nuclear fusion is from:
(a) 107 to 109 K
(b) 10-9 to 10-7 K
(c) 105 to 109
(d) 105 to 107 K
Answer:
(a) 107 to 109 K

Question 20.
1 Rd is equal to _____.
(a) 106 decay / second
(b) 1 decay / second
(c) 3.7 × 1010 becquerel
(d) 1.6 × 1012 decay / second.
Answer:
(a) 106 decay / second

Question 21.
An element ZXA successively undergoes three α decays and four β decays and gets converted an element Y are respectively _____.
(a) Z−6YA−12
(b) Z+2YA−12
(c) Z−2YA−12
(d) Z−10YA−12.
Answer:
(c) Z−2YA−12

Question 22.
In the nuclear reaction 88Ra226 → X + 2He4X is:
(a) 90Th234
(b) 91Pa234
(c) 86Rn222
(d) 88Rn226
Answer:
(d) 88Rn226

Question 23.
Which one of the following is used in the treatment of skin diseases _____.
(a) Na24
(b) I31
(c) Fe59
(d) P32.
Answer:
(d) P32.

Question 24.
Anaemia can be diagnosed by _____.
(a) 15P31
(b) 15P32
(c) 26P59
(d) 11P24.
Answer:
(c) 26P59

Question 25.
Which is used as a coolant?
(a) Graphite
(b) Liquid sodium
(c) Boron
(d) Cadmium
Answer:
(b) Liquid sodium

Question 26.
The energy released per fission is _____.
(a) 220 MeV
(b) 300 MeV
(c) 250 MeV
(d) 200 MeV.
Answer:
(d) 200 MeV.

Question 27.
In the reaction 1N14 + 0n1 → X + 1H1 X is:
(a) 15P30
(b) 6C14
(c) 6C12
(d) 11Na23
Answer:
(c) 6C12

Question 28.
Natural uranium consists of _____.
(a) 99.72 % of U-238
(b) 0.28 % of U-238
(c) 0.72 % of U-238
(d) 99.28 % of U-238.
Answer:
(d) 99.28 % of U-238.

Question 29.
The number of power reactors in India is _____.
(a) 14
(b) 12
(c) 7
(d) 2.
Answer:
(a) 14

Question 30.
In the nucleus of 11Na23 the number of protons and neutrons are:
(a) 12, 11
(b) 10, 12
(c) 11, 12
(d) 11, 23
Answer:
(c) 11, 12

Question 31.
The moderator used in nuclear reactor is _____.
(a) cadmium
(b) boron carbide
(c) heavy water
(d) uranium (92U235).
Answer:
(c) heavy water

Question 32.
The first nuclear reactor was built at _____.
(a) Kalpakkam, India
(b) Hiroshima, Japan
(c) Chicago, USA
(d) Trombay, Bombay.
Answer:
(c) Chicago, USA

Question 33.
Which of the following is used in the treatment of skin cancer?
(a) Radio Cobalt
(b) Radio gold
(c) Radio Cobalt and radio gold
(d) none of the above
Answer:
(c) Radio Cobalt and radio gold

Question 34.
The explosion of an atom bomb is based on the principle of _____.
(a) uncontrolled fission reaction
(b) fusion reaction
(c) controlled fission reaction
(d) none of the above.
Answer:
(a) uncontrolled fission reaction

Question 35.
The reactor in which no moderator used is _____.
(a) fast breeder reactor
(b) pressurised water reactor
(c) pressurised heavy water reactor
(d) boiled water reactor.
Answer:
(a) fast breeder reactor

Question 36.
The number of neutrons present in 92U235is:
(a) 133
(b) 143
(c) 43
(d) 243
Answer:
(b) 143

Question 37.
In fast breeder, the coolant system used is _____.
(a) heavy water
(b) light water
(c) liquid sodium
(d) boiled water.
Answer:
(c) liquid sodium

Question 38.
The only reactor in the world which uses U-233 as fuel is _____.
(a) Zerlina
(b) Purnima
(c) Kamini
(d) Tires.
Answer:
(c) Kamini

Question 39.
The temperature of the interior of Sun is about _____.
(a) 1.4 × 107 K
(b) 108 K
(C) 14 × 107 K
(d) 600 K.
Answer:
(a) 1.4 × 107 K

Question 40.
Total energy radiated by Sun is about _____.
(a) 3.6 × 1028 Js-1
(b) 3.8 × 1028 Js-1
(c) 3.8 × 1026 Js-1
(d) 3.8 × 1023 Js-1.
Answer:
(c) 3.8 × 1026 Js-1


II. Fill in the blanks

Question 1.
Cathode rays are discovered by _____.
Answer:
J.J. Thomson.

Question 2.
Positive rays discovered by _____.
Answer:
Goldstein.

Question 3.
The chargeless particles are called neutron, it was discovered by _____.
Answer:
James Chadwick.

Question 4.
Ernest Rutherford explained that the mass of an atom is concentrated in its central part called _____.
Answer:
Nucleus.

Question 5.
The radioactive elements emit harmful radiations are ____, ____, ____ rays.
Answer:
alpha, beta, gamma.

Question 6.
_____ is an spontaneous process.
Answer:
Natural radioactivity.

Question 7.
The element whose atomic number is more than 83 undergoes _____.
Answer:
spontaneous process.

Question 8.
______ radioactive material is present in the ore of pitchblende.
Answer:
Uranium.

Question 9.
_____ are the example of artificial (or) man-made radioactive elements.
Answer:
Boron, Aluminium.

Question 10.
The element whose atomic number is less than 83 undergoes _____.
Answer:
induced radioactivity.

Question 11.
______ is an controlled manner.
Answer:
Artificial radioactivity.

Question 12.
Spontaneous radioactivity is also known as _____.
Answer:
Natural radioactivity.

Question 13.
One Curie is equal to _____ disintegrations per second.
Answer:
3.7 × 1010

Question 14.
One Rutherford (Rd) is equal to ______ disintegrations per second.
Answer:
106

Question 15.
The radioactive displacement law is framed by _____.
Answer:
Soddy and Fajan.

Question 16.
During the α decay process, the atomic number is ______ by 2 and the mass number is decreases by _____.
Answer:
decreases, 4.

Question 17.
In β-decay the atomic number increases by ____ unit and mass number _____.
Answer:
One, remains the same.

Question 18.
In α radiation, the charge of each alpha particle is _____.
Answer:
+2e.

Question 19.
In γ radiation, the charge of each gamma particle is _____.
Answer:
Zero.

Question 20.
In radioactive radiation, which one is travel with the speed of light _____.
Answer:
Gamma radiation.

Question 21.
zYA→z−2YA−4+X; Then X is _____.
Answer:
2He4 (α decay).

Question 22.
zYA→zYA+X; Then X is _____.
Answer:
γ decay.

Question 23.
The average energy released in each fission process in about _____.
Answer:
3.2 × 10-11 J.

Question 24.
Fissionable material is a radioactive element, which undergoes fission in a sustained manner when it absorbs a _____.
Answer:
Neutron.

Question 25.
_____ isotope is used to detect the presence of block in blood vessels and also used for the effective functioning of the heart.
Answer:
Na24 – Radio sodium.

Question 26.
_____ is used to cure goitre.
Answer:
Radio Iodine – I131

Question 27.
_____ is used to diagnose anaemia and also to provide treatment for the same.
Answer:
Radio – iron (Fe59).

Question 28.
Radio cobalt (Co60) and radio gold (Au198) are used in the treatment of _____.
Answer:
Skin cancer.

Question 29.
_____ are used to sterilize the surgical devices as they can kill the germs and microbes.
Answer:
Radiations.

Question 30.
The age of the earth, fossils, old paintings and monuments can be determined by _____. technique.
Answer:
Radiocarbon dating.

Question 31.
When the body is exposed to about 600 R, it leads to _____.
Answer:
Death.

Question 32.
Radioactive materials should be kept in a thick – walled container of _____.
Answer:
Lead.

Question 33.
_____ is used to remove the heat produced in the reactor core, to produce steam.
Answer:
Coolant.

Question 34.
The abbreviation of BARC is _____.
Answer:
Bhabha Atomic Research Centre.

Question 35.
India’s 1st nuclear power station is _____.
Answer:
Tarapur Atomic Power Station.

Question 36.
The first nuclear reactor built in India was _____.
Answer:
Apsara.

Question 37.
The total nuclear power operating sites in India is _____.
Answer:
7

Question 38.
The energy released in a nuclear fission process is about ______
Answer:
200 Mev.

Question 39.
The number of 0n1 released on an average per fission is _____.
Answer:
2.5.

Question 40.
A hydrogen bomb is based on the principle of _____.
Answer:
Nuclear fusion.

III. Match the following

Question 1.

1. Natural radioactivity(a) 3.7 × 1010decay/second
2. Artificial radioactivity(b) spontaneous process
3. 1 curie(c) 106decay/second
4. 1 Rd (Rutherford)(d) induced process

Answer:
1. (b) spontaneous process
2. (d) induced process
3. (a) 3.7 × 1010 decay / second
4. (c) 106 decay / second

Question 2.

1. Charge of each α particle(a) γ ray
2. Charge of each β particle(b) +2e
3. Penetration power is maximum(c) α ray
4. Ionisation power is maximum(d) zero

Answer:
1. (b) +2e
2. (d) zero
3. (a) γ ray
4. (e) α ray

Question 3.

1. Deuterium(a) −1e0
2. Protium(b) 1H3
3. Tritium(c) 2H4
4. α – decay(d) 1H1
5. β – decay(e) 1H2

Answer:
1. (e) 1H2
2. (d) 1H1
3. (b) 1H3
4. (c) 2H4
5. (a) −1e0

Question 4.

1. Uranium core bomb(a) fusion bomb
2. Plutonium core bomb(b) fission bomb
3. Hydrogen bomb(c) Nagasaki
4. Atom bomb(d) Hiroshima

Answer:
1. (d) Hiroshima
2. (c) Nagasaki
3. (a) fusion bomb
4. (b) fission bomb

Question 5.

1. Radio iron (Fe59)(a) treatment of skin diseases
2. Radio phosphorous (P32)(b) smoke detector
3. Radio gold (Au198)(c) diagnose anaemia
4. An isotope of Americium (Am241)(d) treatment of skin cancer

Answer:
1. (c) diagnose anaemia
2. (a) treatment of skin diseases
3. (d) treatment of skin cancer
4. (b) smoke detector

IV. Arrange the following in the correct sequence

Question 1.
Arrange α, β, γ rays in ascending order, on the basis of their penetrating power?
Answer:
Ascending order:

  • Alpha (α)
  • Beta (β)
  • Gamma (γ)

Question 2.
Arrange in ascending and descending order, on the basis of their Ionisation power.
Alpha (α), Beta (β), Gamma (γ)
Answer:

  1. Ascending order: Gamma (γ), Beta (β), Alpha (α)
  2. Descending order: Alpha (α), Beta (β), Gamma (γ)

Question 3.
Arrange in ascending and descending order, on the basis of their biological effect.
Alpha (α), Gamma (γ), Beta (β)
Answer:

  1. Ascending order: Alpha (α), Beta (β), Gamma (γ)
  2. Descending order: Gamma (γ), Beta (β), Alpha (α).

V. Numerical Problems

Question 1.
92U238 emits 8α particles and 6β particles. What is the neutron / proton ratio in the product nucleus?
Solution:

Question 2.
The element with atomic number 84 and mass number 218 change to another element with atomic number 84 and mass number 214. The number of α and β particles emitted are respectively?
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 8
Number of alpha decay, x = 1
Number of beta decay, y = 2.

Question 3.
The number of α and β particles emitted in the nuclear reaction 90Th228⟶83Bi12are respectively.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics 9
Number of α decay, x = 4
Number of β decay, y = 1.

VI. Assertion and Reason Type Questions

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If Assertion is true, but the reason is false.
(d) If Assertion is false, but the reason is true.
(e) If the Assertion and reason both are false.

Question 1.
Assertion: All the radioactive element are ultimately converted in lead.
Reason: All the elements above lead are unstable.
Answer:
(c) If Assertion is true, but the reason is false.
Explanation: When they are converted into a lead, the emission is stopped because the nucleus of lead is stable (or lead is most stable elements in radioactive series)

Question 2.
Assertion: Among the alpha, beta and gamma-ray a particle has maximum penetrating power.
Reason: The alpha particle is heavier than beta and gamma rays.
Answer:
(e) If the Assertion and reason both are false.
Explanation: The penetrating power is maximum in case of gamma rays because gamma rays are electromagnetic radiation of very small wavelength.

Question 3.
Assertion: The ionising power of β – particle is less compared to α – particles but their penetrating power is more.
Reason: The mass of β-particle is less than the mass of α-particle
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: β – particle being emitted with very high speed compared to α – particle. Due to this reason, their loss of energy is very slow and they can penetrate the medium through a sufficient depth.

Question 4.
Assertion: Neutrons penetrate matter more readily as compared to protons.
Reason: Neutrons are slightly more massive than protons.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Neutron is about 0.1 % more massive than a proton. But the unique thing about the neutron is that while it is heavy, it has no charge (it is neutral). This lack of charge gives it the ability to penetrate matter without interacting as quickly as the beta particles or alpha particles.

Question 5.
Assertion: zXA undergoes a decays and the daughter product is z−2YA−4
Reason: In α – decay, the mass number decreases by 4 and atomic number decreases by.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: zXA⟶z−2XA−4+2He4(α decay)

Question 6.
Assertion: Moderator is used to slowing down the high energy neutrons to provide slow neutrons.
Reason: Cadmium rods are used as control rods.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Graphites and heavy water are commonly used moderators. This helps in moderator to slow down the fast neutrons.

Question 7.
Assertion: Alpha, beta and gamma radiations are emitted.
Reason: Nuclear fission process can be performed at room temperature.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: At room temperature, the nuclear fission process can perform breaking up of heavier nucleus into two smaller nuclei. In this process to emitted the alpha, beta and gamma radiations.

Question 8.
Assertion: An enormous amount of energy is released which is called stellar energy.
Reason: Fusion reaction that takes place in the cores of the Sun and other stars.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy.

Question 9.
Assertion: Artificial radioactivity is a controlled process.
Reason: It is a spontaneous process – natural radioactivity.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Artificial radioactivity is a controlled process. It is an induced process and man-made radioactivity.

Question 10.
Assertion: Gamma rays, penetrates through materials most effectively.
Reason: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation. This is a reason they can penetrate through materials most effectively.

VII. Answer the following questions

Question 1.
Define ‘Radioactivity’.
Answer:
The phenomenon of nuclear decay of certain elements with the emission of radiations like alpha, beta, and gamma rays is called ‘radioactivity’.

Question 2.
By whom radioactivity is detected in pitchblende?
Answer:
Marie curie and Purie curie.

Question 3.
Define ‘Artificial Radioactivity’.
Answer:
The phenomenon by which even light elements are made radioactive, by artificial or induced methods, is called ‘Artificial radioactivity’ or ‘Man – made radioactivity’.

Question 4.
Define ‘One curie’.
Answer:
It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of 1 g of radium 226.
Curie = 3.7 × 1010 disintegrations per second.

Question 5.
In which elements artifical radioactivity is induced?
Answer:
Boron and aluminum

Question 6.
What is alpha decay (α decay)? give an example.
Answer:
A nuclear reaction in which an unstable parent nucleus emits an alpha particle and forms a stable daughter nucleus is called ‘alpha decay’.
E.g. Decay of uranium (U238) to thorium (Th234) with the emission of an alpha particle.
92U238→90Th234+2He4 (α – decay).

Question 7.
What is beta decay (β decay)? Give an example?
Answer:
A nuclear reaction, in which an unstable parent nucleus emits a beta particle and forms a stable daughter nucleus, is called ‘beta decay’.
E.g. Beta decay of phosphorous.
15P32→16S32+−1e0 (β – decay)

Question 8.
What is gamma decay (γ decay)?
Answer:
In a γ – decay, only the energy level of the nucleus changes. The atomic number and mass number of the radioactive nucleus remain the same.

Question 9.
State the value of Roentgen in terms of Coulomb.
Answer:
Roentgen = 2.58 × 10-4 Coulomb in / kg of air.

Question 10.
Define ‘nuclear fission’ Give an example.
Answer:
The process of breaking (splitting) up of a heavier nucleus into two smaller nuclei with the release of a large amount of energy and a few neutrons are called ‘nuclear fission’.
E.g. Nuclear fission of a uranium nucleus (U235)
92U235+0n1→56Ba141+36Kr92+30n1+Q( energy )

Question 11.
Define ‘Nuclear fusion’ Give an example.
Answer:
The process in which two light nuclei combine to form a heavier nucleus is termed as ‘Nuclear fusion’.
E.g. 1H2+1H2→2He4+Q( Energy )

Question 12.
Write down the types of the nuclear reactor.
Answer:
Breeder reactor, fast breeder reactor, pressurized water reactor, pressurized heavy water reactor, boiling water reactor, water – cooled reactor, gas – cooled reactor, fusion reactor and thermal reactor are some types of nuclear reactors, which are used in different places worldwide.

Question 13.
What is the safe limit of receiving radioactive radiations?
Answer:
100 m R per week

VIII. Answer in the details:

Question 1.
Explain the principle and working of an atom bomb?
Answer:
Atom bomb:
(i) The atom bomb is based on the principle of the uncontrolled chain reaction. In an uncontrolled chain reaction, the number of neutrons and the number of fission reactions multiply almost in a geometrical progression.

(ii) This releases a huge amount of energy in a very small time interval and leads to an explosion.

Structure:
(i) An atom bomb consists of a piece of fissile material whose mass is subcritical. This piece has a cylindrical void.

(ii) It has a cylindrical fissile material which can fit into this void and its mass is also subcritical. When the bomb has to be exploded, this cylinder is injected into the void using a conventional explosive.

(iii) The two pieces of fissile material join to form the supercritical mass, which leads to an explosion. During this explosion, a tremendous amount of energy in the form of heat, light and radiation is released.

(iv) A region of very high temperature and pressure is formed in a fraction of a second along with the emission of hazardous radiation like y rays, which adversely affect the living creatures. This type of atom bombs was exploded in 1945 at Hiroshima and Nagasaki in Japan during World War II.

Question 2.
State and define the units of radioactivity.
Answer:
Curie : It is the traditional unit of radioactivity. It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of lg of radium 226. 1 curie = 3.7 × 1010 disintegrations per second.

Rutherford (Rd) : It is another unit of radioactivity. It is defined as the quantity of a radioactive substance, which produces 106 disintegrations in one second.
1 Rd = 106 disintegrations per second.

Becquerel (Bq) : It is the SI unit of radioactivity is becquerel. It is defined as the quantity of one disintegration per second.

Roentgen (R) : It is the radiation exposure of γ and x-rays is measured by another unit called roentgen. One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 3.
Write down the features of nuclear fission and nuclear fusion.
Answer:

Nuclear FissionNuclear Fusion
1. The process of breaking up (splitting) of a heavy nucleus into two smaller nuclei is called ‘nuclear fission’.1. Nuclear fusion is the combination of two lighter nuclei to form a heavier nucleus.
2. Can be performed at room temperature.2. Extremely high temperature and pressure are needed.
3. Alpha, beta and gamma radiations are emitted.3. Alpha rays, positrons, and neutrinos are emitted.
4. Fission leads to emission of gamma radiation. This triggers the mutation in the human gene and causes genetic transform diseases.4. Only light and heat energy are emitted.

Question 4.
Write down the medical and industrial application of radioisotopes?
Answer:

  1. Radio sodium (Na24) is used for the effective functioning of the heart.
  2. Radio – Iodine (I131) is used to cure goitre.
  3. Radio – Iron is (Fe59) is used to diagnose anaemia and also to provide treatment for the same.
  4. Radio Phosphorous (P32) is used in the treatment of skin diseases.
  5. Radio Cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
  6. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.
  7. Radio cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
  8. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.

Question 5.
Write a note about stellar energy.
Answer:
The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy. Where does this high energy come from? All-stars contain a large amount of hydrogen. The surface temperature of the stars is very high which is sufficient to induce fusion of the hydrogen nuclei.

Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as ‘stellar energy’. Thus, nuclear fusion or thermonuclear reaction is the source of light and heat energy in the Sun and other stars.

IX. Additional HOT Questions

Question 1.
Why is neutron so effective as bombarding particle?
Answer:
A neutron carries no charge. It easily penetrates even a heavy nucleus without being repelled or attracted by nucleus and electrons. So it serves as an ideal projectile for starting a nuclear reaction.

Question 2.
Is there any difference between electron and a beta particle.
Answer:
Basically, there is no difference between an electron and a beta particle. β particle is the name given to an electron emitted from the nucleus.

Question 3.
Why are the control rods made of cadmium?
Answer:
Cadmium has high cross – section for the absorption of neutrons.

Question 4.
Name two radioactive elements that are not found in observable quantities why is it so?
Answer:
Tritium and Plutonium are two radioactive elements that are not found in observable quantities in the universe.
It is because half-life period of each of two elements is very short compared to the age of the universe.


Tamilnadu, ssc, Physics, science, Chapter 5, Acoustics, Samacheer kalvi,

I. Choose the best answer.

Question 1.

Which of the following is correct?

(a) Rate of change of charge is electrical power.

(b) Rate of change of charge is current.

(c) Rate of change of energy is current.

(d) Rate of change of current is charge.

Answer:

(b) Rate of change of charge is current.

Question 2.

SI unit of resistance is:

(a) mho

(b) joule

(c) ohm

(d) ohm meter

Answer:

(c) ohm

Question 3.

In a simple circuit, why does the bulb glow when you close the switch?

(a) The switch produces electricity.

(b) Closing the switch completes the circuit.

(c) Closing the switch breaks the circuit.

(d) The bulb is getting charged.

Answer:

(b) Closing the switch completes the circuit

Question 4.

Kilowatt hour is the unit of:

(a) resistivity

(b) conductivity

(c) electrical energy

(d) electrical power

Answer:

(c) electrical energy

II. Fill in the blanks.

When a circuit is open, ……….. cannot pass through it.

The ratio of the potential difference to the current is known as ……….

The wiring in a house consists of ………… circuits.

The power of an electric device is a product of ……… and ………..

LED stands for ………..

Answer:

current

resistance

parallel

potential difference, current

Light Emitting Diode

III. State whether the following statements are true or false: If false correct the statement.

Ohm’s law states the relationship between power and voltage.

MCB is used to protect house hold electrical appliances.

The SI unit for electric current is the coulomb.

One unit of electrical energy consumed is equal to 1000 kilowatt hour.

The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances.

Answer:

False – Ohm’s law states that the relationship between current and voltage.

True

False – The SI unit for electric current is ampere.

False – One unit of electrical energy consumed is equal to 1 kilowatt hour.

False – The effective resistance of three resistors connected in series is greater than the 

highest of the “individual resistance.

IV. Match the items in column-1 to the items in column-ll.

Answer:

(i) – (e)

(ii) – (a)

(iii) – (b)

(iv) – (c)

(v) – (d)

V. Assertion and reason type Questions.

Mark the correct choice as

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

1. Assertion: Electric appliances with a metallic body have three wire connections.

Reason: Three pin connections reduce heating of the connecting wires.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

2. Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.

Reason: The current flows towards the point of the highest potential.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

3. Assertion: LED bulbs are far better than incandescent bulbs.

Reason: LED bulbs consume less power than incandescent bulbs.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

VI. Very short answer questions.

Question 1.

Define the unit of current.

Answer:

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,

1 ampere = 1coulomb1second

Question 2.

What happens to the resistance, as the conductor is made thicker?

Answer:

If the conductor is made a thicker area of cross-section of conduction increases that will decrease the resistance.

Question 3.

Why is tungsten metal used in bulbs, but not in fuse wires?

Answer:

Tungsten metal is used in bulbs because its melting point is the greatest.

But it is not used in fuse wires. When a current more than 5A flows tungsten wire will be melted. Hence tungsten is not used in fuse wire.

Question 4.

Name any two devices, which are working on the heating effect of the electric current.

Answer:

The heating effect of electric current is used in many home appliances such as electric iron and electric toaster.

VII. Short Answer Questions.

Question 1.

Define electric potential and potential difference.

Answer:

Electric Potential: It is the amount of work done in moving unit positive charge from infinity to that point against the electric force.

Electric potential

Potential difference : It is the amount of work done in moving a unit positive charge from one point to another against the electric force.

Potential difference VA – VB = WA−WBQ

Question 2.

What is the role of the earth wire in domestic circuits?

Answer:

The earth wire provides a low resistance path to the electric current.

The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of the metallic electric appliance.

Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.

Question 3.

State Ohm’s law.

Answer:

According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor.

I ∝ V. Hence, = 1V = constant.

The value of this proportionality constant is found to 1R

Therefore, I = (1R) V

V = IR

Question 4.

Distinguish between the resistivity and conductivity of a conductor.

Answer:

Question 5.

What connection is used in domestic appliances and why?

Answer:

In the domestic appliance, it is used as a parallel connection to avoid short circuit and breakage.

It has an alternative current (AC). Not DC current as it is from cables, so high potential flows through this.

One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

VIII. Long answer Questions.

Question 1.

With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected:

(a) in series and

(b) in parallel

Answer:

(a) Resistors in series : A series circuit connects the components one after the other to form a ‘single loop’. A series circuit has only one loop through which current can pass. If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work. Series circuits are commonly used in devices such as flashlights. Thus, if resistors are connected end to end, so that the same current passes through each of them, then they are said to be connected in series.

Let, three resistances R1, R2 and R3 be connected in series. Let the current flowing through theorem be I. According to Ohm’s Law, the potential differences V1,V2 and V3 across R1, R2 and R3 respectively, are given by:

V1 = I R1 ………. (1)

V2 = I R2 ……… (2)

v3 = I R3 ………. (3)

The sum of the potential differences across the ends of each resistor is given by:

V = V1 + V2 + V3

Using equations (1), (2) and (3), we get

V = I R1 + I R2 + I R3 …….. (4)

The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit. Let, the effective resistance of the series-combination of the resistors, be RS.

Then,

V = I RS ……….(5)

Combining equations (4) and (5), we get,

I RS = I R1 + I R2 + I R3

RS = R1 + R2 + R3 ……….. (6)

Thus, you can understand that when a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances. When ‘n’ resistors of equal resistance R are connected in series, the equivalent resistance is ‘n R’.

i.e., RS = n R

The equivalent resistance in a series combination is greater than the highest of the individual resistances.

(b) Resistors in Parallel : A parallel circuit has two or more loops through which current can pass. If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s). The wiring in a house consists of parallel circuits.

Consider that three resistors R1, R2 and R3 are connected across two common points A and B The potential difference across each resistance is the same and equal to the potential difference between A and B. This is me sured using the voltmeter. The current I arriving at A divides into three branches I1, I2 and I3 passing through R1, R2 and R3 respectively.

According to the Ohm’s law, you have,

The total current through the circuit is given by

I = I1 + I2 + I3

Using equations (1), (2) and (3), you get

I = VR1 + VR2 + VR3 ……… (4)

Let the effective resistance of the parallel combination of resistors be RP Then,

I = VRP ……… (5)

Combining equations (4) and (5), you have

Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance. When ‘n’ resistors of equal resistances R are connected in parallel, the equivalent resistance is Rn

The equivalent resistance in a parallel combination is less than the lowest of the individual resistances.

Question 2.

(a) What is meant by electric current? Give its direction?

(b) Name and define its unit.

(c) Which instrument is used to measure the electric current? How should it be r connected in a circuit?

Answer:

(a) (i) Electric current is often termed as ‘current’ and it is represented by the symbol ‘I’. It is defined as the rate of flow of charges in a conductor.

(ii) The electric current represents the number of charges flowing in any cross-section of a conductor (say a metal wire) in unit time.

(b) The SI unit of electric current is ampere (A).

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,

1 ampere = 1 coulomb 1 second 

(c) (i) The ammeter is used to measure the current.

(ii) An Ammeter is connected in series with the circuit.

(iii) The Ammeter is a low impedance device connecting it in parallel with the circuit would cause a short circuit, damaging the Ammeter or the circuit.

Question 3.

(a) State Joule’s law of heating.

(b) An alloy of nickel and chromium is used as the heating element. Why?

(c) How does a fuse wire protect electrical appliances?

Answer:

(a) Joule’s law of heating states that the heat produced in any resistor is:

directly proportional to the square of the current passing through the resistor.

directly proportional to the resistance of the resistor.

directly proportional to the time for which the current is passing through the resistor.

(b) Because,

It has high resistivity,

It has a high melting point,

It is not easily oxidized.

(c) When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage.

Question 4.

Explain about domestic electric circuits, (circuit diagram not required)

Answer:

Electricity is distributed through the domestic electric circuits wired by the electricians.

The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.

The important components of the main-box are: (i) a fuse box and (ii) a meter. The meter is used to record the consumption of electrical energy.

The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).

The function of the fuse wire or an MCB is to protect the household electrical appliances from overloading due to excess current.

Question 5.

(a) What are the advantages of LED TV over the normal TV?

(b) List the merits of LED bulb.

Answer:

(a) Advantages of LED TV:

It has brighter picture quality.

It is thinner in size.

It uses less power and consumes very less energy.

Its life span is more.

It is more reliable.

(b) Advantages of LED bulb:

As there is no filament, there is no loss of energy in the form of heat.

It is cooler than the incandescent bulb.

In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.

It is not harmful to the environment.

A wide range of colours is possible here.

It is cost-efficient and energy efficient.

Mercury and other toxic materials are not required. One way of overcoming the energy crisis is to use more LED bulbs.

IX. Numerical problems.

Question 1.

An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?

Answer:

(i) When heating is maximum, the power

P1 = 420 W

Applied voltage V = 220 V

P = VI

Current I = PV

I = 420220 = 1.909 A

I = 1.909 A

(ii) When heating is minimum

Power P2 = 180 W

Applied voltage V = 220 V

P = VI

∴ Current I = PV

I = 180220 = 0.8181 A

I = 0.8181 A

Question 2.

A 100-watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.

Solution:

100 W = 100 joules per second

1 watt hours = 3600 joules

The electric bulb is lighted for 5 hours daily,

100 W × 5 = 500 watt hours

500 watt hours = 1800000 joules

1 kWh = 3600000 joules

Units consumed per day = 18000003600000 = 0.5 units

Untis consumed in month = 0.5 × 31 = 15.5 units …. (1)

Now, Sum of power of four 60 watt bulbs = 240 W

240 W × 5 hours = 1200 watt hours

1200 watt hours = 4320000 joules

Energy consumed per day = 43200003600000 = 1.2 units

Energy consumed in a month = 1.2 × 31 = 37.2 units …. (2)

Total energy consumed in a month = 15.5 + 37.2 = 52.7 units

1 unit = 1 kWh

The energy consumed in the month of January = 52.7 kWh.

Question 3.

A torch bulb is rated at 3 V and 600 mA. Calculate it’s

(a) power

(b) resistance

(c) energy consumed if it is used for 4 hour.

Answer:

Voltage V = 3 V

Current I = 600 mA

(a) Power = VI

= 3 × 600 × 10-3

= 1800 × 10-3

= 1.8 W

(c) Time = 4h

Energy consumed E = P × t

E = 1.8 × 4

= 7.2 W

Question 4.

A piece of wire having a resistance R is cut into five equal parts.

(a) How will the resistance of each part of the wire change compared with the original resistance?

(b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?

(c) What will be ratio of the effective resistance in series connection to that of the parallel connection?

Answer:

(a) Original resistance

R = lA

∴ R α l

After cutting length of each piece

r = l5

New resistance

R’ = l′A

R’ α l’

R : R’ = 5 : 1

(b) When five parts of the wire are placed in parallel.

Effective Resistance

Resistance of the combinations

RP = R25

(c) When resistance are connected in series

RS = R

When resistance are connected in parallel

RP = R25

RS : RP = R : R25

= 25 R : R_1

RS : RP = 25 : 1

X. HOT Questions.

Question 1.

Two resistors when connected in parallel give the resultant resistance of 2 ohm; but when connected in series the effective resistance becomes 9 ohm. Calculate the value of each resistance.

Answer:

Let the resistance be R, and R2 when two resistances are connected in series

RS = R1 + R2

= 9

R1 + R2 = 9 ……….(1)

When two resistance are connected in parallel

Using (1) equation (2) becomes

9R1R2

R1 R2 = 18 ……….(3)

(R1 – R2)² = (R1 + R2)² – 4R1R2

= (9)² – 4 × 18

= 81 – 72 = 9

∴ (R1 – R2) = √9 = 3 ………(4)

From (1)

R1 + R2 = 9

2R1 = 12

∴ R1 = 122 = 6 ohm

From (1)

R2 = 9 – R1

= 9 – 6 = 3Ω

The values of resistances are

R1 = 6 ohm

R2 = 3 ohm

Question 2.

How many electrons are passing per second in a circuit in which there is a current of 5 A?

Solution:

Current I = 5A

Time (t) = 1 second

Charge of electron e = 1.6 × 10-19 C

I=qt=net

Number of electron, n = Ite = 5×11.6×10−19

n = 3.125 × 1019.

Question 3.

A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.

Answer:

Resistance R = 10Ω

Let l be the length of the wire R ∝ 1

When the length is increased to three times,

l’ = 3l

∴ New Resistance

R’ ∝ l’ ∝ 3l

∴ RR′ = l3l = 13

∴ R’ = 3R

New resistance = 3 times the original resistance.

Important Questions and Answers

I. Choose the best answer.

Question 1.

Electric current is defined as the rate of flow of:

(a) energy

(b) power

(c) mass

(d) charge

Answer:

(d) charge

Question 2.

The S.I. unit of electric current is _____.

(a) Volt

(b) Power

(c) Ampere

(d) newton.

Answer:

(c) Ampere

Question 3.

The unit of electric current is:

(a) ampere

(b) volt

(c) watt

(d) kilo-watt

Answer:

(a) ampere

Question 4.

The work done in moving a charge of 2 C across two points in a circuit is 2 J. What is the potential difference between the points?

(a) 1 V

(b) 10 V

(c) 100 V

(d) 0.

Answer:

(a) 1 V

Question 5.

The amount of work done to move a unit charge from one point to the other is:

(a) resistance

(b) current

(c) Potential

(d) none of the above

Answer:

(c) Potential

Question 6.

Ohm’s law gives the relative between potential difference and:

(a) emf

(b) temperature

(c) resistance

(d) current

Answer:

(d) current

Question 7.

The unit of resistance is _____.

(a) volt

(b) volt ampere-1

(c) ampere

(d) Joule.

Answer:

(b) volt ampere-1

Question 8.

The symbol of battery is:

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 14

Answer:

(b)

Question 9.

Electrical resistivity for a given material is ______.

(a) zero

(b) constant

(c) both (a) and (b)

(d) only (b).

Answer:

(b) constant

Question 10.

The potential difference required to pass a current 0.2 A in a wire of resistance 20 ohm is:

(a) 100 V

(b) 4 V

(c) 0.01 V

(d) 40 V

Answer:

(b) 4 V

Question 11.

The unit of electrical conductivity ______.

(a) ohm-1 metre

(b) ohm-1 metre-1

(c) Volt Ampere-1

(d) ohm.

Answer:

(b) ohm-1 metre-1

Question 12.

Kilowatt-hour is the unit of:

(a) potential difference

(b) electric power

(c) electric energy

(d) charge

Answer:

(c) electric energy

Question 13.

The resistivity of a material is 4 × 10-8 Ωm and its conductivity ______.

(a) 25 × 10-8 mho m-1

(b) 0.25 × 10-8 mho m-1

(c) 25 × 108 mho m-1

(d) 0.25 × 108 mho m-1.

Answer:

(d) 0.25 × 108 mho m-1.

Question 14.

The commonly used safely fuse wire is made of:

(a) nickel

(b) lead

(c) an alloy of tin and lead

(d) copper

Answer:

(c) an alloy of tin and lead

Question 15.

The value of one horse power is:

(a) 746 kW

(b) 746 W

(c) 3.6 × 105 W

(d) 3.6 × 106 kW

Answer:

(b) 746 W

Question 16.

When ‘n’ number of resistors are connected in series, the effecive resistance for series is ______.

(a) nR

(b) nR

(c) Rn

(d) none of these.

Answer:

(a) nR

Question 17.

Name the physical quantity which is measured in kW:

(a) electric energy

(b) electric power

(c) electric current

(d) electric potential

Answer:

(b) electric power

Question 18.

What is the amount of current, when 20 C of charges flows in 4 s through a conductor? [l = qv]

(a) 5 A

(b) 80 A

(c) 4 A

(d) 2 A

Answer:

(a) 5 A

Question 19.

Nichrome is ______.

(a) a conductor

(b) an insulator

(c) an alloy

(d) none of these.

Answer:

(c) an alloy

Question 20.

The main source of biomass energy is:

(a) coal

(b) heat energy

(c) thermal energy

(d) cow-dung

Answer:

(d) cow-dung

Question 21.

The value of one ampere is:

(a) 1second1coulomb

(b) 1 coulomb × sec

(c) 1coulomb1second

(d) 1 coulomb

Answer:

(c) 1coulomb1second

Question 22.

The heat produced in an electric heater of resistance 2 Ω is connected to an electric source, when a current of 6 A flows for 5 minutes _____.

(a) 216 × 102 J

(b) 2160 J

(c) 900 J

(d) 150 J.

Answer:

(a) 216 × 102 J

Hint: Formula H = I2Rt Joule.

Question 23.

The value of 1joule1coulomb

(a) 1 kWh

(b) 1 Wh

(c) ampere

(d) volt

Answer:

(d) volt

Question 24.

The mathematical from of Ohm’s law is given by:

(a) V = IR

(b) I = VR

(c) R = IV

(d) I = RV

Answer:

(a) V = IR

Question 25.

One kilowatt hour is _____.

(a) 3.6 × 106 J

(b) 36 × 106 J

(c) 3.6 × 105 J

(d) 36 × 105 J.

Answer:

(a) 3.6 × 106 J

Question 26.

If the length and radius of a conductor is doubled then its specific resistance will:

(a) be doubled

(b) be halved

(c) be tripled

(d) remain the same

Answer:

(d) remain the same

Question 27.

The value of resistivity of nichrome is:

(a) 1.5 × 106 Ωm

(b) 1.5 × 10-6 Ωm

(c) 5.1 × 106 Ωm

(d) 5.1 × 10-6 Ωm

Answer:

(b) 1.5 × 10-6 Ωm

Question 28.

Due to short circuit, effective resistance in the circuit becomes _____.

(a) large

(b) very small

(c) very large

(d) zero.

Answer:

(b) very small

Question 29.

if a conductor has a length of 1 m and area of 1 m² then its resistivity is equal to its:

(a) resistance

(b) conductance

(c) length

(d) conductivity

Answer:

(a) resistance

Question 30.

When resistors are connected in parallel, potential difference across each resistor will be:

(a) different

(b) same

(c) vary

(d) none of the above

Answer:

(b) same

Question 31.

LED TV screen was developed by James P. Mitchell in _____.

(a) 1797

(b) 1977

(c) 2009

(d) 1987.

Answer:

(b) 1977

Question 32.

Heat developed across a conductor H =

(a) IRt

(b) VR

(c) I2Rt

(d) I2R

Answer:

(c) I2Rt

Question 33.

Expression for electric energy is:

(a) W = VI

(b) W = VIt

(c) W = Vt

(d) W = VIt

Answer:

(b) W = VIt

Question 34.

In our home, fuse box consists of:

(a) fuse wire

(b) MCB

(c) fuse wire or MCB

(d) switches

Answer:

(c) fuse wire or MCB

Question 35.

Which of the following is a semi conductor device?

(a) LED bulb

(b) fuse

(c) MCB

(d) switch

Answer:

(a) LED bulb

II. Fill in the blanks.

1. The flow of charge: Electric current. A continuous closed path of an electric current is ………. The unit of charge: Coulomb then-current ……….

2. Electric current I: Charge (Q)/ ………. while electric potential V is ……….

3. A resistor of resistance R: Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 15 Then variable resistance and ……….. rheostat ……….

4. In series connection of resistors: ………. Then for parallel connection of resistors: ………..

5. The transformation of energy in Electric oven: ………… Electric cell ………..

6. The expression obtained from Ohm’s law ………… joule’s law

7. The unit of electric power …………. then electric energy ………..

8. The equivalent of 1 volt ………. then for 1 ohm ………..

9. The tap-key is used to ……… and ……….. an electric circuit.

10. The opposition to flow of current is called ………. and its unit is ………..

11. The heat developed in a conductor is directly proportional to the square of ………… and ………. of flow.

12. The S.I unit of electric current is ……….

13. The S.l unit of resistance is ……….

14. ……….. is the S.l unit of potential difference.

15. From Ohm’s law VI =

16. If a current 2A flows through conductor having a potential difference of 6 V then its resistance is ……….

17. If R is the resistance of a conductor then its conductance is G = ……….

18. Conductivity is ……… for ……….. than insulators.

19. When resistors are connected in series the equivalent resistance is …………. than the highest resistance of individual resistors.

20. In series connection ……….. is less as effective resistance is more.

21. Tungsten is used as heating element because its resistance is ……….

22. Tungsten is used as filament in the electric bulb because its melting points is …………

23. If a current of 6A flows through a 5Ω resistance for 10 minutes than heat developed in the resistance is ……….

24. When a current of IA flows through a conductor having potential difference of IV, the electric power is ………..

25. 746 watt is equivalent to ………..

26. In displays are used ………….

Answer:

1. Electric circuit, Ampere

2. Time (t), Work done (W)/charge (Q)

3. Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 16

4. Current is same, Potential difference is same

5. electrical into heat energy, chemical into electrical energy

6. V = IR, H = I2Rt

7. Kilowatt, want hour

8. 1joule1coulomb, 1volt1ampere

9. open, close

10. resistance, ohm

11. current, time

12. ampere

13. Ohm 15. R

14. Volt

15. R

16. 3 ohm

17. IR

18. more, conductors

19. greater

20. Current

21. high

22. high

23. 108 kJ

24. 1 W

25. 1 horse power

26. LED bulbs

III. State whether the following statements are true or false: If false correct the statement.

1. Current is the rate of flow of charges

2. The symbol of diode is

3. Potential = chargetime

4. Mathematical form of ohm’s law is V = IR

5. Nichrome is used in electric bulb.

6. The unit of conductance is mho.

7. The equivalent resistance in a parallel combination is less than the lowest of the individual resistance.

8. In parallel connection the effective resistance is RP = R1+R2R1R2

9. Heat produced in a conductor is H = l²Rt

10. 1 kWh = 3.6 J.

11. An MCB is a switching device.

12. LED means Light Emitting Diode.

Answer:

1. True

2. True

3. False – Potential = Workdone(W)Charge(Q)

4. True

5. False – Nichrome is used in heating device.

6. True

7. True

8. False – In parallel connection, the effective resistance is RP = R1R2R1+R2

9. True

10. False – 1 kWh = 3.6 × 106 J

11. True

12. True

IV. Match the items in column-1 to the items in column-ll.

Question 1.

Match the Column I with Column II.

Answer:

(i) – (d)

(ii) – (c)

(iii) – (a)

(iv) – (b)

Question 2.

Match the components with symbols

Answer:

(i) – (d)

(ii) – (c)

(iii) – (b)

(iv) – (a)

Question 3.

Match the Column I with Column II.

Answer:

(i) – (c)

(ii) – (d)

(iii) – (b)

(iv) – (a)

Question 4.

Match the Column I with Column II.

Answer:

(i) – (d)

(ii) – (c)

(iii) – (b)

(iv) – (a)

Question 5.

Match the column I with column II.

Answer:

(i) – (e)

(ii) – (d)

(iii) – (a)

(iv) – (b)

Question 6.

Match the column I with column II:

Answer:

(i) – (d)

(ii) – (e)

(iii) – (a)

(iv) – (b)

Question 7.

Match the column I with column II:

Answer:

(i) – (e)

(ii) – (a)

(iii) – (d)

(iv) – (c)

V. Assertion and reason type questions.

Question 1.

Assertion: In a series system, equivalent resistance is the sum of the individual resistance.

Reason: The current that passes through each resistor is the same.

(a) Assertion is true but Reason is false.

(b) Assertion is true and Reason doesn’t explains Assertion,

(c) Both Assertion and Reason are false.

(d) Assertion is true and Reason explains Assertion

Answer:

(d) Assertion is true and Reason explains Assertion

Question 2.

Assertion: In a parallel system, the total current is equal to the sum of the current through each resistor.

Reason: The potential difference across each resistor is the same.

(a) Assertion is true and Reason explains Assertion.

(b) Assertion is true and Reason doesn’t explains Assertion.

(c) Both Assertion and Reason are false

(d) Assertion is true but Reason is false.

Answer:

(a) Assertion is true and Reason explains Assertion.

Question 3.

Assertion: The unit of power watt is not frequently used in practice. Reason: it cannot be converted into Joule.

(a) Both Assertion and Reason are false.

(b) Assertion is true but Reason is false.

(c) Both Assertion and Reason are true and Reason explains Assertion.

(d) Both Assertion and Reason are true and Reason doesn’t explains Assertion.

Answer:

(b) Assertion is true but Reason is false.

Question 4.

Assertion: A wire carrying a current has electric field around d.

Reason: A wire carrying current is stays electrically neutral.

(a) If both the assertion and the reason a re true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 5.

Assertion: In order to pass current through electric circuit, it must be closed.

Reason: In our home, the switch is ON, then the current flows through the bulb. So, the bulb glows.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 6.

Assertion: Resistance of a material opposes the flow of charges.

Reason: It is different for different materials.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 7.

Assertion: Electrical conductivity is the reciprocal of electrical resistivity. Reason: The unit of conductivity is Ohm.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(c) If the assertion is true, but the reason is false.

Question 8.

Assertion: One end of the earthing wire is connected to a body of the electrical appliance and its other end is connected to a metal tube that is burried into the Earth.

Reason: The earth wire provides low resistance path to the electric current.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 9.

Assertion: The passage of electric current through a wire results in the production of heat.

Reason: The heating effect is used in electric heater electric iron etc.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 10.

Assertion: One kilowatt hour is known as one unit of electrical energy.

Reason: 1 kWh = 3.6 × 106J

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

VI. Very short answer type Questions

Question 1.

If a charge of QC flows through a conductor in time t second then what is the value of current?

Answer:

Charge = Chargetime

I = Qt

Question 2.

What is the electric circuit?

Answer:

An electric circuit is a closed conducting loop.

(or)

path, which has a network of electrical components through which electrons are able to flow. This path is made using electrical wires so as to connect an electric appliance to a source of electric charges (battery).

Question 3.

If the length of a wire is doubled and its cross-section is also doubled than what happens its resistance?

Answer:

Resistance of a wire

R = plA ………. (1)

l’ = 2l

A’ = 2A

∴ New Resistance

R’ = l × 2l2A

R’ = plA …….. (2)

Form (1) and (2) it is found that resistance remains unchanged.

Question 4.

What is the unit of resistance and resistivity of a conductor?

Answer:

(i) The unit of resistance is ohm.

(ii) The unit of resistivity is ohm meter.

Question 5.

Define the unit of resistance? (or) Define one ohm.

Answer:

The resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends.

 Ohm =1 volt 1 ampere .

Question 6.

What is the unit of conductivity?

Answer:

The unit of conductivity is mho meter.

Question 7.

What is the value of one kilo watt hour?

Answer:

One kilowatt hour = 1000 W hr

1 kWh = 3.6 × 106J

Question 8.

What is the heating effect of electric current?

Answer:

The passage of electric current through a wire resulting in the production of heat. This phenomenon is called the heating effect of current. This heating effect of current is used in devices like electric heater, electric iron, etc.

VII. Short Answer Questions.

Question 1.

What is an electric circuit?

Answer:

An electric circuit is a closed conducting loop (or) path, which has a network of electrical components through which electrons are able to flow.

Question 2.

Draw a circuit diagram to represent a simple electric circuit.

Answer:

Question 3.

What is the direction of conventional current?

Answer:

By convention, the direction of current is taken as the direction of flow of positive charge (or) opposite to the direction of flow of electrons.

Question 4.

Define electric potential.

Answer:

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

Question 5.

What is meant by overloading?

Answer:

(i) Overloading happens when a large number of appliances are connected in series to the same source of electric power. This leads to a flow of excess current in the electric circuit.

(ii) When the amount of current passing through a wire exceeds the maximum permissible limit, the wires get heated to such an extent that a fire may be caused. This is known as overloading.

Question 6.

What is meany by short circuit?

Answer:

When a live wire comes in contact with a neutral wire, it causes a ‘short circuit’.

This happens when the insulation of the wires get damaged due to temperature changes or some external force.

Due to a short circuit, the effective resistance in the circuit becomes very small, which leads to the flow of a large current through the wires.

This results in heating of wires to such an extent that a fire may be caused in the building.

Question 7.

Draw an electric circuit to understand Ohm’s law.

Answer:

Question 8.

Define resistance of a conductor.

Answer:

The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.

Question 9.

Define Resistance. Give its unit and conductance.

Answer:

The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.

Conductance: It is defined as the reciprocal of its resistance (R). Hence, the conductance ‘G’ of a conductor is given by

G=1R

Its unit is ohm-1. It is also represented as ‘mho’.

Question 10.

Define electrical resistivity of a material.

Answer:

The electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section. Its unit is ohm metre.

Question 11.

What is meant by electrical conductivity?

Answer:

The reciprocal of electrical resistivity of a material is called its electrical conductivity.

σ = 1p

Question 12.

Mention the differences between the combination of resistances in series and parallel.

Answer:

Question 13.

Write short notes about filament in electric bulbs.

Answer:

In electric bulbs, a small wire is used, known as filament. The filament is made up of a material whose melting point is very high. When current passes through this wire, heat is produced in the filament. When the filament is heated, it glows and gives out light. Tungsten is the commonly used material to make the filament in bulbs.

Question 14.

What is meant by electric power?

Answer:

The electric power is the product of the electric current and the potential difference due to which the current passes in a circuit.

Question 15.

What is meant by overloading of an electric circuit?

Answer:

When the amount of current passing through a wire exceeds the maximum permissible limit, the wires get heated to such an extent that a fire may be caused. This is known as overloading.

Question 16.

What is meant by LED bulb?

Answer:

An LED bulb is a semiconductor device that emits visible light when an electric current passes through it. The colour of the emitted light will depend on the type of materials used.

Question 17.

W hat is meant by seven segment display? state its uses.

Answer:

(i) A ‘Seven Segment Display’ is the display device used to give an output in the form of numbers or text.

(ii) It is used in digital meters, digital clocks, microwave ovens, etc.

Question 18.

What do you know about LED television.

Answer:

LED Television is one of the most important applications of Light Emitting Diodes. An LED TV is actually an LCD TV (Liquid Crystal Display) with LED display. An LED display uses LEDs for backlight and an array of LEDs act as pixels. LEDs emitting white light are used in monochrome (black and white) TV; Red, Green and Blue (RGB) LEDs are used in colour television.

Question 19.

What is fuse wire?

Answer:

The function of fuse wire or an MCB is to protect the household electrical appliances from excess current due to overloading or a short circuit.

Question 20.

Draw a group between potential difference and current.

Answer:

Question 21.

Write short note about short circuit?

Answer:

When a live wire comes in contact with a neutral wire, it causes a ‘short ‘ circuit’. This happens when the insulation of the wires get damaged due to temperature changes or some external force. Due to a short circuit, the effective resistance in the circuit becomes very small, which leads to the flow of a large current through the wires. This results in heating of wires to such an extent that a fire may be caused in the building.

VIII. Long answer questions.

Question 1.

Tabulate various components used in electrical circuit and their uses?

Answer:

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 29

Question 2.

Explain series connection of parallel resistors.

Answer:

If we consider the connection of a set of parallel resistors that are connected in series, you get a series – parallel circuit. Let R1 and R2 be connected in parallel to give an effective resistance of Rp1. Similarly, let R3, and R4 be connected in parallel to give an effective resistance of Rp2. Then, both of these parallel segments are connected in series. Using equation we get

We get = R1 + R2 + R3 ……….(1)

Finally, using equation (1), the net effective resistance is given by

Rtotal = RP1 + RP2

Question 3.

Explain parallel connection of series resistors.

Answer:

If you consider a connection of a set of series resistors connected in a parallel circuit, you get a parallel-series circuit. Let R1 and R2 be connected in series to give an effective resistance of RS1. Similarly, let R3 and R4 be connected in series to give an effective resistance of RS2. Then, both of these serial segments are connected in parallel.

Using equation

RS = R1 + R2 + R3

We get RS1 = R1 + R2, RS2 = R3 + R4

Finally, using equation

1RP = 1R1 + 1R2 + 1R3

the net effective resistance is given by

1Rtotal = 1RS1 + 1RS2

Question 4.

Explain applications of heating effect.

Answer:

Electric Heating Device: The heating effect of electric current is used in many home appliances such as electric iron, electric toaster, electric oven, electric heater, geyser, etc. In these appliances Nichrome, which is an alloy of Nickel and Chromium is used as the heating element. Because:

(i) it has high resistivity,

(ii) it has a highmelting point,

(Hi) it is not easily oxidized.

Fuse Wire : The fuse wire is connected in series, in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage. The fuse wire is made up of a material whose melting point is relatively low.

Filament in bulbs : In electric bulbs, a small wire is used, known as filament. The filament is made up of a material whose melting point is very high. When current passes through this wire, heat is produced in the filament. When the filament is heated, it glows and gives out light. Tungsten is the commonly used material to make the filament in bulbs.

Question 5.

Write short notes about

(i) LED bulb

(ii) Seven segment display

Answer:

(i) An LED bulb is a semiconductor device that emits visible light when an electric current passes through it. The colour of the emitted light will depend on the type of materials used. With the help of the chemical compounds like Gallium Arsenide and Gallium Phosphide, the manufacturer can produce LED bulbs that radiates red, green, yellow and orange colours. Displays in digital watches and calculators, traffic signals, street lights, decorative lights, etc., are some examples for the use of LEDs.

(ii) A ‘Seven Segment Display’ is the display device used to give an output in the form of numbers or text. It is used in digital meters, digital clocks, micro wave ovens, etc. It consists of 7 segments of LEDs in the form of the digit 8. These seven LEDs are named as a, b, c, d, e, f and g. An extra 8th LED is used to display a dot.

IX. Numerical problems.

Question 1.

An electric iron draws a current of 0.5 A when the voltage is 220 volts. Calculate the amount of electric charge flowing through it in one hour.

Answer:

I = Qt

Charge Q = ?

Charge I = 0.5 A

Time = 1 hour

= 60 × 60 s

Q = It

= 0.5 × 3600

= 1800 C

The amount of electric charge flowing = 1800 C

Question 2.

A current of 5A flows through a heater for 10 minutes. Calculate the amount of electric charge flowing through the electric circuit.

Answer:

Q = I × t

Current I = 5A

Time t = 10 × 60 = 600 s

= 5 × 600 = 3000 C

Question 3.

A torch bulb draws a current 0.6 A, when glowing from a source of 6 V. Calculate the resistance of the bulb when glowing.

Answer:

V = IR

Current I = 0.6A

Potential V = 6V

R = ?

R = VI = 60.6 = 10Ω

The resistance of the bulb = 10Ω

Question 4.

Find the potential difference required to pass a current of 0.2 A in a wire of resistant 20Ω.

Answer:

V = IR

Current I = 0.2A

Resistance R = 20Ω

Potential difference V = ?

V = 0.2 × 20 = 4V

The potential difference in a wire = 4V

Question 5.

Calculate the amount of work done in moving charge of 25 C across two points having potential difference of 20V.

Answer:

W = QV

Charge Q = 25 C

Potential difference V = 20 V

W = 25 × 20 = 500 J

Question 6.

Three resistances are connected in an electrical circuit as shown in the circuit diagram. Determine the potential difference across resistance R2.

Answer:

For series connection, the effective resistance

RS = R1 + R2 + R3

= 1 + 2 + 3 = 6Ω

Total potential = 12V

Amount of current flowing through the circuit

I = VRS = 12V6Ω = 2A

Potential difference across resistance R2

V = I × R2

= 2 × 2 = 4V

The potential difference across the resistance R2 = 4 V

Question 7.

In the given network, find the equivalent resistance between A and B.

Answer:

An equivalent of the given network is drawn in the relevant parts as follows: Resistance of the combination R1 and R2 is

RS = 5 + 5 = 10Ω

Resistance of the combination R1, R2 and R3 is

The resistance of series combination RP1 and R4 is RS1 = 5 + 5 = 10Ω

Resistance of the combination RS1 and R5 is

Resistance of the series combination RP2 and R6 is

RS2 = 5 + 5 = 10Ω

Resistance of the combination RS2 and R7 is

Resistance of the series combination RP3 and R8 is

RP3 = 5 + 5 = 10Ω

Resistance of the combination RS3 and R9 is

Resistance of the series combination RP4 and R6 is

RS4 = 5 + 5 = 10Ω

Resistance of the combination RS4 and R8 is

∴ Resistance between A and B is 5Ω

Question 8.

For a given circuit calculate

(i) the total effective resistance of the circuit.

(ii) the total current in the circuit

(iii) the current through each resistor.

Answer:

For parallel connection, the effective resistance

(i) Total effective resistance of the circuit

RP = 0.588Ω

(ii) Total current in the circuit

I = 10.2 A

(iii) Current through R1 = 6A

Current through R2 = 3A

Current through R3 = 1.2A

Question 9.

An electric iron a rating of 750 W, 220 V.

(i) Calculate current passing through it and

(ii) Its resistance when in use.

Answer:

P = VI

I = PV

Ohm’s law V = IR

(i) The amount of current passing

I = 3.4A

(ii) Resistance

R = 64.7Ω

Question 10.

Following graph was plotted between V and I values. What would be the values of VI ratios when the potential difference is 0.8 V and 1.2 V?

Answer:

V1 = 0.8V, I1 = 32A; V2 = 1.2V; I2 = 48A

Question 11.

Three resistors of 2Ω, 4Ω and 8Ω are connected in parallel with a battery of 3 V. Calculate

(i) Current through each resistor and

(ii) Total current in the circuit.

Answer:

Potential difference across each resistor is same.

(i) Current through each resistor:

(ii) Total current in the circuit: I = I1 + I2 + I3

I = 1.5 + 0.75 + 0. 375

I = 2.625 A

Question 12.

Two bulbs of 40 W and 60 W are connected in series to an external potential difference. Which bulb will glow brighter? Why?

Answer:

Let the external potential difference be 230 V

For 40 W bulb resistance is R

For 60 W bulb resistance is R

According to Ohm’s law

I = VR

Current flowing through 40 W bulb is

2301322.5 = 0.1739 A

Current flowing through 60 W bulb is

230881.6 = 0.2608 A

When bulbs are connected in series effective resistance is

RS = R1+ R2 = 1322.5 + 881.6

RS = 2204.1Ω

Net current

I = 2302204.1 = 0.1043 A

Using power equation P = I²R

For 40 W bulb P = I²R

= (0.1043)² × 1322.5

= (0.01087) × 1322.5 = 14.386 W

For 60 W bulb P = I²R

= (0.1043)² × 881.6

= (0.01087) × 881.6 = 9.5904 W

In a series system, higher the resistance, higher the brightness so, 40 W bulb glows brighter.

Question 13.

A wire is bent into a circle. The effective resistance across the diameter is 8Ω. Find the resistant if the wire.

Answer:

RP = 8Ω = resistance across diameter

1R + 1R = 18

2R = 18

R = 16Ω

x is the resistance of the wire.

x = 16 + 16 = 32Ω

x = 32Ω


X. HOT Questions.

Question 1.

A 60 W bulb is connected in parallel with a room heater. This combination is connected across the mains. If 60 W bulb is replaced by a 100 W bulb what happens to the heat produced by the heater? Given reason.

Answer:

Heat produced by the heater will be same. When the bulb and a heater are connected in parallel and this combination is connected across the mains, potential difference across each is the same equal to the voltage V of the mains irrespective of the resistance of the bulb.

If R is the resistance of the heater then heat produced by the heater will be V2R in both cases. Hence heat produces by heater will not be changed.

Question 2.

Two bulbs 60 W and 100 W are connected in series and this combination is connected to a d.c power supply. Will the potential difference across 60 W bulb be higher than that across 100 W bulb?

Answer:

60 W bulb has a higher resistance than the resistance across 100 W bulb since the power developed is P = V2R.

Potential difference across a bulb will be proportional to resistance. Hence potential difference across 60 W bulb is higher than that across 100 W bulb.

Question 3.

Super conductors has lowest resistance. Is it true. Give reason.

Answer:

True. When the temperature of super conductor is reduced to zero or near by zero its resistance becomes zero.

Question 4.

A constant voltage is applied between two ends of a uniform conducting wire. If both the length and radius of the wire is doubled then what happens to the heat produced in the wire?

Answer:

We know that resistance of a conducting wire is R = plA

Plπr2

If length l and radius r are doubled, then resistance will become half. But heat produced H = V2R.

Hence, heat produced per second will become thrice.

Question 5.

Calculate the effective resistance between A and B.

Answer:

Electricity 51

The electrical circuit can be redrawn as

Electricity 52

The resistance R1 and R2 are in series

RS = R1 + R2 = 2 + 2 = 4Ω

The resistance RS and R3 are in parallel

Electricity 53

∴ Effective resistance Reff= 1.33Ω

Question 6.

Two wires of same material and length have resistances 5Ω and 10Ω respectively. Calculate the ratio of radii of the two wires.

Answer:

Resistance

Electricity 54

∴ r1 : r2 = √2 : 1

Question 7.

An iron box of 400 W power is used daily for 30 minutes. If the cost per unit is 75 paise, find the weekly expense on using the iron box.

Answer:

Power P = 400 W

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 55

Energy consumed = Power × Time

= 400 × 12 = 200 Wh

Energy consumed in one week= 200 × 7

= 1400 Wh = 1.4 kWh

= 1.4 unit

∴ Total cost per week = 1.4 × 0.75 = Rs 1.05

Weekly expense = Rs 1.05


Tamilnadu, ssc, Physics, science, chapter 4, Electricity,  Samacheer kalvi,

1. Choose the best answer.

Question 1.

Which of the following is correct?

(a) Rate of change of charge is electrical power.

(b) Rate of change of charge is current.

(c) Rate of change of energy is current.

(d) Rate of change of current is charge.

Answer:

(b) Rate of change of charge is current.

Question 2.

SI unit of resistance is:

(a) mho

(b) joule

(c) ohm

(d) ohm meter

Answer:

(c) ohm

In a simple circuit, why does the bulb glow when you close the switch?

(a) The switch produces electricity.

(b) Closing the switch completes the circuit.

(c) Closing the switch breaks the circuit.

(d) The bulb is getting charged.

Answer:

(b) Closing the switch completes the circuit

Question 4.

Kilowatt hour is the unit of:

(a) resistivity

(b) conductivity

(c) electrical energy

(d) electrical power

Answer:

(c) electrical energy

II. Fill in the blanks.

When a circuit is open, ……….. cannot pass through it.

The ratio of the potential difference to the current is known as ……….

The wiring in a house consists of ………… circuits.

The power of an electric device is a product of ……… and ………..

LED stands for ………..

Answer:

III. State whether the following statements are true or false: If false correct the statement.

Ohm’s law states the relationship between power and voltage.

MCB is used to protect house hold electrical appliances.

The SI unit for electric current is the coulomb.

One unit of electrical energy consumed is equal to 1000 kilowatt hour.

The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances.

Answer:

False – Ohm’s law states that the relationship between current and voltage.

True

False – The SI unit for electric current is ampere.

False – One unit of electrical energy consumed is equal to 1 kilowatt hour.

False – The effective resistance of three resistors connected in series is greater than the highest of the “individual resistance.

IV. Match the items in column-1 to the items in column-ll.

Answer:

(i) – (e)

(ii) – (a)

(iii) – (b)

(iv) – (c)

(v) – (d)

V. Assertion and reason type Questions.

Mark the correct choice as

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

1. Assertion: Electric appliances with a metallic body have three wire connections.

Reason: Three pin connections reduce heating of the connecting wires.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

2. Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.

Reason: The current flows towards the point of the highest potential.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

3. Assertion: LED bulbs are far better than incandescent bulbs.

Reason: LED bulbs consume less power than incandescent bulbs.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Very short answer questions.

Question 1.

Define the unit of current.

Answer:

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,

1 ampere =

1coulomb1second

Question 2.

What happens to the resistance, as the conductor is made thicker?

Answer:

If the conductor is made a thicker area of cross-section of conduction increases that will decrease the resistance.

Question 3.

Why is tungsten metal used in bulbs, but not in fuse wires?

Answer:

Tungsten metal is used in bulbs because its melting point is the greatest.

But it is not used in fuse wires. When a current more than 5A flows tungsten wire will be melted. Hence tungsten is not used in fuse wire.

Question 4.

Name any two devices, which are working on the heating effect of the electric current.

Answer:

The heating effect of electric current is used in many home appliances such as electric iron and electric toaster.

VII. Short Answer Questions.

Question 1.

Define electric potential and potential difference.

Answer:

Electric Potential: It is the amount of work done in moving unit positive charge from infinity to that point against the electric force.

Electric potential

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 2

Potential difference : It is the amount of work done in moving a unit positive charge from one point to another against the electric force.

Potential difference VA – VB =

WA

Question 2.

What is the role of the earth wire in domestic circuits?

Answer:

The earth wire provides a low resistance path to the electric current.

The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of the metallic electric appliance.

Thus, the earth wire serves as a protective conductor, which saves us from electric shocks.

Question 3.

State Ohm’s law.

Answer:

According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor.

I ∝ V. Hence, =

1V

= constant.

The value of this proportionality constant is found to

1R

Therefore, I = (

) V

V = IR

Question 4.

Distinguish between the resistivity and conductivity of a conductor.

Answer:

Question 5.

What connection is used in domestic appliances and why?

Answer:

In the domestic appliance, it is used as a parallel connection to avoid short circuit and breakage.

It has an alternative current (AC). Not DC current as it is from cables, so high potential flows through this.

One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.

Long answer Questions.

Question 1.

With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected:

(a) in series and

(b) in parallel

Answer:

(a) Resistors in series : A series circuit connects the components one after the other to form a ‘single loop’. A series circuit has only one loop through which current can pass. If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work. Series circuits are commonly used in devices such as flashlights. Thus, if resistors are connected end to end, so that the same current passes through each of them, then they are said to be connected in series.

Let, three resistances R1, R2 and R3 be connected in series. Let the current flowing through theorem be I. According to Ohm’s Law, the potential differences V1,V2 and V3 across R1, R2 and R3 respectively, are given by:

V1 = I R1 ………. (1)

V2 = I R2 ……… (2)

v3 = I R3 ………. (3)

The sum of the potential differences across the ends of each resistor is given by:

V = V1 + V2 + V3

Using equations (1), (2) and (3), we get

V = I R1 + I R2 + I R3 …….. (4)

The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit. Let, the effective resistance of the series-combination of the resistors, be RS.

Then,

V = I RS ……….(5)

Combining equations (4) and (5), we get,

I RS = I R1 + I R2 + I R3

RS = R1 + R2 + R3 ……….. (6)

Thus, you can understand that when a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances. When ‘n’ resistors of equal resistance R are connected in series, the equivalent resistance is ‘n R’.

i.e., RS = n R

The equivalent resistance in a series combination is greater than the highest of the individual resistances.

(b) Resistors in Parallel : A parallel circuit has two or more loops through which current can pass. If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s). The wiring in a house consists of parallel circuits.

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 5

Consider that three resistors R1, R2 and R3 are connected across two common points A and B The potential difference across each resistance is the same and equal to the potential difference between A and B. This is me sured using the voltmeter. The current I arriving at A divides into three branches I1, I2 and I3 passing through R1, R2 and R3 respectively.

According to the Ohm’s law, you have,

The total current through the circuit is given by

I = I1 + I2 + I3

Using equations (1), (2) and (3), you get

I =

V

+

+

R3

……… (4)

Let the effective resistance of the parallel combination of resistors be RP Then,

I =

VRP

……… (5)

Combining equations (4) and (5), you have

Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance. When ‘n’ resistors of equal resistances R are connected in parallel, the equivalent resistance is

The equivalent resistance in a parallel combination is less than the lowest of the individual resistances.

Question 2.

(a) What is meant by electric current? Give its direction?

(b) Name and define its unit.

(c) Which instrument is used to measure the electric current? How should it be r connected in a circuit?

Answer:

(a) (i) Electric current is often termed as ‘current’ and it is represented by the symbol ‘I’. It is defined as the rate of flow of charges in a conductor.

(ii) The electric current represents the number of charges flowing in any cross-section of a conductor (say a metal wire) in unit time.

(b) The SI unit of electric current is ampere (A).

The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence,

1 ampere =

1 coulomb 1 second 

(c) (i) The ammeter is used to measure the current.

(ii) An Ammeter is connected in series with the circuit.

(iii) The Ammeter is a low impedance device connecting it in parallel with the circuit would cause a short circuit, damaging the Ammeter or the circuit.

Question 3.

(a) State Joule’s law of heating.

(b) An alloy of nickel and chromium is used as the heating element. Why?

(c) How does a fuse wire protect electrical appliances?

Answer:

(a) Joule’s law of heating states that the heat produced in any resistor is:

directly proportional to the square of the current passing through the resistor.

directly proportional to the resistance of the resistor.

directly proportional to the time for which the current is passing through the resistor.

(b) Because,

It has high resistivity,

It has a high melting point,

It is not easily oxidized.

(c) When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage.

Question 4.

Explain about domestic electric circuits, (circuit diagram not required)

Answer:

Electricity is distributed through the domestic electric circuits wired by the electricians.

The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer.

The important components of the main-box are: (i) a fuse box and (ii) a meter. The meter is used to record the consumption of electrical energy.

The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).

The function of the fuse wire or an MCB is to protect the household electrical appliances from overloading due to excess current.

Question 5.

(a) What are the advantages of LED TV over the normal TV?

(b) List the merits of LED bulb.

Answer:

(a) Advantages of LED TV:

It has brighter picture quality.

It is thinner in size.

It uses less power and consumes very less energy.

Its life span is more.

It is more reliable.

(b) Advantages of LED bulb:

As there is no filament, there is no loss of energy in the form of heat.

It is cooler than the incandescent bulb.

In comparison with the fluorescent light, the LED bulbs have significantly low power requirement.

It is not harmful to the environment.

A wide range of colours is possible here.

It is cost-efficient and energy efficient.

Mercury and other toxic materials are not required. One way of overcoming the energy crisis is to use more LED bulbs.

IX. Numerical problems.

Question 1.

An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?

Answer:

(i) When heating is maximum, the power

P1 = 420 W

Applied voltage V = 220 V

P = VI

Current I =

PV

I =

= 1.909 A

I = 1.909 A

(ii) When heating is minimum

Power P2 = 180 W

Applied voltage V = 220 V

P = VI

∴ Current I =

PV

I =

= 0.8181 A

I = 0.8181 A

Question 2.

A 100-watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.

Solution:

100 W = 100 joules per second

1 watt hours = 3600 joules

The electric bulb is lighted for 5 hours daily,

100 W × 5 = 500 watt hours

500 watt hours = 1800000 joules

1 kWh = 3600000 joules

Units consumed per day =

18000003600000

= 0.5 units

Untis consumed in month = 0.5 × 31 = 15.5 units …. (1)

Now, Sum of power of four 60 watt bulbs = 240 W

240 W × 5 hours = 1200 watt hours

1200 watt hours = 4320000 joules

Energy consumed per day =

43200003600000

= 1.2 units

Energy consumed in a month = 1.2 × 31 = 37.2 units …. (2)

Total energy consumed in a month = 15.5 + 37.2 = 52.7 units

1 unit = 1 kWh

The energy consumed in the month of January = 52.7 kWh.

Question 3.

A torch bulb is rated at 3 V and 600 mA. Calculate it’s

(a) power

(b) resistance

(c) energy consumed if it is used for 4 hour.

Answer:

Voltage V = 3 V

Current I = 600 mA

(a) Power = VI

= 3 × 600 × 10-3

= 1800 × 10-3

= 1.8 W

(c) Time = 4h

Energy consumed E = P × t

E = 1.8 × 4

= 7.2 W

Question 4.

A piece of wire having a resistance R is cut into five equal parts.

(a) How will the resistance of each part of the wire change compared with the original resistance?

(b) If the five parts of the wire are placed in parallel, how will the resistance of the combination change?

(c) What will be ratio of the effective resistance in series connection to that of the parallel connection?

Answer:

(a) Original resistance

R =

lA

∴ R α l

After cutting length of each piece

r =

l5

New resistance

R’ =

R’ α l’

R : R’ = 5 : 1

(b) When five parts of the wire are placed in parallel.

Effective Resistance

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 11

Resistance of the combinations

RP =

R25

(c) When resistance are connected in series

RS = R

When resistance are connected in parallel

RP =

R25

RS : RP = R :

R25

= 25 R : R_1

RS : RP = 25 : 1

X. HOT Questions.

Question 1.

Two resistors when connected in parallel give the resultant resistance of 2 ohm; but when connected in series the effective resistance becomes 9 ohm. Calculate the value of each resistance.

Answer:

Let the resistance be R, and R2 when two resistances are connected in series

RS = R1 + R2

= 9

R1 + R2 = 9 ……….(1)

When two resistance are connected in parallel

Using (1) equation (2) becomes

9R2

R1 R2 = 18 ……….(3)

(R1 – R2)² = (R1 + R2)² – 4R1R2

= (9)² – 4 × 18

= 81 – 72 = 9

∴ (R1 – R2) = √9 = 3 ………(4)

From (1)

R1 + R2 = 9

2R1 = 12

∴ R1 =

122

= 6 ohm

From (1)

R2 = 9 – R1

= 9 – 6 = 3Ω

The values of resistances are

R1 = 6 ohm

R2 = 3 ohm

Question 2.

How many electrons are passing per second in a circuit in which there is a current of 5 A?

Solution:

Current I = 5A

Time (t) = 1 second

Charge of electron e = 1.6 × 10-19 C

I=

Number of electron, n =

=

=

=

∴ R’ = 3R

New resistance = 3 times the original resistance.

Important Questions and Answers

I. Choose the best answer.

Question 1.

Electric current is defined as the rate of flow of:

(a) energy

(b) power

(c) mass

(d) charge

Answer:

(d) charge

Question 2.

The S.I. unit of electric current is _____.

(a) Volt

(b) Power

(c) Ampere

(d) newton.

Answer:

(c) Ampere

Question 3.

The unit of electric current is:

(a) ampere

(b) volt

(c) watt

(d) kilo-watt

Answer:

(a) ampere

Question 4.

The work done in moving a charge of 2 C across two points in a circuit is 2 J. What is the potential difference between the points?

(a) 1 V

(b) 10 V

(c) 100 V

(d) 0.

Answer:

(a) 1 V

Question 5.

The amount of work done to move a unit charge from one point to the other is:

(a) resistance

(b) current

(c) Potential

(d) none of the above

Answer:

(c) Potential

Question 6.

Ohm’s law gives the relative between potential difference and:

(a) emf

(b) temperature

(c) resistance

(d) current

Answer:

(d) current

Question 7.

The unit of resistance is _____.

(a) volt

(b) volt ampere-1

(c) ampere

(d) Joule.

Answer:

(b) volt ampere-1

Question 8.

The symbol of battery is:

Answer:

(b)

Question 9.

Electrical resistivity for a given material is ______.

(a) zero

(b) constant

(c) both (a) and (b)

(d) only (b).

Answer:

(b) constant

Question 10.

The potential difference required to pass a current 0.2 A in a wire of resistance 20 ohm is:

(a) 100 V

(b) 4 V

(c) 0.01 V

(d) 40 V

Answer:

(b) 4 V

Question 11.

The unit of electrical conductivity ______.

(a) ohm-1 metre

(b) ohm-1 metre-1

(c) Volt Ampere-1

(d) ohm.

Answer:

(b) ohm-1 metre-1

Question 12.

Kilowatt-hour is the unit of:

(a) potential difference

(b) electric power

(c) electric energy

(d) charge

Answer:

(c) electric energy

Question 13.

The resistivity of a material is 4 × 10-8 Ωm and its conductivity ______.

(a) 25 × 10-8 mho m-1

(b) 0.25 × 10-8 mho m-1

(c) 25 × 108 mho m-1

(d) 0.25 × 108 mho m-1.

Answer:

(d) 0.25 × 108 mho m-1.

Question 14.

The commonly used safely fuse wire is made of:

(a) nickel

(b) lead

(c) an alloy of tin and lead

(d) copper

Answer:

(c) an alloy of tin and lead

Question 15.

The value of one horse power is:

(a) 746 kW

(b) 746 W

(c) 3.6 × 105 W

(d) 3.6 × 106 kW

Answer:

(b) 746 W

Question 16.

When ‘n’ number of resistors are connected in series, the effecive resistance for series is ______.

(a) nR

(b)

nR

(c)

(d) none of these.

Answer:

(a) nR

Question 17.

Name the physical quantity which is measured in kW:

(a) electric energy

(b) electric power

(c) electric current

(d) electric potential

Answer:

(b) electric power

Question 18.

What is the amount of current, when 20 C of charges flows in 4 s through a conductor? [l =

qv

]

(a) 5 A

(b) 80 A

(c) 4 A

(d) 2 A

Answer:

(a) 5 A

Question 19.

Nichrome is ______.

(a) a conductor

(b) an insulator

(c) an alloy

(d) none of these.

Answer:

(c) an alloy

Question 20.

The main source of biomass energy is:

(a) coal

(b) heat energy

(c) thermal energy

(d) cow-dung

Answer:

(d) cow-dung

Question 21.

The value of one ampere is:

(a)

1second1coulomb

(b) 1 coulomb × sec

(c)

1coulomb1second

(d) 1 coulomb

Answer:

(c)

1coulomb1second

Question 22.

The heat produced in an electric heater of resistance 2 Ω is connected to an electric source, when a current of 6 A flows for 5 minutes _____.

(a) 216 × 102 J

(b) 2160 J

(c) 900 J

(d) 150 J.

Answer:

(a) 216 × 102 J

Hint: Formula H = I2Rt Joule.

Question 23.

The value of

1joule1coulomb

(a) 1 kWh

(b) 1 Wh

(c) ampere

(d) volt

Answer:

(d) volt

Question 24.

The mathematical from of Ohm’s law is given by:

(a) V = IR

(b) I = VR

(c) R =

IV

(d) I =

Answer:

(a) V = IR

Question 25.

One kilowatt hour is _____.

(a) 3.6 × 106 J

(b) 36 × 106 J

(c) 3.6 × 105 J

(d) 36 × 105 J.

Answer:

(a) 3.6 × 106 J

Question 26.

If the length and radius of a conductor is doubled then its specific resistance will:

(a) be doubled

(b) be halved

(c) be tripled

(d) remain the same

Answer:

(d) remain the same

Question 27.

The value of resistivity of nichrome is:

(a) 1.5 × 106 Ωm

(b) 1.5 × 10-6 Ωm

(c) 5.1 × 106 Ωm

(d) 5.1 × 10-6 Ωm

Answer:

(b) 1.5 × 10-6 Ωm

Question 28.

Due to short circuit, effective resistance in the circuit becomes _____.

(a) large

(b) very small

(c) very large

(d) zero.

Answer:

(b) very small

Question 29.

if a conductor has a length of 1 m and area of 1 m² then its resistivity is equal to its:

(a) resistance

(b) conductance

(c) length

(d) conductivity

Answer:

(a) resistance

Question 30.

When resistors are connected in parallel, potential difference across each resistor will be:

(a) different

(b) same

(c) vary

(d) none of the above

Answer:

(b) same

Question 31.

LED TV screen was developed by James P. Mitchell in _____.

(a) 1797

(b) 1977

(c) 2009

(d) 1987.

Answer:

(b) 1977

Question 32.

Heat developed across a conductor H =

(a) IRt

(b) VR

(c) I2Rt

(d) I2R

Answer:

(c) I2Rt

Question 33.

Expression for electric energy is:

(a) W =

VI

(b) W = VIt

(c) W = Vt

(d) W =

VIt

Answer:

(b) W = VIt

Question 34.

In our home, fuse box consists of:

(a) fuse wire

(b) MCB

(c) fuse wire or MCB

(d) switches

Answer:

(c) fuse wire or MCB

Question 35.

Which of the following is a semi conductor device?

(a) LED bulb

(b) fuse

(c) MCB

(d) switch

Answer:

(a) LED bulb


II. Fill in the blanks.

1. The flow of charge: Electric current. A continuous closed path of an electric current is ………. The unit of charge: Coulomb then-current ……….

2. Electric current I: Charge (Q)/ ………. while electric potential V is ……….

3. A resistor of resistance R:

 Then variable resistance and ……….. rheostat ……….

4. In series connection of resistors: ………. Then for parallel connection of resistors: ………..

5. The transformation of energy in Electric oven: ………… Electric cell ………..

6. The expression obtained from Ohm’s law ………… joule’s law

7. The unit of electric power …………. then electric energy ………..

8. The equivalent of 1 volt ………. then for 1 ohm ………..

9. The tap-key is used to ……… and ……….. an electric circuit.

10. The opposition to flow of current is called ………. and its unit is ………..

11. The heat developed in a conductor is directly proportional to the square of ………… and ………. of flow.

12. The S.I unit of electric current is ……….

13. The S.l unit of resistance is ……….

14. ……….. is the S.l unit of potential difference.

15. From Ohm’s law

VI

=

16. If a current 2A flows through conductor having a potential difference of 6 V then its resistance is ……….

17. If R is the resistance of a conductor then its conductance is G = ……….

18. Conductivity is ……… for ……….. than insulators.

19. When resistors are connected in series the equivalent resistance is …………. than the highest resistance of individual resistors.

20. In series connection ……….. is less as effective resistance is more.

21. Tungsten is used as heating element because its resistance is ……….

22. Tungsten is used as filament in the electric bulb because its melting points is …………

23. If a current of 6A flows through a 5Ω resistance for 10 minutes than heat developed in the resistance is ……….

24. When a current of IA flows through a conductor having potential difference of IV, the electric power is ………..

25. 746 watt is equivalent to ………..

26. In displays are used ………….

Answer:

1. Electric circuit, Ampere

2. Time (t), Work done (W)/charge (Q)

3.

4. Current is same, Potential difference is same

5. electrical into heat energy, chemical into electrical energy

6. V = IR, H = I2Rt

7. Kilowatt, want hour

8. 1joule, 1coulomb, 1volt, 1ampere

9. open, close

10. resistance, ohm

11. current, time

12. ampere

13. Ohm 15. R

14. Volt

15. R

16. 3 ohm

17.

IR

18. more, conductors

19. greater

20. Current

21. high

22. high

23. 108 kJ

24. 1 W

25. 1 horse power

26. LED bulbs

III. State whether the following statements are true or false: If false correct the statement.

1. Current is the rate of flow of charges

2. The symbol of diode is

3. Potential =

chargetime

4. Mathematical form of ohm’s law is V = IR

5. Nichrome is used in electric bulb.

6. The unit of conductance is mho.

7. The equivalent resistance in a parallel combination is less than the lowest of the individual resistance.

8. In parallel connection the effective resistance is RP =

R1

9. True

10. False – 1 kWh = 3.6 × 106 J

11. True

12. True

IV. Match the items in column-1 to the items in column-ll.

Question 1.

Match the Column I with Column II.

Answer:

(i) – (d)

(ii) – (c)

(iii) – (a)

(iv) – (b)

Question 2.

Match the components with symbol

Answer:

(i) – (d)

(ii) – (c)

(iii) – (b)

(iv) – (a)

Question 3.

Match the Column I with Column II.

Samacheer Kalvi 10th Science Guide Chapter 4 Electricity 20

Answer:

(i) – (c)

(ii) – (d)

(iii) – (b)

(iv) – (a)

Question 4.

Match the Column I with Column II.

Answer:

(i) – (d)

(ii) – (c)

(iii) – (b)

(iv) – (a)

Question 5.

Match the column I with column II.

Answer:

(i) – (e)

(ii) – (d)

(iii) – (a)

(iv) – (b)

Question 6.

Match the column I with column II:

Answer:

(i) – (d)

(ii) – (e)

(iii) – (a)

(iv) – (b)

Question 7.

Match the column I with column II:

Answer:

(i) – (e)

(ii) – (a)

(iii) – (d)

(iv) – (c)

V. Assertion and reason type questions.

Question 1.

Assertion: In a series system, equivalent resistance is the sum of the individual resistance.

Reason: The current that passes through each resistor is the same.

(a) Assertion is true but Reason is false.

(b) Assertion is true and Reason doesn’t explains Assertion,

(c) Both Assertion and Reason are false.

(d) Assertion is true and Reason explains Assertion

Answer:

(d) Assertion is true and Reason explains Assertion

Question 2.

Assertion: In a parallel system, the total current is equal to the sum of the current through each resistor.

Reason: The potential difference across each resistor is the same.

(a) Assertion is true and Reason explains Assertion.

(b) Assertion is true and Reason doesn’t explains Assertion.

(c) Both Assertion and Reason are false

(d) Assertion is true but Reason is false.

Answer:

(a) Assertion is true and Reason explains Assertion.

Question 3.

Assertion: The unit of power watt is not frequently used in practice. Reason: it cannot be converted into Joule.

(a) Both Assertion and Reason are false.

(b) Assertion is true but Reason is false.

(c) Both Assertion and Reason are true and Reason explains Assertion.

(d) Both Assertion and Reason are true and Reason doesn’t explains Assertion.

Answer:

(b) Assertion is true but Reason is false.

Question 4.

Assertion: A wire carrying a current has electric field around d.

Reason: A wire carrying current is stays electrically neutral.

(a) If both the assertion and the reason a re true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 5.

Assertion: In order to pass current through electric circuit, it must be closed.

Reason: In our home, the switch is ON, then the current flows through the bulb. So, the bulb glows.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 6.

Assertion: Resistance of a material opposes the flow of charges.

Reason: It is different for different materials.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 7.

Assertion: Electrical conductivity is the reciprocal of electrical resistivity. Reason: The unit of conductivity is Ohm.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(c) If the assertion is true, but the reason is false.

Question 8.

Assertion: One end of the earthing wire is connected to a body of the electrical appliance and its other end is connected to a metal tube that is burried into the Earth.

Reason: The earth wire provides low resistance path to the electric current.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 9.

Assertion: The passage of electric current through a wire results in the production of heat.

Reason: The heating effect is used in electric heater electric iron etc.

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 10.

Assertion: One kilowatt hour is known as one unit of electrical energy.

Reason: 1 kWh = 3.6 × 106J

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

(c) If the assertion is true, but the reason is false.

(d) If the assertion is false, but the reason is true.

Answer:

(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

VI. Very short answer type Questions

Question 1.

If a charge of QC flows through a conductor in time t second then what is the value of current?

Answer:

Charge =

Chargetime

I =

Qt

Question 2.

What is the electric circuit?

Answer:

An electric circuit is a closed conducting loop.

(or)

path, which has a network of electrical components through which electrons are able to flow. This path is made using electrical wires so as to connect an electric appliance to a source of electric charges (battery).

Question 3.

If the length of a wire is doubled and its cross-section is also doubled than what happens its resistance?

Answer:

Resistance of a wire

R =

plA

………. (1)

l’ = 2l

A’ = 2A

∴ New Resistance

R’ = l ×

2l2A

R’ =

…….. (2)

Form (1) and (2) it is found that resistance remains unchanged.

Question 4.

What is the unit of resistance and resistivity of a conductor?

Answer:

(i) The unit of resistance is ohm.

(ii) The unit of resistivity is ohm meter.

Question 5.

Define the unit of resistance? (or) Define one ohm.

Answer:

The resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends.

 Ohm =

.

Question 6.

What is the unit of conductivity?

Answer:

The unit of conductivity is mho meter.

Question 7.

What is the value of one kilo watt hour?

Answer:

One kilowatt hour = 1000 W hr

1 kWh = 3.6 × 106J

Question 8.

What is the heating effect of electric current?

Answer:

The passage of electric current through a wire resulting in the production of heat. This phenomenon is called the heating effect of current. This heating effect of current is used in devices like electric heater, electric iron, etc.

VII. Short Answer Questions.

Question 1.

What is an electric circuit?

Answer:

An electric circuit is a closed conducting loop (or) path, which has a network of electrical components through which electrons are able to flow.

Question 2.

Draw a circuit diagram to represent a simple electric circuit.

Answer:

Question 3.

What is the direction of conventional current?

Answer:

By convention, the direction of current is taken as the direction of flow of positive charge (or) opposite to the direction of flow of electrons.

Question 4.

Define electric potential.

Answer:

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

Question 5.

What is meant by overloading?

Answer:

(i) Overloading happens when a large number of appliances are connected in series to the same source of electric power. This leads to a flow of excess current in the electric circuit.

(ii) When the amount of current passing through a wire exceeds the maximum permissible limit, the wires get heated to such an extent that a fire may be caused. This is known as overloading.

Question 6.

What is meany by short circuit?

Answer:

When a live wire comes in contact with a neutral wire, it causes a ‘short circuit’.

This happens when the insulation of the wires get damaged due to temperature changes or some external force.

Due to a short circuit, the effective resistance in the circuit becomes very small, which leads to the flow of a large current through the wires.

This results in heating of wires to such an extent that a fire may be caused in the building.

Question 7.

Draw an electric circuit to understand Ohm’s law.

Answer:

Question 8.

Define resistance of a conductor.

Answer:

The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.

Question 9.

Define Resistance. Give its unit and conductance.

Answer:

The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.

Conductance: It is defined as the reciprocal of its resistance (R). Hence, the conductance ‘G’ of a conductor is given by

G=

Its unit is ohm-1. It is also represented as ‘mho’.

Question 10.

Define electrical resistivity of a material.

Answer:

The electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section. Its unit is ohm metre.

Question 11.

What is meant by electrical conductivity?

Answer:

The reciprocal of electrical resistivity of a material is called its electrical conductivity.

σ =

1p

Question 12.

Mention the differences between the combination of resistances in series and parallel.

Answer:

Question 13.

Write short notes about filament in electric bulbs.

Answer:

In electric bulbs, a small wire is used, known as filament. The filament is made up of a material whose melting point is very high. When current passes through this wire, heat is produced in the filament. When the filament is heated, it glows and gives out light. Tungsten is the commonly used material to make the filament in bulbs.

Question 14.

What is meant by electric power?

Answer:

The electric power is the product of the electric current and the potential difference due to which the current passes in a circuit.

Question 15.

What is meant by overloading of an electric circuit?

Answer:

When the amount of current passing through a wire exceeds the maximum permissible limit, the wires get heated to such an extent that a fire may be caused. This is known as overloading.

Question 16.

What is meant by LED bulb?

Answer:

An LED bulb is a semiconductor device that emits visible light when an electric current passes through it. The colour of the emitted light will depend on the type of materials used.

Question 17.

W hat is meant by seven segment display? state its uses.

Answer:

(i) A ‘Seven Segment Display’ is the display device used to give an output in the form of numbers or text.

(ii) It is used in digital meters, digital clocks, microwave ovens, etc.

Question 18.

What do you know about LED television.

Answer:

LED Television is one of the most important applications of Light Emitting Diodes. An LED TV is actually an LCD TV (Liquid Crystal Display) with LED display. An LED display uses LEDs for backlight and an array of LEDs act as pixels. LEDs emitting white light are used in monochrome (black and white) TV; Red, Green and Blue (RGB) LEDs are used in colour television.

Question 19.

What is fuse wire?

Answer:

The function of fuse wire or an MCB is to protect the household electrical appliances from excess current due to overloading or a short circuit.

Question 20.

Draw a group between potential difference and current.

Answer:

Question 21.

Write short note about short circuit?

Answer:

When a live wire comes in contact with a neutral wire, it causes a ‘short ‘ circuit’. This happens when the insulation of the wires get damaged due to temperature changes or some external force. Due to a short circuit, the effective resistance in the circuit becomes very small, which leads to the flow of a large current through the wires. This results in heating of wires to such an extent that a fire may be caused in the building.

VIII. Long answer questions.

Question 1.

Tabulate various components used in electrical circuit and their uses?

Answer:

Question 2.

Explain series connection of parallel resistors.

Answer:

If we consider the connection of a set of parallel resistors that are connected in series, you get a series – parallel circuit. Let R1 and R2 be connected in parallel to give an effective resistance of Rp1. Similarly, let R3, and R4 be connected in parallel to give an effective resistance of Rp2. Then, both of these parallel segments are connected in series. Using equation we get

We get = R1 + R2 + R3 ……….(1)

Finally, using equation (1), the net effective resistance is given by

Rtotal = RP1 + RP2

Question 3.

Explain parallel connection of series resistors.

Answer:

If you consider a connection of a set of series resistors connected in a parallel circuit, you get a parallel-series circuit. Let R1 and R2 be connected in series to give an effective resistance of RS1. Similarly, let R3 and R4 be connected in series to give an effective resistance of RS2. Then, both of these serial segments are connected in parallel.

Using equation

RS = R1 + R2 + R3

We get RS1 = R1 + R2, RS2 = R3 + R4

Finally, using equation

1

+

RS2

Question 4.

Explain applications of heating effect.

Answer:

Electric Heating Device: The heating effect of electric current is used in many home appliances such as electric iron, electric toaster, electric oven, electric heater, geyser, etc. In these appliances Nichrome, which is an alloy of Nickel and Chromium is used as the heating element. Because:

(i) it has high resistivity,

(ii) it has a highmelting point,

(Hi) it is not easily oxidized.

Fuse Wire : The fuse wire is connected in series, in an electric circuit. When a large current passes through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage. The fuse wire is made up of a material whose melting point is relatively low.

Filament in bulbs : In electric bulbs, a small wire is used, known as filament. The filament is made up of a material whose melting point is very high. When current passes through this wire, heat is produced in the filament. When the filament is heated, it glows and gives out light. Tungsten is the commonly used material to make the filament in bulbs.

Question 5.

Write short notes about

(i) LED bulb

(ii) Seven segment display

Answer:

(i) An LED bulb is a semiconductor device that emits visible light when an electric current passes through it. The colour of the emitted light will depend on the type of materials used. With the help of the chemical compounds like Gallium Arsenide and Gallium Phosphide, the manufacturer can produce LED bulbs that radiates red, green, yellow and orange colours. Displays in digital watches and calculators, traffic signals, street lights, decorative lights, etc., are some examples for the use of LEDs.

(ii) A ‘Seven Segment Display’ is the display device used to give an output in the form of numbers or text. It is used in digital meters, digital clocks, micro wave ovens, etc. It consists of 7 segments of LEDs in the form of the digit 8. These seven LEDs are named as a, b, c, d, e, f and g. An extra 8th LED is used to display a dot.

IX. Numerical problems.

Question 1.

An electric iron draws a current of 0.5 A when the voltage is 220 volts. Calculate the amount of electric charge flowing through it in one hour.

Answer:

I =

Qt

Charge Q = ?

Charge I = 0.5 A

Time = 1 hour

= 60 × 60 s

Q = It

= 0.5 × 3600

= 1800 C

The amount of electric charge flowing = 1800 C

Question 2.

A current of 5A flows through a heater for 10 minutes. Calculate the amount of electric charge flowing through the electric circuit.

Answer:

Q = I × t

Current I = 5A

Time t = 10 × 60 = 600 s

= 5 × 600 = 3000 C

Question 3.

A torch bulb draws a current 0.6 A, when glowing from a source of 6 V. Calculate the resistance of the bulb when glowing.

Answer:

V = IR

Current I = 0.6A

Potential V = 6V

R = ?

R =

VI

=

= 10Ω

The resistance of the bulb = 10Ω

Question 4.

Find the potential difference required to pass a current of 0.2 A in a wire of resistant 20Ω.

Answer:

V = IR

Current I = 0.2A

Resistance R = 20Ω

Potential difference V = ?

V = 0.2 × 20 = 4V

The potential difference in a wire = 4V

Question 5.

Calculate the amount of work done in moving charge of 25 C across two points having potential difference of 20V.

Answer:

W = QV

Charge Q = 25 C

Potential difference V = 20 V

W = 25 × 20 = 500 J

Question 6.

Three resistances are connected in an electrical circuit as shown in the circuit diagram. Determine the potential difference across resistance R2.

Answer:

For series connection, the effective resistance

RS = R1 + R2 + R3

= 1 + 2 + 3 = 6Ω

Total potential = 12V

Amount of current flowing through the circuit

I =

V

=

12V6Ω

= 2A

Potential difference across resistance R2

V = I × R2

= 2 × 2 = 4V

The potential difference across the resistance R2 = 4 V

Question 7.

In the given network, find the equivalent resistance between A and B.

Answer:

An equivalent of the given network is drawn in the relevant parts as follows: Resistance of the combination R1 and R2 is

RS = 5 + 5 = 10Ω

Resistance of the combination R1, R2 and R3 is

The resistance of series combination RP1 and R4 is RS1 = 5 + 5 = 10Ω

Resistance of the combination RS1 and R5 is

Resistance of the series combination RP2 and R6 is

RS2 = 5 + 5 = 10Ω

Resistance of the combination RS2 and R7 is

Resistance of the series combination RP3 and R8 is

RP3 = 5 + 5 = 10Ω

Resistance of the combination RS3 and R9 is

Resistance of the series combination RP4 and R6 is

RS4 = 5 + 5 = 10Ω

Resistance of the combination RS4 and R8 is

∴ Resistance between A and B is 5Ω

Question 8.

For a given circuit calculate

(i) the total effective resistance of the circuit.

(ii) the total current in the circuit

(iii) the current through each resistor.

Answer:

For parallel connection, the effective resistance

(i) Total effective resistance of the circuit

RP = 0.588Ω

(ii) Total current in the circuit

I = 10.2 A

(iii) Current through R1 = 6A

Current through R2 = 3A

Current through R3 = 1.2A

Question 9.

An electric iron a rating of 750 W, 220 V.

(i) Calculate current passing through it and

(ii) Its resistance when in use.

Answer:

P = VI

I =

PV

Ohm’s law V = IR

(i) The amount of current passing

I = 3.4A

(ii) Resistance

R = 64.7Ω

Question 10.

Following graph was plotted between V and I values. What would be the values of

VI

ratios when the potential difference is 0.8 V and 1.2 V?

Answer:

V1 = 0.8V, I1 = 32A; V2 = 1.2V; I2 = 48A

Question 11.

Three resistors of 2Ω, 4Ω and 8Ω are connected in parallel with a battery of 3 V. Calculate

(i) Current through each resistor and

(ii) Total current in the circuit.

Answer:

Potential difference across each resistor is same.

(i) Current through each resistor:

(ii) Total current in the circuit: I = I1 + I2 + I3

I = 1.5 + 0.75 + 0. 375

I = 2.625 A

Question 12.

Two bulbs of 40 W and 60 W are connected in series to an external potential difference. Which bulb will glow brighter? Why?

Answer:

Let the external potential difference be 230 V

For 40 W bulb resistance is R

For 60 W bulb resistance is R

According to Ohm’s law

I =

VR

Current flowing through 40 W bulb is

2301322.5

= 0.1739 A

Current flowing through 60 W bulb is

230881.6

= 0.2608 A

When bulbs are connected in series effective resistance is

RS = R1+ R2 = 1322.5 + 881.6

RS = 2204.1Ω

Net current

I =

2302204.1

= 0.1043 A

Using power equation P = I²R

For 40 W bulb P = I²R

= (0.1043)² × 1322.5

= (0.01087) × 1322.5 = 14.386 W

For 60 W bulb P = I²R

= (0.1043)² × 881.6

= (0.01087) × 881.6 = 9.5904 W

In a series system, higher the resistance, higher the brightness so, 40 W bulb glows brighter.

Question 13.

A wire is bent into a circle. The effective resistance across the diameter is 8Ω. Find the resistant if the wire.

Answer:

RP = 8Ω = resistance across diameter

1R

+

R = 16Ω

x is the resistance of the wire.

x = 16 + 16 = 32Ω

x = 32Ω

. HOT Questions.

Question 1.

A 60 W bulb is connected in parallel with a room heater. This combination is connected across the mains. If 60 W bulb is replaced by a 100 W bulb what happens to the heat produced by the heater? Given reason.

Answer:

Heat produced by the heater will be same. When the bulb and a heater are connected in parallel and this combination is connected across the mains, potential difference across each is the same equal to the voltage V of the mains irrespective of the resistance of the bulb.

If R is the resistance of the heater then heat produced by the heater will be

V2

in both cases. Hence heat produces by heater will not be changed.

Question 2.

Two bulbs 60 W and 100 W are connected in series and this combination is connected to a d.c power supply. Will the potential difference across 60 W bulb be higher than that across 100 W bulb?

Answer:

60 W bulb has a higher resistance than the resistance across 100 W bulb since the power developed is P = V2R

Potential difference across a bulb will be proportional to resistance. Hence potential difference across 60 W bulb is higher than that across 100 W bulb.

Question 3.

Super conductors has lowest resistance. Is it true. Give reason.

Answer:

True. When the temperature of super conductor is reduced to zero or near by zero its resistance becomes zero.

Question 4.

A constant voltage is applied between two ends of a uniform conducting wire. If both the length and radius of the wire is doubled then what happens to the heat produced in the wire?

Answer:

We know that resistance of a conducting wire is R = plAr2

If length l and radius r are doubled, then resistance will become half. But heat produced H = V2

Hence, heat produced per second will become thrice.

Question 5.

Calculate the effective resistance between A and B.

Answer:

The electrical circuit can be redrawn as

The resistance R1 and R2 are in series

RS = R1 + R2 = 2 + 2 = 4Ω

The resistance RS and R3 are in parallel

∴ Effective resistance Reff= 1.33Ω

Question 6.

Two wires of same material and length have resistances 5Ω and 10Ω respectively. Calculate the ratio of radii of the two wires.

Answer:

Resistance

∴ r1 : r2 = √2 : 1

Question 7.

An iron box of 400 W power is used daily for 30 minutes. If the cost per unit is 75 paise, find the weekly expense on using the iron box.

Answer:

Power P = 400 W

Energy consumed = Power × Time

= 400 ×

12

= 200 Wh

Energy consumed in one week= 200 × 7

= 1400 Wh = 1.4 kWh

= 1.4 unit

∴ Total cost per week = 1.4 × 0.75 = Rs 1.05

Weekly expense = Rs 1.05


Chapter 6, light, science, physics, Tamilnadu board,

I. Choose the correct answer.

Question 1.

A ray of light passes from one medium to another medium. Refraction takes place when angle of incidence is …………..

(a) 0°

(b) 45°

(c) 90°

Answer:

(b) 45°

Question 2.

…………. is used as reflectors in torchlight.

(a) Concave mirror

(b) Plane mirror

(c) Convex mirror

Answer:

(a) Concave mirror

Question 3.

We can create enlarged, virtual images with ………………

(a) Concave mirror

(b) Plane mirror

(c) Convex mirror

Answer:

(a) Concave mirror

Question 4.

When the reflecting surface is curved outwards the mirror formed will be

(a) concave mirror

(b) convex mirror

(c) plane mirror

Answer:

(b) convex mirror

Question 5.

When a beam of white light passes through a prism it gets

(a) reflected

(b) only deviated

(c) deviated and dispersed

Answer:

(a) Reflected

Question 6.

The speed of light is maximum in

(a) vacuum

(b) glass

(c) diamond

Answer:

(a) vacuum


II. True or False – If false give the correct answer.

  1. The angle of deviation depends on the refractive index of the glass – True.
  2. If a ray of light passes obliquely from one medium to another, it does not suffer any deviation – False.
    Correct Statement: When light travels from one medium to another, it suffers deviation.
  3. The convex mirror always produces a virtual, diminished and erect image of the object – True.
  4. When an object is at the centre of curvature of concave mirror the image formed will be virtual and erect – False.
    Correct Statement: The image formed is real, inverted and same size of the object.
  5. The reason for brilliance of diamonds is total internal reflection of light – True.

III. Fill in the blanks.

  1.  In going from a rarer to denser medium, the ray of light bends …………….
  2. The mirror used in search light is ………………
  3. The angle of deviation of light ray in a prism depends on the angle of ……………..
  4. The radius of curvature of a concave mirror whose focal length is 5 cm is …………….
  5. Large ………… mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Answer:

  1. towards normal
  2. concave mirror
  3. prism and angle of incident
  4. 10 cm
  5. concave

IV. Match the following.

1. Ratio of height of image to height of object(a) Concave Mirror
2. Used in hairpin bends in mountains(b) Total Internal Reflection
3. Coin inside water appearing slightly raised(c) Magnification
4. Mirage(d) convex Mirror
5. Used as Dentist’s mirror(e) Refraction

Answer:

  1. (c)
  2. (d)
  3. (e)
  4. (b)
  5. (a)

V. Assertion & Reason.

Mark the correct choice as:

(a) If both assertion and reason are true and reason is the correct explanation.

(b) If both assertion and reason are true and reason is not the correct explanation.

(c) If assertion is true but reason is false.

(d) If assertion is false but reason is true.

Question 1.

Assertion: For observing the traffic at a hairpin bend in mountain paths a plane mirror is

preferred over convex mirror and concave mirror.

Reason : A convex mirror has a much larger field of view than a plane mirror or a concave mirror.

Answer:

(d) If assertion is false but reason is true.


Question 2.

Assertion : Incident ray is directed towards the centre of curvature of spherical mirror.

After reflection it retraces its path.

Reason : Angle of incidence i = Angle of reflection r = 0°.

Answer:

(b) If both assertion and reason are true and reason is not the correct explanation.


VI. Answer very briefly.

Question 1.

According to cartesian sign convention, which mirror and which lens has negative focal length?

Answer:

Concave mirror is having a negative focal length.

Question 2.

Name the mirror(s) that can give (i) an erect and enlarged image, (II) same sized, inverted image.

Answer:

Concave mirror

Question 3.

If an object is placed at the focus of a concave mirror, where is the image formed? Image will be formed at infinity as real and inverted.

Question 4.

Why does a ray of light bend when it travels from one medium to another?

Answer:

The bending of light rays when they pass obliquely from one medium to another medium is called refraction of light.

Light rays get deviated from their original path while entering from one transparent medium to another medium of different optical density. This deviation (change in direction) in the path of light is due to the change in velocity of light in the different medium. The velocity of light depends on the nature of the medium in which it travels.

Question 5.

What is speed of light in vacuum?

Answer:

The speed of light in vacuum is known to be almost exactly 300,000 km per second. In 1665 the Danish astronomer Ole Roemer first estimated the speed of light by observing one of the twelve moons of the planet Jupiter.

Question 6.

Concave mirrors are used by dentists to examine teeth. Why?

Answer:

As a dentist’s head mirror: You would have seen a circular mirror attached to a band tied to the forehead of the dentist/ENT specialist. A parallel beam of light is made to fall on the concave mirror; this mirror focuses the light beam on a small area of the body (such as teeth, throat etc.).


VII. Answer briefly.

Question 1.

(a) Complete the diagram to show how a concave mirror forms the image of the object.

(b) What is the nature of the image?

Solution:

(a)

(b) magnified, real and inverted.

Question 2.

Pick out the concave and convex mirrors from the following and tabulate them

Answer:

Rear-view mirror, Dentist’s mirror, Torch-light mirror, Mirrors in shopping malls, Make-up mirror.

Concave mirrorConvex mirror
Dentist’s mirrorRear view mirror
Torch light mirrorMirrors in shopping malls
Make-up mirror

Question 3.

State the direction of incident ray which after reflection from a spherical mirror retraces its path. Give reason for your answer.

Answer:

When incident ray is directed towards the centre of curvature, at all the points of spherical mirror, the ray is always normal. Therefore, angle of incidence i = Angle of reflection r = 0°.

Question 4.

What is meant by magnification? Write its expression. What is its sign for real image and virtual image?

Answer:

Magnification produced by a spherical mirror gives how many times the image of an object is magnified with respect to the object size.

It can be defined as the ratio of the height of the image (hi) to the height of the object (ho).

Magnification = m = 

hi

ho

=

 height of the image 

 height of the object 

  • for real image it is negative,
  • for virtual image it is positive.

Question 5.

Write the spherical mirror formula and explain the meaning of each symbol used in it.

Answer:

The expression relating the distance of the object u, distance of image v and focal length/of a spherical mirror is called the mirror equation. It is given as:

Mirror formula: 

f = u+v

Here, f – focal length of spherical mirror; u – distance of the objective; v – distance of the image.


VIII. Answer in detail.

Question 1.

(a) Draw ray diagrams to show how the image is formed, using a concave mirror when the position of object is

  1.  at C
  2.  between C and F
  3.  between F and P of the mirror.

Answer:

(1) At the centre of curvature C

(2) Between C and F

(3) Between the focus F and the Pole P of the mirror.

(b) Mention the position and nature of image in each case.

Answer:

Position of objectPosition of ImageNature of Image
At the centre of curvature CAt CReal and Inverted
Between C and FBeyond CReal and inverted
Between F and PBehind the mirrorVirtual and Erect

Question 2.

Explain with diagrams how refraction of incident light takes place from

(a) rarer to denser medium

(b) denser to rarer medium

(c) normal to the surface separating the two media.

Answer:

(a) rarer to denser medium:

When a ray of light travels from optically rarer medium to optically denser medium, it bends towards the normal.

(b) denser to rarer medium:

When a ray of light travels from an optically denser medium to an optically rarer medium it bends away from the normal.

(c) normal to the surface separating the two media:

A ray of light incident normally on a denser medium, goes without any deviation.


IX. Numerical problems.

Question 1.

A concave mirror produces three times magnified real image of an object placed at 7 cm in front of it. Where is the image located? (21 cm in front of the mirror)

Answer:

Given: hi = 3ho

u = 7 cm

Solution:

  1. m = 
  2. h
  3. i
  4. h
  5. O
  6. 3ho
  7. h
  8. O
  9.  ∴ m = 3.
  10. m = – 
  11. v
  12. u

  13. v = – m × u
    = – 3 × 7 cm = – 21 cm.
    Hence, Real, inverted and magnified image will be formed at 21 cm in front of the mirror.

Question 2.

Light enters from air into a glass plate having refractive index 1.5. What is the speed of light in glass? (2 × 108 ms– 1)

Given: Refractive index (µ) = 1.5

Speed of light in vacuum (c) = 2 × 108ms– 1

Speed of light in glass (υ) = ?

Solution:

∴ Speed of light in glass = 1.3 × 108 ms< sup>- 1


Question 3.

The speed of light in water is 2.25 × 108ms– 1. If the speed of light in vacuum is 3 × 108ms– 1, calculate the refractive index of water.

Given:

Speed of light in water (υ) = 2.25 × 108ms– 1

Speed of light in vacuum (c) = 3 × 108ms– 1

Solution:

μ = 

c

υ

 ⇒ μ = 

10

8

ms

−1

2.25×

10

8

ms

−1

∴ μ = 1.33


X. HOTS.

Question 1.

Light ray emerges from water into air. Draw a ray diagram indicating the change in its path in water.

Answer:

When a ray of light travels from dense medium to rarer medium [from water medium to air medium], light ray moves away from the normal.

∴ Angle of incidence < Angle of refraction.

Question 2.

When a ray of light passes from air into glass, is the angle of refraction greater than or less than the angle of incidence?

Answer:

When a ray of light travels from rare (air) medium to dense [glass] medium, light ray moves towards the normal.

∴ Angle of refraction < Angle of incidence.

Question 3.

What do you conclude about the speed of light in diamond if the refractive index of diamond is 2.41? way from the mirror. Does the image become smaller or larger? What do you observe?

Answer:

µ = 2.41; Ca = 3 × 108ms [Velocity of light]

µ = 

 Speed of light in air 

 Speed of light in diamond 

=

C

a

C

d

Cd

C

a

μ

=

10

8

2.41

 = 1. 245 × 108 m/s

∴ Speed of light decreases when the light ray travels from air to diamond.


CHAPTER 7, heat, 9th std, science , physics, tamilnadu board,

I. Choose the correct answer:

Question 1.

Calorie is the unit of

(a) heat

(b) work

(c) temperature

(d) food

Answer:

(a) heat

Question 2.

SI unit of temperature is

(a) fahrenheit

(b) joule

(c) Celsius

(d) kelvin

Answer:

(d) kelvin

Question 3.

Two cylindrical rods of same length have the area of cross-section in the ratio 2:1. If both the rods are made up of same material, which of them conduct heat faster?

(a) Both rods

(b) Rod-2

(c) Rod-1

(d) None of them

Answer:

(c) Rod-1


Question 4.

In which mode of transfer of heat, molecules pass on heat energy to neighbouring molecules without actually moving from their positions?

(a) Radiation

(b) Conduction

(c) Convection

(d) Both B and C

Answer:

(b) Conduction


Question 5.

A device in which the loss of heat due to conduction, convection and radiation is minimized is

(a) Solar cell

(b) Solar cooker

(c) Thermometer

(d) Thermos flask

Answer:

  1.  (a) Heat
  2.  (d) kelvin
  3.  (c) Rod-1
  4.  (b) Conduction
  5.  (d) Thermos flask

II. Fill in the blanks.

  1. The fastest mode of heat transfer is ……………….
  2. During day time, air blows from ………… to …………….
  3. Liquids and gases are generally …………….. conductors of heat.
  4. The fixed temperature at which matter changes state from solid to liquid is called ………………

Answer:

  1. radiation
  2. land, sea
  3. bad
  4. melting point

III. Assertion and Reason type questions.

Mark the correct choice as:

(a) If both assertion and reason are true and reason is the correct explanation of assertion.

(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

(c) If assertion is true but reason is false.

(d) If assertion is false but reason is true.

Question 1.

Assertion(A): Food can be cooked faster in copper bottom vessels.

Reason (R): Copper is the best conductor of heat.

Answer:

(a) If both assertion and reason are true and reason is the correct explanation of assertion.

Question 2.

Assertion(A): Maximum sunlight reaches earth’s surface during the afternoon time.

Reason(R): Heat from the sun reaches earth’s surface by radiation.

Answer:

(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

Question 3.

Assertion(A): When water is heated up to 100°C, there is no raise in temperature until all water gets converted into water vapour.

Reason(R): Boiling point of water is 10°C.

Answer:

(c) If assertion is true but reason is false.


IV. Answer briefly.

Question 1.

Define conduction.

Answer:

The process of transfer of heat in solids from a region of higher temperature to a region of lower temperature without the actual movement of molecules is called conduction.

Question 2.

Ice is kept in a double-walled container. Why?

Answer:

Ice is kept in a double-walled container so as to prevent melting of ice from the heat absorbed present in the immediate surroundings. The vacuum present in between the two walls prevents the transfer of heat from the first to the second wall and hence the ice remains in the solid form for a longer time period.

Question 3.

How does the water kept in an earthen pot remain cool?

Answer:

An earthen pot consists of small pores from which the water inside the pot constantly seeps out and gets evaporated due to the presence of high temperature around it. The evaporation process requires heat which is acquired from the surface of the pot, hence making the water and the pot cooler. 

Question 4.

Differentiate convection and radiation.

Answer:

S.No.ConvectionRadiation
1.Flow of heat through a fluid from places of higher temperature to places of lower temperature by movement of the fluid itself.Flow of heat from one place to another by means of electromagnetic waves.
2.Convection needs matter to be present.Radiation can occur even in vacuum.
3.Convection seen in daily life:Hot air balloons, breeze, wind, chimney.Radiation in daily life:White or light coloured cloths, highly polished surface of airplane, helps to reflect most of the heat radiation from the sun.

Question 5.

Why do people prefer wearing white clothes during summer?

Answer:

People prefer white or light coloured clothes during summer as they are good reflectors of heat and hence, they keep us cool.

Question 6.

What is specific heat capacity?

Answer:

  • The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 kg of the substance by 1°C or 1 K.
  • Q = mC∆T, where Q is the quantity of heat required to raise the temperature and m is the mass of the body and AT is the change in temperature of the body.
  • The SI unit of specific heat capacity is Jkg– 1 K– 1. The most commonly used units of specific heat capacity are J/kg°C and J/g°C.

Question 7.

Define thermal capacity.

Answer:

  • Heat capacity or thermal capacity is defined as the amount of heat energy required to raise the temperature of a body by 1°C. It is denoted by ‘C’.
  • C = Q/t, where C’ is the heat capacity, ‘Q’ is the quantity of heat required and ‘f’ is rise in temperature.
  • SI unit of heat capacity is J/K. It is also expressed in cal/°C, kcal/°C or J/°C.

Question 8.

Define specific latent heat capacity.

Answer:

  • Specific latent heat is the amount of heat energy absorbed or liberated by unit mass of a substance during change of state without causing any change in temperature.
  • Specific latent heat is given as L = Q/m, where ‘Q’ is the amount of heat energy absorbed or liberated and ‘m’ is mass of a substance during its change of phase at a constant temperature.
  • The SI unit of specific latent heat is J/kg.

V. Answer in detail.

Question 1.

Explain convection in daily life.

Convection in daily life:

(i) Hot air balloons: Air molecules at the bottom of the balloon get heated by a heat source and rise. As the warm air rises, cold air is pushed downward and it is also heated. When the hot air is trapped inside the balloon, it rises.

(ii) Breeze: During day time, the air in contact with the land becomes hot and rises. Now the cool air over the surface of the sea replaces it. It is called sea breeze. During night time, air above the sea is warmer. As the warmer air over the surface of the sea rises, cooler air above the land moves towards the sea.

(iii) Winds: Air flows from area of high pressure to area of low pressure. The warm air molecules over hot surface rise and create low pressure. So, cooler air with high pressure flows towards low pressure area. This causes wind flow.

(iv) Chimneys: Tall chimneys are kept in kitchen and industrial furnaces. As the hot gases and smoke are lighter, they rise up in the atmosphere.

Question 2.

What are the changes of state in water? Explain.

Answer:

Any matter around us can be in three forms: solid, liquid and gas, called as states of matter. Depending upon the temperature, pressure and transfer of heat, matter is converted from one state to another and is known as change of state in matter. There are different such processes in the change of state in matter.

For example;

  • Water molecules are in liquid state at normal temperature. When water is heated to 100°C, it becomes steam or vapour which is a gaseous state of matter. The process by which a
    liquid is converted to vapour by absorbing heat is called boiling or vaporization.
  • The temperature at which a liquid changes its state to gas is called boiling point.
  • On reducing the temperature of the steam it becomes water again. The process by which a vapour is converted to liquid by releasing heat is called condensation. On reducing the temperature of water further to 0°C, it becomes ice which is a solid state of water.
  • The process by which a liquid is converted to solid by releasing heat is called freezing. The
    temperature by which a liquid changes its state to solid is called freezing point. Ice on
    heating, becomes water again by absorbing heat, a process known as melting.
  • Dry ice changes directly to gaseous state without becoming liquid. This process is called
    sublimation.

Thus, water changes its state when there is a change in temperature.


Question 3.

How can you experimentally prove that water is a bad conductor of heat? How is it possible to heat water easily while cooking.

Experiment to prove that water is a bad conductor of heat:

Answer:

Take a glass tube and drop an ice cube wrapped in wire gauze in it.

Now fill 3/4th of this tube with water and place it above the burner as shown in the figure.

You can observe that the water boils at the edge and the ice present in the bottom of the tube has not melted indicating that heat has not reached the bottom where the ice cube is present. This proves that water is a bad conductor of heat.

It is easy to heat water easily or quickly while cooking. This is because, while cooking the vessel or pan is usually covered with a lid.

This leads to three things;

  1. Radiation from the hot water is reflected back into the pan rather than being emitted
  2. Free convection is effectively eliminated, and
  3. Evaporative cooling’ is also eliminated.
    This in turn allows the water to be heated more easily.

VI. Problems:

Question 1.

What is the heat in joules required to raise the temperature of 25 grams of water from 0°C to 100°C? What is the heat in Calories? (Specific heat of water = 4.18 J/g°C).

Solution:

Given m = 25 g, ∆T = (100 – 0) = 100°C

Or in terms of Kelvin (373.15 – 273.15) = 100K,

C = 4.18J/g°C

Heat energy required, Q = m × C × ∆T = 25 × 4.18 × 100 = 10450 J


Question 2.

What could be the final temperature of a mixture of 100 g of water at 90°C and 600g of water at 20°C.

Solution:

To find final temperature: ∆Q = mc

lOOg of water originally at 90°C will loose an amount of heat,

∆Q = mc ∆T

∆Q = 100 × c × (90 – T)

The same amount of heat will be absorbed by 600g of water originally at 20°C to raise its temperature to T.

∆Q = 600 × c × (T – 30)

600C (T – 20°) = 100C (90° – T)

6T – 120° = 90° – T

6T + T = 120° + 90°

7T = 210° ⇒ T = 210/7

T = 30°C


Question 3.

How much heat energy is required to change 2 kg of ice at 0°C into water at 20°C? (Specific latent heat of fusion of water = 3,34,000J/kg, Specific heat capacity of water = 4200Jkg– 1 K– 1 ).

Solution:

Total heat = Heat required to convert 2Kg of ice into water at 0°C + Heat required to convert 2Kg of water at 0°C to 2Kg of water at 20°C

Heat = m (hfw) + mc∆T

Here, m(mass of ice) = 2Kg

hfw (specific latent heat of water) = 3,34,000J/Kg

C (specific heat capacity of water) = 4200JKg– 1K– 1

AT (Temperature difference) = 20°C

Therefore, Heat required = (2 × 334000) + (2 × 4200) (20 – 0)

= 668000 + 8400 (20)

= 668000+ 168000

Heat required = 8,36,000 J


10th Science, Laws of Motion, Textual, Solved Problems, physics, Tamil nadu, board,

Question 1.
Calculate the velocity of a moving body of mass 5 kg whose linear momentum is 2.5 kg ms-1.
Solution:
Linear momentum = mass × velocity
Velocity = linearmomentummass
V = 2.50.5 = 0.5 ms-1.

Question 2.
A door is pushed, at a point, whose distance from the hinges is 90 cm, with a force of 40 N. Calculate the moment of the force about the hinges.
Solution:
Formula: The moment of a force M = F × d
Given: F = 40 N and d = 90 cm = 0.9 m.
Hence, moment of the force = 40 × 0.9 = 36 Nm.

Question 3.
At what height from the centre of the Earth the acceleration due to gravity will be 14th of its value as at the Earth.
Solution:

Data: Height from the centre of the Earth, R’ = R + h
The acceleration due to gravity at that height, g’ = g4
Formula:

From the centre of the Earth, the obect is placed at twice the radius of the earth.


10th Science Laws of Motion Textbook Evaluation,

I. Choose the correct answer.

1. The inertia of a body depends on _____ .
(a) weight of the object
(b) acceleration due to gravity of the planet
(c) mass of the object
(d) Both a & b.
Answer:
(c) mass of the object

Question 2.
Impulse is equals to:
(a) rate of change of momentum
(b) rate of force and time
(c) change of momentum
(d) rate of change of mass
Answer:
(c) change of momentum

Question 3.
Newton’s III law is applicable to ______ .
(a) for a body is at rest
(b) for a body in motion
(c) both a & b
(d) only for bodies with equal masses.
Answer:
(c) both a & b

Question 4.
Plotting a graph for momentum on the X-axis and time on Y-axis. The slope of the momentum-time graph gives:
(a) Impulsive force
(b) Acceleration
(c) Force
(d) Rate of force
Answer:
(c) Force

Question 5.
In which of the following sport the turning of the effect of force used?
(a) swimming
(b) tennis
(c) cycling
(d) hockey.
Answer:
(c) cycling


Question 6.
The unit of ‘g’ is ms-2. It can be also expressed as _____ .
(a) cm s-1
(b) N kg-1
(c) Nm2 kg-1
(d) cm2 s-2
Answer:
(b) N kg

Question 7.
One kilogram force equals to:
(a) 9.8 dyne
(b) 9.8 × 104 N
(c) 98 × 104 dyne
(d) 980 dyne
Answer:
(c) 98 × 104 dyne

Question 8.
The mass of a body is measured on planet Earth as M kg. When it is taken to a planet of radius half that of the Earth then its value will be ____ kg.
(a) 4M
(b) 2M
(c) M4
(d) M.
Answer:
(c) M4

Question 9.
If the Earth shrinks to 50% of its real radius its mass remaining the same, the weight of a body on the Earth will _____ .
(a) decrease by 50%
(b) increase by 50%
(c) decrease by 25%
(d) increase by 300%.
Answer:
(c) decrease by 25%

Question 10.
To project the rockets which of the following
principle(s) is /(are) required?
(a) Newton’s third law of motion
(b) Newton’s law of gravitation
(c) Law of conservation of linear momentum
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)


II. Fill in the blanks.

Question 1.
To produce a displacement _____ is required.
Answer:
force.

Question 2.
Passengers lean forward when the sudden brake is applied in a moving vehicle. This can be explained by ____.
Answer:
the inertia of motion

Question 3.
By convention, the clockwise moments are taken as _____ and the anticlockwise moments are taken as _____.
Answer:
negative, positive.

Question 4.
______ is used to change the speed of the car.
Answer:
Acceleration.

Question 5.
A man of mass 100 kg has a weight of _____ at the surface of the Earth.
Answer:
980 N.

III. State whether the following statements are true or false. Correct the statement if it is false:

Question 1.
The linear momentum of a system of particles is always conserved.
Answer:
False.
Correct Statement: The linear momentum of a system of particles is always conserved if no external force acts.

Question 2.
The apparent weight of a person is always equal to his actual weight.
Answer:
True.

Question 3.
Weight of a body is greater at the equator and less at the polar region.
Answer:
False.
Correct Statement: Weight of a body is lesser at the equator and greater at the polar region.

Question 4.
Turning a nut with a spanner having a short handle is so easy than one with a long handle.
Answer:
False.
Correct Statement: Turning a nut with a spanner having a short handle is so harder than one with a long handle.

Question 5.
There is no gravity in the orbiting space station around the Earth. So the astronauts feel weightlessness.
Answer:
False.
Correct Statement: There is a gravity in the orbiting space station around the earth. Since space station and astronauts have equal acceleration. Both the astronauts and space station are in the state of weightlessness.


IV. Match the following.

Question 1.

Column IColumn II
1. Newton’s I law(a) Propulsion of a rocket
2. Newton’s II law(b) Stable equilibrium of a body
3. Newton’s III law(c) Law of force
4. Law of conservation of linear momentum(d) Flying nature of a bird

Answer:
1. (b) Stable equilibrium of a body
2. (c) Law of force
3. (d) Flying nature of a bird
4. (a) Propulsion of a rocket.


V. Assertion & Reasoning

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: The sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Reason: The principle of conservation of momentum is valid if the external force on the system is zero.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 2.

  1. Assertion: The value of ‘g’ decreases as height and depth increase from the surface of the Earth.
  2. Reason: ‘g’ depends on the mass of the object and the Earth.

Answer:
(c) The assertion is true, but the reason is false.


VI. Answer briefly.

Question 1.
Define inertia. Give its classification.
Answer:
Inertia: The inherent property of a body to resist any change in its state of rest or the state of uniform motion, unless it is influenced upon by an external unbalanced force, is known as ‘inertia’.
Types of Inertia

  • Inertia of rest
  • Inertia of motion
  • Inertia of direction

Question 2.
Classify the types of force based on their application.
Answer:

  1. Like parallel forces
  2. Unlike parallel forces

Question 3.
If a 5 N and a 15 N forces are acting opposite to one another. Find the resultant force and the direction of action of the resultant force.
Solution:
The two forces are unlike parallel forces
Samacheer Kalvi 10th Science Solutions Chapter 1 Laws of Motion 2
Let P = 5N, Q = 15N
Resultant force (R) = P – Q = 5 + (-15) = -10N
R = -10N.
The resultant force acting along the direction of “Q”.

Question 4.
Differentiate mass and weight.
Answer:

MassWeight
The quantity of matter contained in the bodyThe gravitation force exerted on it due to the Earth’s gravity alone.
Scalar quantityVector quantity
Unit: KgUnit: N
Constant at all the placesVariable with respect to gravity.

Question 5.
Define moment of a couple.
Answer:
Rotating effect of a couple is known as moment of a couple.
Moment of a couple = Force × perpendicular distance between the line of action of forces
M = F × S

Question 6.
State the principle of moments.
Answer:
When a number of like or unlike parallel forces act on a rigid body and the body is in equilibrium, then the algebraic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anticlockwise direction.

Question 7.
State Newton’s second law.
Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.

Question 8.
Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
Answer:
This is because turning effect to tighten the screws depends upon the perpendicular distance of the applied force from the axis of rotation is power arm. Larger the power armless is the force required to turn the screws. So spanner is provided with a long handle.

Question 9.
While catching a cricket ball the fielder lowers his hands backwards. Why?
Answer:
(i) When the fielder lowers his hands backwards, he increases the value of time of collision and so retardation is decreased.
(ii) Hence retarding force becomes lesser than before and the palm of the fielder is not hurt very much.

Question 10.
How does an astronaut float in a space shuttle?
Answer:
An astronaut float in a space shuttle because both are in the state of weightlessness. Both are experiencing equal acceleration towards earth as free fall bodies. Astronauts are not floating but falling freely.


VII. Solve the given problems.

Question 1.
Two bodies have a mass ratio of 3 : 4. The force applied to the bigger mass produces an acceleration of 12 ms-2. What could be the acceleration of the other body, if the same force acts on it?
Solution:
Mass ratio of
the bodies = 3 : 4 and same force is (m1 : m2) acting on the body and a2 = 12 ms-2
∴ m1a1 = m2a2
m1m2=a2a1⇒34=a2a1
a1=43×12=16ms−2

Question 2.
A ball of mass 1 kg moving with a speed of 10 ms-1 rebounds after a perfectly elastic collision with the floor. Calculate the change in linear momentum of the ball.
Solution:
Mass of a ball = 1 kg
Velocity of the bail before collision,
u = 10 m/s
Velocity of the ball after collision,
v = – u
= -10 m/s
Change in momentum,
P = m(v – u)
P = 1(-10 – 10)
= -20 kg m/s.

Question 3.
A mechanic unscrews a nut by applying a force of 140 N with a spanner of length 40 cm. What should be the length of the spanner if a force of 40 N is applied to unscrew the same nut?
Solution:
Given F1 = 140 N, d1 = 40 cm, F2 = 40 N, d2= ?
In, both the cases, moment of forces applied are equal
F1d1 = F2d2
d2=(F1F2)d1d2=40×14040=140cm

Question 4.
The ratio of masses of two planets is 2 : 3 and the ratio of their radii are 4 : 7 Find the ratio of their accelerations due to gravity.
Solution:
The ratio of masses of two planets m1 : m2 = 2 : 3
The ratio of radii of two planets R1 : R2 = 4 : 7
Formula:
Samacheer Kalvi 10th Science Solutions Chapter 1 Laws of Motion 3


VIII. Answer in detail.

Question 1.
What are the types of inertia? Give an example for each type.
Answer:
Types of Inertia:
(i) Inertia of rest: The resistance of a body to change its state of rest is called inertia of rest. Eg: When you vigorously shake the branches of a tree, some of the leaves and fruits are detached and they fall down.

(ii) Inertia of motion: The resistance of a body to change its state of motion is called inertia of motion. Eg: An athlete runs some distance before jumping. Because, this will help him jump longer and higher.

(iii) Inertia of direction: The resistance of a body to change its direction of motion is called inertia of direction. Eg: When you make a sharp turn while driving a car, you tend to lean sideways.

Question 2.
State Newton’s laws of motion.
Answer:
Newton’s First Law: Everybody continues to be in its state of rest or the state of uniform motion along a straight line unless it is acted upon by some external force.

Newtons Second law: The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force.

Newtons Third Law: For every action, there is an equal and opposite reaction. They always act on two different bodies.

Question 3.
Deduce the equation of a force using Newton’s second law of motion.
Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.
Let, ‘m’ be the mass of a moving body, moving along a straight line with an initial speed ‘u’ After a time interval of ‘t’, the velocity of the body changes to ‘v’ due to the impact of an unbalanced external force F.
Initial momentum of the body (Pi) = mu
Final momentum of the body (Pf) = mv
Change in momentum ∆p = Pf – Pi = mv – mu
By Newton’s second law of motion,
Force, F ∝ rate of change of momentum
F ∝ change in momentum / time
F∝mv−mutF=km(v−u)t
Here, k is the proportionality constant, k = 1 in all systems of units.
Hence, F=m(v−u)t
Since, acceleration = change in velocity / time,
a = (v – u)/t.
Hence, we have F = m × a
Force = mass × acceleration

  • No external force is required to maintain the motion of a body moving with uniform velocity.
  • When the net force acting on a body is not equal to zero, then definitely the velocity of the body will change.
  • Thus, change in momentum takes place in the direction of the force. The change may take place either in magnitude or in direction or in both.

Question 4.
State and prove the law of conservation of linear momentum.
Answer:
(i) There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
(ii) Let us prove the law of conservation of linear momentum with the following illustration:
Samacheer Kalvi 10th Science Solutions Chapter 1 Laws of Motion 4
(iii) Let two bodies A and B having masses m1 and m2 move with initial velocity u1and u2 in a straight line.
(iv) Let the velocity of the first body be higher than that of the second body. i.e., u1 > u2.
(v) During an interval of time t second, they tend to have a collision. After the impact, both of them move along the same straight line with a velocity v1 and v2respectively.
Force on body B due to A,
FB=m2[v2−u2]t
Force on body A due to B,
FA=m1[v1−u1]t
By Newton’s III law of motion, Action force = Reaction force
FA = -FB
m1[v1−u1]tm1v1+m2v2=m2[v2−u2]t=m1u1+m2u2
The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
Hence the law of conservation linear momentum is proved.

Question 5.
Describe rocket propulsion.
Answer:
Propulsion of rockets is based on the law of conservation of linear momentum as well as Newton’s III law of motion. Rockets are filled with a fuel (either liquid or solid) in the propellant tank. When the rocket is fired, this fuel is burnt and a hot gas is ejected with a high speed from the nozzle of the rocket, producing a huge momentum. To balance this momentum, an equal and opposite reaction force is produced in the combustion chamber, which makes the rocket project forward.

While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out. Since, there is no net external force acting on it, the linear momentum of the system is conserved. The mass of the rocket decreases with altitude, which results in the gradual increase in velocity of the rocket. At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity.


Question 6.
State the universal law of gravitation and derive its mathematical expression.
Answer:
This law states that every particle of matter in this universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centres of these masses. The direction of the force acts along the line joining the masses.

The force between the masses is always attractive and it does not depend on the medium where they are placed.
Let, m1 and m2 be the masses of two bodies A and B placed r metre apart in space.
Force F ∝ m1 × m2
F ∝ 1/r2
On combining the above two expressions
F∝m1×m2r2F=Gm1m2r2
Where G is the universal gravitational constant.
Its value in SI unit is 6.674 × 10-11 Nm2 kg-2.

Question 7.
Give the applications of the universal law of gravitation.
Answer:
Application of Newton’s law of gravitation are:
(i) Dimensions of the heavenly bodies can be measured using the gravitation law. Mass of the Earth, radius of the Earth, acceleration due to gravity, etc., can be calculated with a higher accuracy.
(ii) Helps in discovering new stars and planets.
(Hi) Helps to explain germination of roots is due to the property of geotropism which is the property of a root responding to the gravity.
(iv) One of the irregularities in the motion of stars is called ‘Wobble’ lead to the disturbance in the motion of a planet nearby. In this condition the mass of the star can be calculated using the law of gravitation.
(v) Helps to predict the path of the astronomical bodies.


IX. HOT Questions.

Question 1.
Two blocks of masses 8 kg and 2 kg respectively, lie on a smooth horizontal surface in contact with one other. They are pushed by a horizontally applied force of 15 N. Calculate the force exerted on the 2 kg mass.
Solution:
Samacheer Kalvi 10th Science Solutions Chapter 1 Laws of Motion 6
Given: m1 = 8 kg, m2 = 2 kg, F = 15 N
F = mtotal, F = (m1 + m2) a = (8 + 2) a = 10 a
15 = 10 a
⇒ a = 1510=32 ms-2
Force exerted by mass of 8 kg
F = ma = 8×32 = 12 N.

Question 2.
A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta. (Ratio of momenta = 1 : 2)
Solution:
Given: Let m1, m2 are the masses of truck and bike.
m1 = 4m2
Here kinetic energies of both truck and bike are same
m1v214m2v21v1v2v2=m2v22=m2v22=12=2v1
Ratio of momenta: p1p2=m1v1m2v2=4m2m2⋅v12v1 = 2
P1 : P2 = 2 : 1.

Question 3.
“Wearing a helmet and fastening the seat belt is highly recommended for the safe journey”.Justify your answer using Newton’s laws of motion.
Answer:
During the motion of car and two wheelers, when the brakes are applied, the vehicles slow down but our body tends to continue in the same state of motion due to inertia. So this may cause injury to passengers. Hence they are advised to wear a helmet and seat belt.


10th Science Laws of Motion Additional Questions

I. Choose the correct answer.

Question 1.
A cricketer catches a ball of mass 150 gm in 0.1s and which is moving with a speed of 20 ms-1, then he experiences the force of ____ _.
(a) 300 N
(b) 30 N
(c) 3 N
(d) 0.3 N.
Answer:
(b) 30 N
Hint: Impulse = change in momentum
F.∆t = mv – mu
F=mv−muΔt
=150×10−3×200.1=30N

Question 2.
SI unit of force is:
(a) Dyne
(b) newton
(c) kgms-1
(d) Joule
Answer:
(b) newton

Question 3.
A coin is dropped in a lift. It takes time t1to reach the floor, when the lift is stationary, it takes time t2, when the lift is moving up with constant acceleration, then ____ _.
(a) t1 > t2
(b) t1 < t2
(c) t1 = t2
(d) None.
Answer:
(a) t1 > t2

Question 4.
An unbalanced force acts on a body, the body:
(a) must remain at rest
(b) must be accelerated
(c) must move with uniform velocity
(d) move with uniform motion
Answer:
(b) must be accelerated

Question 5.
A satellite in its orbit around the earth is weightless on account of its _____.
(a) velocity
(b) momentum
(c) angular momentum
(d) acceleration.
Answer:
(c) angular momentum

Question 6.
When two or more forces acting on a body and the body does not change its position, then the forces are:
(a) imbalanced
(b) mechanical force
(c) balanced forces
(d) none
Answer:
(c) balanced forces

Question 7.
What would be the acceleration due to gravity at another planet, whose mass and radius core twice that of earth?
(a) g
(b) g2
(c) g4
Answer:

Answer:
(b) g2
Hint: We know that g=GMR2
g1g2=(GMR2)(G.2M4R2)
g1g2=2⇒g2=12g1

Question 8.
At sea level, the value of “g” is maximum at _____.
(a) the poles
(b) the equator
(c) 45° south latitude
(d) 45° north of longitude.
Answer:
(a) the poles

Question 9.
An object cannot change the state of rest or motion, until a force is applied. This inability of the object is called:
(a) inertia
(b) mass
(c) weight
(d) acceleration
Answer:
(a) inertia

Question 10.
The ability of a body to maintain its state of rest or motion is called ______.
(a) mechanics
(b) kinematics
(c) kinetics
(d) Inertia.
Answer:
(d) Inertia.

Question 11.
_____ deals with the bodies, which are at rest under the action of forces.
(a) Statics
(b) Dynamics
(c) Kinematics
(d) Kinetics.
Answer:
(a) Statics

Question 12.
A motor car starts from rest and moves after 5 seconds. If its velocity is 200 m/s then its acceleration is:
(a) 100 m/s²
(b) 40 m/s²
(c) 20 m/s²
(d) 80 m/s²
Answer:
(b) 40 m/s²

Question 13.
_____ deals with the motion of bodies considering the cause of motion.
(a) Force
(b) Dynamics
(c) Statics
(d) Kinetics.
Answer:
(d) Kinetics

Question 14.
Linear momentum = _____
(a) mass × velocity
(b) mass × distance
(c) distance × time
(d)  mass  velocity .
Answer:
(a) mass × velocity

Question 15.
The inability of the body to change its state is:
(a) force
(b) momentum
(c) acceleration
(d) inertia
Answer:
(d) inertia

Question 16.
Two or more forces of equal or unequal magnitude acting along the same direction, parallel to each other are called _____.
(a) like parallel forces
(b) unlike parallel forces
(c) resultant force
(d) balanced force.
Answer:
(a) like parallel forces

Question 17.
The axis of the fixed edge about which the thing is rotated is called as _____ .
(a) axis of rotation
(b) fixed axis rotation
(c) point of rotation
(d) Fixed point.
Answer:
(a) axis of rotation

Question 18.
When a net force acts on an object, the object will be accelerated in the direction of force with an acceleration proportional to:
(a) force on the object
(b) velocity
(c) mass
(d) inertia
Answer:
(a) force on the object

Question 19.
Rotating effect of a couple is known as ______ .
(a) product of forces
(b) the momentum of a couple
(c) mass
(d) momentum.
Answer:
(b) momentum of a couple

Question 20.
The amount of force required to produce an acceleration of 1 ms-2 in a body of mass _____ is called unit force.
(a) 10 kg
(b) 100 kg
(c) 1 kg
(d) 0 kg.
Answer:
(c) 1 kg

Question 21.
The acceleration of a body is due to:
(a) balance force
(b) electrostatic force
(c) unbalanced force
(d) conservative force
Answer:
(c) unbalanced force

Question 22.
Universal gravitational constant ______ .
(a) G = 6.684 × 10-10 Nm2 kg-1
(b) G = 7.4 × 1010 Nm2
(c) G = 6.623 × 1011 Nm2 kg-1
(d) G = 6.674 × 10-11 Nm2 kg-2
Answer:
(d) G = 6.674 × 10-11 Nm2 kg-2

Question 23.
Mean value of the acceleration due to gravity is ______ .
(a) 10.1 ms-2
(b) 8.8 ms-2
(c) 9.8 ms-2
(d) 9.8 ms.
Answer:
(c) 9.8 ms-2

Question 24.
The unit of weight is:
(a) kg
(b) g
(c) Newton
(d) ms-1
Answer:
(c) Newton

Question 25.
The value of accelaration due to gravity on the surface of the moon is _____ .
(a) 1.75 ms-1
(b) 3.8 ms-2
(c) 1.625 ms-2
(d) 1.625 ms-1
Answer:
(c) 1.625 ms-2

Question 26.
The unit of weight is _____ .
(a) kg m
(b) kg
(c) newton
(d) kg m-1
Answer:
(c) newton

Question 27.
The weight of a body is _____ poles than at the equatorial region.
(a) more
(b) less
(c) zero
(d) one.
Answer:
(a) more

Question 28.
In a collision between a heavier body and a lighter body, which body experiences greater force?
(a) heavier body
(b) lighter body
(c) both the body experience same force
(d) both body exchange acceleration
Answer:
(c) both the body experience same force.


II. Fill in the blanks.

Question 1.
Turning a tap is an example of ____
Answer:
couple.

Question 2.
Torque is a _______ quantity.
Answer:
vector.

Question 3.
1 gf is equal to _____ dyne.
Answer:
980.

Question 4.
The resultant force acting on a body is equal to zero then the body will be in ______
Answer:
equilibrium.

Question 5.
The force equal to resultant but opposite in direction is ______
Answer:
equilibrate.

Question 6.
The product of force and time is ______
Answer:
impulse.

Question 7.
The force between the masses is always ______
Answer:
attractive.

Question 8.
The quantity of matter contained in the object is known as _____
Answer:
mass.

Question 9.
The magnitude of the universal gravitational constant is _____.
Answer:
6.674 × 10-11 Nm2 kg-2

Question 10.
Propulsion of rockets is based on the ____ and ____
Answer:
Law of conservation of linear momentum & Newton’s third law.

Question 11.
Parallel unequal forces are acting in ______ directions.
Answer:
Opposite.

Question 12.
Torque and force are the ____ quantities.
Answer:
vector.

Question 13.
The unit of moment of a couple is _____ .
Answer:
newton metre (Nm).

Question 14.
A _____ enables you to manoeuvre a car easily by transferring a _______ to the wheels with less effort.
Answer:
steering wheel, torque.

Question 15.
_____ is required to produce the acceleration of a body.
Answer:
Force.

Question 16.
The acceleration is produced along the radius is called ______
Answer:
centripetal acceleration.

Question 17.
______ is equal to the magnitude of change in momentum.
Answer:
impulse.

Question 18.
A large force acting for a very short interval of time is called as ______ .
Answer:
impulse Force.

Question 19.
Mass of the earth _____
Answer:
5.972 × 1024 kg.

Question 20.
The relation between acceleration due to gravity (g) and the universal gravitational constant (G) is _____ .
Answer:
g=GMR2.


III. State whether the following statements are true or false, correct the statement if it is false.

Question 1.
Rest and motion are interrelated terms.
Answer:
True.

Question 2.
In the C.G.S. system, the unit of linear momentum is kg ms-1.
Answer:
False.
Correct Statement: In the C.G.S. system, the unit of linear momentum is g cms-1.

Question 3.
An external force is required to maintain the motion of a body moving with uniform velocity.
Answer:
False.
Correct Statement: No external force is required to maintain the motion of a body moving with uniform velocity.

Question 4.
The amount of force required for a body of mass 1 gram produces an acceleration of 1 cm s-2
Answer:
True.

Question 5.
By Newton’s III – law of motion, the action force is not equal to the reaction force.
Answer:
False.
Correct Statement: By Newton’s III – law of motion, the action force is equal to the reaction force.

Question 6.
The value of acceleration due to gravity (g) is not the same at all the points on the surface of the earth.
Answer:
True.

Question 7.
The value of acceleration due to gravity on the surface of the moon is 1.625 ms-2.
Answer:
True.

Question 8.
The regularities in the motion of stars are called ‘wobble’.
Answer:
False.
Correct Statement: The irregularities in the motion of stars is called ‘wobble’.

Question 9.
Mechanics is divided into kinematics and kinetics.
Answer:
False.
Correct Statement: Mechanics is divided into statics and dynamics.

Question 10.
Application of Newton’s law of gravitation helps to predict the path of the astronomical bodies.
Answer:
True.


IV. Match the following.

Question 1.

1. Linear momentum(a) Mass and acceleration
2. Force(b) Change in momentum
3. Moment of force(c) GM/R2
4. Impulse(d) Mass and velocity
5. Acceleration due to gravity(e) Force and perpendicular distance

Answer:
1. (d) Mass and velocity
2. (a) Mass and acceleration
3. (e) Force and perpendicular distance
4. (b) Change in momentum
5. (c) GM/R2

Question 2.

1. Kinetics(a) Causes the motion
2. Kinematics(b) In equilibrium
3. Balanced force(c) The motion of bodies without cause
4. Unbalanced force(d) The motion of bodies with cause

Answer:
1. (d) The Motion of bodies with cause
2. (c) The Motion of bodies without cause
3. (b) In equilibrium
4. (a) Causes the motion


V. Assertion & Reasoning

Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.

Question 1.
Assertion: At poles value of acceleration due to gravity (g) is greater than that of the equator.
Reason: Earth rotates on its axis in addition to revolving around the sun.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 2.
Assertion: The force exerted by the earth on an apple is more than that exerted by apple on the earth.
Reason: The force on apple exerts on the earth is determined by the mass of the apple only.
Answer:
(d) The assertion is false but the reason is true

Question 3.
Assertion: A freely falling body is in the state of weightlessness
Reason: A body becomes conscious of its weight only when it is opposed
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 4.
Assertion: Newton’s third law of motion is applicable only when bodies are in motion.
Reason: Newton’s third law applies to all types of forces, e.g. gravitational, electric or magnetic forces.
Answer:
(d) The assertion is false but the reason is true.

Question 5.
Assertion: The apparent weight of the person is zero, in which condition or state is known as weightless.
Reason: When the person in a lift moves down with an acceleration (a) is equal to the acceleration due to gravity (g)
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 6.
Assertion: A gear is a circular wheel with teeth around its rim.
Reason: It helps to change the speed of rotation of a wheel by changing the force and helps to transmit power.
Answer:
(c) The assertion is true, but the reason is false.

Question 7.
Assertion: Mass of a body is defined as the gravitational force exerted on it due to earth’s gravity alone
Reason: Weight = mass × acceleration due to gravity.
Answer:
(d) The assertion is false, but the reason is true.

Question 8.
Assertion: Weight is a vector quantity.
Reason: Direction of weight is always towards the centre of the earth.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.

Question 9.
Assertion: Resultant force is equal to the vector sum of all the forces.
Reason: A system cannot be brought to equilibrium by applying another force.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.


VI. Answer briefly

Question 1.
Define Linear momentum.
Answer:
The product of mass and velocity of a moving body gives the magnitude of linear momentum. It acts in the direction of the velocity of the object. Linear momentum is a vector quantity.
Linear Momentum = mass × velocity.
Unit of momentum in SI system is Kg ms-1and in C.G.S system its unit is g cm s-1

Question 2.
What is the resultant force?
Answer:
When several forces act simultaneously on the same body, then the combined effect of the multiple forces can be represented by a single force, which is termed as ‘resultant force’.

Question 3.
What is meant by equilibrant?
Answer:
A system can be brought to equilibrium by applying another force, which is equal to the resultant force in magnitude, but opposite in direction. Such force is called as ‘Equilibrant’.

Question 4.
Explain the Newton third law of motion with examples.
Answer:
‘For every action, there is an equal and opposite reaction. They always act in two different bodies’.
Example: When you fire a bullet, the gun recoils backwards and the bullet is moving forward (Action) and the gun equalises this forward action by moving backwards (Reaction).

Question 5.
How did the change in momentum achieve?
Answer:
Change in momentum can be achieved in two ways. They are:

  • A large force acting for a short period of time and
  • A smaller force acting for a longer period of time.

Question 6.
Define impulse.
Answer:
A large force acting for a very short interval of time is called as ‘Impulsive force’.When a force F acts on a body for a period of time t, then the product of force and time is known as ‘impulse’ represented by ‘J’
Impulse, J = F × t …….. (1)
By Newton’s second law
F = ∆p / t (A refers to change)
∆p = F × t ………. (2)
From (1) and (2)
J = ∆p
Impulse is also equal to the magnitude of change in momentum.
Its unit is kg ms-1 or Ns.

Question 7.
What is meant by free fall?
Answer:

  1. When the person in a lift moves down with an acceleration (a) equal to the acceleration due to gravity (g), i.e., when a = g, this motion is called as ‘free fall’.
  2. The apparent weight (R = m (g – g) = 0) of the person is zero. This condition or state refers to the state of weightlessness.

Question 8.
Define weightlessness.
Answer:
Whenever a body or a person falls freely under the action of Earth’s gravitational force alone, it appears to have zero weight. This state is referred to as ‘weightlessness’.

Question 9.
Explain the various causes of the apparent weight of a person in a moving lift.
Answer:

Case 1: Lift is moving upward with an acceleration ‘a’Case 2: Lift is moving downward with an acceleration ‘a’Case 3: Lift is at rest.Case 4: Lift is falling down freely
R – W = Fnet= ma
⇒ R = W + ma
⇒ R = mg + ma
⇒ R = m(g + a)
W – R = Fnet= ma
⇒ R = W – ma
⇒ R = mg – ma
⇒ R = m(g – a)
Here,the acceleration is zero
a = 0
R = W
R = mg
Here,the acceleration is equal to g
a = g
R = m(g – g) = 0
R > WR < WR = WR = 0
Apparent weight is greater than the actual weight.Apparent weight is lesser than the actual weight.Apparent weight is equal to the actual weight.Apparent weight is equal to zero.

Question 10.
Explain the variation of acceleration due to gravity.
Answer:
Variation of acceleration due to gravity (g):

  1. Since, g depends on the geometric radius of the Earth, (g ∝ 1 / R2), its value changes from one place to another on the surface of the Earth.
  2. The geometric radius of the Earth is maximum in the equatorial region and minimum in the polar region, the value of g is maximum in the polar region and minimum at the equatorial region.
  3. When you move to a higher altitude from the surface of the Earth, the value of g reduces.
  4. when you move deep below the surface of the Earth, the value of g reduces. Value of g is zero at the centre of the Earth.

Question 11.
Define one newton and one dyne.
Answer:
Definition of 1 newton (N): The amount of force required for a body of mass 1 kg produces an acceleration of 1 ms-2, 1 N = 1 kg ms-2
Definition of 1 dyne: The amount of force required for a body of mass 1 gram produces an acceleration of 1 cms-2, 1 dyne = 1 g cms-2; also 1 N = 105 dyne.

Question 12.
How can you measure the moment of the couple?
Answer:
(i) Rotating effect of a couple is known as the moment of a couple.
(ii) Moment of a couple is measured by the product of any one of the forces and the perpendicular distance between the line of action of two forces. The turning effect of a couple is measured by the magnitude of its moment.
(iii) Moment of a couple = Force × perpendicular distance between the line of action of forces
M = F × S
(iv) The unit of moment of a couple is newton metre (N m) in SI system and dyne cm in the CGS system.
(v) By convention, the direction of moment of a force or couple is taken as positive if the body is rotated in the anti-clockwise direction and negative if it is rotating in the clockwise direction.
They are shown in Figures.

Clock wise moment and Anti clock wise moment

Question 13.
Define Torque.
Answer:
(i) The rotating or turning effect of a force about a fixed point or fixed axis is called the moment of the force about that point or torque (τ).
(ii) τ = F × d
(iii) Torque is a vector quantity.
(iv) Its SI unit is Nm.


VII. Answer in detail.

Question 1.
Explain any three application of Torque.
Answer:
Application of Torque:
(i) Gears: A gear is a circular wheel with teeth around its rim. It helps to change the speed of rotation of a wheel by changing the torque and helps to transmit power.

(ii) Seasaw: Most of you have played on the seesaw. Since there is a difference in the weight of the persons sitting on it, the heavier person lifts the lighter person. When the heavier person comes closer to the pivot point (fulcrum) the distance of the line of action of the force decreases. It causes less amount of torque to act on it. This enables the lighter person to lift the heavier person.

(iii) Steering Wheel: A small steering wheel enables you to manoeuvre a car easily by transferring torque to the wheels with less effort.

Question 2.
State Newton’s third law. Explain it with three examples.
Answer:
Newton’s third law of motion: Newton’s third law states that ‘for every action, there is an equal and opposite reaction. They always act in two different bodies’.
If a body A applies a force FA on a body B, then the body B reacts with force FB on the body A, which is equal to FA in magnitude, but opposite in direction.
FB = -FA
Examples:

  • When birds fly they push the air downwards with their wings (Action) and the air pushes the bird upwards (Reaction).
  • When a person swims he pushes the water using the hands backwards (Action), and the water pushes the swimmer in the forward direction (Reaction).
  • When you fire a bullet, the gun recoils backwards and the bullet is moving forward (Action) and the gun equalises this forward action by moving backwards (Reaction).

Question 3.
Derive the relation between ‘g’ and G. Explain how to determine the mass of earth.
Answer:
(i) Let us compute the magnitude of this force in two ways. Let, M be the mass of the Earth and m be the mass of the body.
(ii) The entire mass of the Earth is assumed to be concentrated at its centre.
(iii) The radius of the Earth is R = 6378 km (= 6400 km approximately). By Newton’s law of gravitation, the force acting on the body is given by
F=GMmR2 ….(1)

(iv) The radius of the body considered is negligible when compared with the Earth’s radius. Now, the same force can be obtained from Newton’s second law of motion.
(v) According to this law, the force acting on the body is given by the product of its mass and acceleration (called weight). Here, acceleration of the body is under the action of gravity hence a = g
F = ma = mg
F = weight = mg ……. (2)
Comparing equations J = F × t and ΔP = F × t, we get
mg=GMmR2 …….. (3)
Acceleration due to gravity
g=GMmR2 ……. (4)
Mass of the Earth (M):
Rearranging the equation (4), the mass of the Earth is obtained as follows:
Mass of the Earth M = g R/ G
Substituting the known values of g, R and G, you can calculate the mass of the Earth as M = 5.972 × 1024 kg.


VIII. Problems.

Question 1.
A cricket ball of mass 0.5 kg strikes a bat normally with a velocity of 30 ms-1 and rebounds with a velocity of 20 ms-1 in the opposite direction, calculate the impulse of the force exerted by the ball on the bat.
Solution:
Impulse = change in momentum = mu – (-mv)
= m (u + v)
= 0.5 (30 + 20)
= 25 Ns

Question 2.
A force exerted on a body of mass 100 g changes its speed by 0.2 ms-1 in each second. Calculate the magnitude of the force.
Given, mass m = 100 g = 0.1 kg and acceleration a = 200 cms-2 = 0.2 ms-2.
Solution:
F = ma = 0.1 × 0.2 = 0.02 N.


CBSE, Syllabus, for, Class 11, 2020 – 2021 (Reduced & Revised), Chemistry,

CBSE Class 11 Chemistry Syllabus (Reduced) 2020-21 (Released on 7th July)

CBSE, Syllabus, for, Class 11, 2020 – 2021 (Reduced & Revised), Physics,

⇒ CBSE Class 11 Physics Syllabus (Reduced) 2020-21 (Released on 7th July) 

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