CUET UG 2024 : VVI Most Important Physics Question with Answers

In this article, we are providing CUET Physics Important Questions 2024. These CUET Physics Questions will help you understand the type of questions asked in CUET Physics sections in the upcoming shift.

Memory-based CUET UG Physics Paper is very beneficial for the candidates whose exam is scheduled in the upcoming shift. In this article, we have provided previous year's asked CUET Physics questions.

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CUET Physics Important Questions 2024

1. Electric Charges and Fields

Multiple Choice Questions

1. Which one of the following is the unit of electric charge?

(a) Coulomb
(b) Newton
(c) Volt
(d) Coulomb/volt

Ans. (a) Coulomb

Explanation: Unit of electric charge is coulomb (C).

2. The electric charge always resides:

(a) at the centre of charged conductor
(b) at the interior of charged conductor
(c) on the outer surface of charged conductor
(d) randomly all over the charged conductor

Ans. (c) on the outer surface of charged conductor

Explanation: Electric charge always resides on the outer surface of charged conductor.

3. One metallic sphere A is given positive charge whereas another identical metallic sphere B of exactly same mass as of A is given equal amount of negative charge. Then:

(a) mass of A and mass of B still remain equals and same
(b) mass of A increases
(c) mass of B decreases
(d) mass of B increases

Ans. (d) mass of B increases

Explanation:

Negative charge means excess of electrons which increases the mass of sphere B, whereas positive charge on sphere A is given by removal of electrons i.e., decrease of mass.

4. Among two discs A and B, first have radius 10 cm and charge 10−6 C and second have radius 30 cm and charge 10−5 C. When they are touched, charge on both qA and qB respectively will, be:

(a) qA = 2.75 mC, qB = 3.15 mC
(b) qA = 1.09 mC, qB = 1.53 mC
(c) qA = qB = 5.5 mC
(d) None of these

Ans. (c) qA = qB = 5.5 mC

Explanation: The charge on disc A is 10−6 C. The charge on disc B is 10 × 10−6 C. The total charge on both = 11 × 10−6 C. When touched, this charge will be distributed equally, i.e., 5.5 × 10−6 C or 5.5 mC on each disc.

5. Relative permittivity of water is 81. If ew and e0 are permittivities of water and vacuum respectively, then:

(a) e0 = 9ew
(b) e0 = 81ew
(c) ew = 9e0
(d) ew = 81e0

Ans. (d) ew = 81e0

Explanation: Relative permittivity,

⇒ ew = 81e0

6. The figure shows a charge +q at point P held in equilibrium in air with the help of four +q charges situated at the vertices of a square. The net electrostatic force on q is given by:

(a) Newton’s law
(b) Coulomb’s law
(c) Principle of superposition
(d) Net electric flux out the position of +q

Ans. (c) Principle of superposition

Explanation: The weight mg of the charge held in air is in equilibrium with net electrostatic force exerted by the four charges situated at the corners. The net electrostatic force is given by the vector sum of the individual forces exerted by the charges at the corners. This is principle of superposition.

7. The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is: (mp = 1836 me )

(a) 2F
(b) F
(c) 1836F
(d) F/1836

Ans. (b) F Explanation: Electrostatic force is given by,

6. The figure shows a charge +q at point P held in equilibrium in air with the help of four +q charges situated at the vertices of a square. The net electrostatic force on q is given by:

(a) Newton’s law
(b) Coulomb’s law
(c) Principle of superposition
(d) Net electric flux out the position of +q

Ans. (c) Principle of superposition

Explanation: The weight mg of the charge held in air is in equilibrium with net electrostatic force exerted by the four charges situated at the corners. The net electrostatic force is given by the vector sum of the individual forces exerted by the charges at the corners. This is principle of superposition.

7. The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is: (mp = 1836 me )

(a) 2F
(b) F
(c) 1836F
(d) F/1836

Ans. (b) F

Explanation: Electrostatic force is given by,

Here, charge and distance are same and force does not depend on the mass. So, force between two protons will be same.

8. Two point charges of +3 mC and +4 mC repel each other with a force of 10 N. If each is given an additional charge of −6 mC, the new force is:

(a) 6 N
(b) 6.5 N
(c) 7.5 N
(d) 7 N

Ans. (c) 7.5 N

Explanation: Given that, q1 = 3 mC, q2 = + 4 mC, F = 10 N
q'1 = + 3 – 6 = – 3 mC, q'2 = + 4 – 6 = – 2 mC

9. The electrostatic attracting force on a small sphere of charge 0.2 mC due to another small sphere of charge −0.3 mC in air is 0.3 N. The distance between the two spheres is:

(a) 43.2 × 10^–6 m
(b) 4.24 × 10^–3 m
(c) 4.24 × 10^–2 m
(d) 4.24 m

Ans. (c) 4.24 × 10^–2 m

Explanation: Given that, q1 = 0.2 mC = 0.2 × 10^–6 C
q2 = – 0.3 mC = – 0.3 × 10^–6 C, F = 0.3 N

10. If charge q is placed at the centre of the line joining two equal charges Q, the system of these charges will be in equilibrium if q is:

(a) −4Q
(b) −Q/4
(c) −Q/2
(d) +Q/2

Ans. (b) −Q/4

Explanation: For equilibrium, net force on Q = 0. Let, the distance between line joining two equal charges Q is x unit.

CUET UG Study Material 2024

CUET UG Study Material 2024
Quantitative Aptitude	Numerical Aptitude Data Interpretation
Reasoning	General Knowledge
Maths	Physics
Chemistry	Biology

11. In figure, two positive charges, q2 and q3 fixed along the Y-axis, exert a net electric force in the +x-direction on a charge fixed along the X-axis. If a positive charge Q is added at (x, 0), the force on q1: [NCERT Exemplar]

(a) shall increase along the positive X-axis.
(b) shall decrease along the positive X-axis.
(c) shall point along the negative X-axis.
(d) shall increase but the direction changes because of the intersection of Q with q2 and q3.

Ans. (a) shall increase along the positive X-axis.

Explanation: As shown in the figure, since positive charge q2 and q3 exert a net force in the +X-direction on the charge q1 fixed along the X-axis, the charge q1 is negative. Obviously, due to addition of positive charge Q at (x, 0), the force on −q shall increase along the positive X-axis.

12. Three charges each equal to 1 mC are placed at the corners of an equilateral triangle. If force between any two charges is F, then the net force on either will be:

Explanation: Each charge experiences two forces inclined at 60°.
Therefore,

13. Three charges are placed at the vertices of an equilateral triangle of side ‘a’ as shown in the following figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is:

Ans. (c) Zero

Explanation: As we know that,

Hence, force experienced by the charge at A in the direction normal to BC is zero.

14. Three equal charges are placed on the three corners of a square. If the force between q1 and q2 is F12 and that between q1 and q3 is F13, the ratio of magnitudes is:

a) 0.5
(b) 1
(c) 1.5
(d) 2

Ans. (d) 2

Explanation: By using Coulomb’s law

15. Two positive point charges are 3 m apart and their combined charge is 20 mC. If the force between them is 0.075 N, then the charges are:

(a) 10 mC, 10 mC
(b) 15 mC, 5 mC
(c) 12 mC, 8 mC
(d) 14 mC, 6 mC

Ans. (b) 15 mC, 5 mC

Explanation: By using Coulomb’s law,

2. Electrostatic Potential and Capacitance

Multiple Choice Questions

1. The electric potential inside a conducting sphere:

(a) increases from centre to surface
(b) decreases from centre to surface
(c) remains constant from centre to surface
(d) is zero at every point inside

Ans. (c) remains constant from centre to surface

Explanation: Electric potential inside a conductor is constant and it is equal to that on the surface of the conductor.

2. One volt is equivalent to:

(a) newton/second
(b) newton/coulomb
(c) joule/coulomb
(d) joule/second

Ans. (c) joule/coulomb

Explanation: Volt is the electrical unit of voltage or potential difference. V is related to charge in potential energy DU as

From above formula, we get 1 volt potential difference when a change of 1 joule of potential is there in moving a charge of 1 coulomb Therefore, 1 volt = 1 Joule/1 coulomb.

3. The potential at a point due to a charge of 4 × 10−7 C located 10 cm away is:

(a) 3.6 × 105 V
(b) 3.6 × 104 V
(c) 4.5 × 104 V
(d) 4.5 × 105 V

Ans. (b) 3.6 × 104 V

Explanation: Given that, q = 4 × 10–7 C, r = 10 cm = 0.1 m

4. A conductor is charged to a potential V by importing a charge q, the variations of q with V is represented by figures:

5. The electric potential at a point in free space due to a charge Q coulomb is Q × 10¹¹ V. The electric field at that point is:

(a) 12pe0 Q × 10²² Vm−1
(b) 4pe0 Q × 10²⁰ Vm−1
(c) 12pe0 Q × 10²⁰ Vm−1
(d) 4pe0 Q × 10²² Vm−1 .

Ans. (d) 4pe0 Q × 10²² Vm−1 .

Explanation:

6. An electric dipole is placed at the centre of a hollow conducting sphere. Which of the following is correct?

(a) Electric field is zero at every point of the sphere
(b) Electric field is not zero anywhere on the sphere
(c) The flux of electric field is not zero through the sphere
(d) None of the above

Ans. (b) Electric field is not zero anywhere on the sphere

Explanation: When electric dipole is held in the sphere, electric field is not zero anywhere on the sphere. However, net electric flux through the sphere is zero.

7. Figure shows the field lines of a positive point charge. The work done by the field in moving a small positive charge from Q to P is:

(a) zero
(b) positive
(c) negative
(d) data insufficient.

Ans. (c) negative Explanation: In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.

8. A test charge is moved from lower potential point to a higher potential point. The potential energy of test charge will:

(a) remains the same
(b) increase
(c) decrease
(d) becomes zero

Ans. (b) increase

Explanation: By using, U = QV ... Q = + 1, U = V At high potential, potential energy will be high and at lower potential, the potential energy will be low.

9. In the given figure, a charge Q is fixed. Another charge q is moved along circular arc MN of radius r around it, from the point M to the point N such that the length of the arc MN = l. The work done in this process is :



Ans. (a) zero

Explanation: As distance of point M and N from C is constant so work done will be zero. As it makes an equipotential surface.

10. The electrostatic potential on the surface of a charged conducting sphere is 100 V. Two statements are made in this regard.
S1: At any point inside the sphere, electric intensity is zero.
S2: At any point inside the sphere, the electrostatic potential is 100 V. Which of the following is a correct statement? [NCERT Exemplar]

(a) S₁ is true but S₂ is false
(b) Both S₁ and S₂ are false
(c) S₁ is true, S₂ is also true and S₁ is the cause of S₂.
(d) S₁ is true, S₂ is also true but the statements are independent

Ans. (c) S₁ is true, S₂ is also true and S₁ is the cause of S₂.

Explanation: Potential at any point inside a charged conducting sphere = Potential on the surface,

11. Can two equipotential surfaces intersect each other?

(a) Yes
(b) No
(c) Sometimes
(d) Only when surfaces intersect at 90°

Ans. (b) No

Explanation: Intersection of two equipotential surfaces at a point will give two directions of electric field intensity at that point, which is not possible.

12. The physical quantity having SI unit NC^–1 m is:

(a) Electric potential
(b) Electric force
(c) Electric field intensity
(d) None of the above

Ans. (a) Electric potential

Explanation: V = E.d = NC^–1. m

13. Four charges each equal to q are placed at the corners of a square of side l. The electric potential at the centre of the square is :

Explanation:
As we know that,

Electric potential due to each charge at the centre of the square is Hence total potential is,

14. If a unit positive charge is taken from one point to another over an equipotential surface, then:

(a) work is done on the charge
(b) work is done by the charge
(c) work done is constant
(d) no work is done

Ans. (d) no work is done

Explanation: On the equipotential surface, electric field is normal to the charged surface (where potential exists). So, no work will be done for it.

15. A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. Let VA, VB, VC be the potentials at points A, B and C respectively. Then:

(a) VC > VB
(b) VB > VC
(c) VA > VB
(d) VA = VC.

Ans. (d) VA = VC.

Explanation: Conducting surface behaves as equipotential surface.

3. Current Electricity.

Multiple Choice Questions

1. When no current is passed through a conductor:

(a) the free electrons do not move
(b) the average speed of a free electron over a large period of time is not zero
(c) the average velocity of a free electron over a large period of time is zero
(d) the average of the velocities of all the free electrons at an instant is nonzero

Ans. (d) the average of the velocities of all the free electrons at an instant is non-zero

Explanation: Average of the velocities of all free electrons at an instant is non-zero, if no current is passed through a conductor.

2. If a current of 0.5 A flows in a 60 W lamp, then the total charge passing through it in two hours will be:

(a) 1800 C
(b) 2400 C
(c) 3000 C
(d) 3600 C

Ans. (d) 3600 C

Explanation: As we know that,

3. Drift velocity of electrons is due to:

(a) motion of conduction electrons due to random collisions.
(b) motion of conduction electrons due to electric field E.
(c) repulsion to the conduction electrons due to inner electrons of ions.
(d) collision of conduction electrons with each other.

Ans. (b) motion of conduction electrons due to electric field E.

Explanation: Motion of conduction electrons due to random collisions has no preferred direction and average to zero. Drift velocity is caused due to motion of conduction electrons due to applied electric field.

4. The relaxation time in conductors:

(a) increases with the increases of temperature
(b) decreases with the increases of temperature
(c) it does not depend on temperature
(d) all of sudden changes at 400 K

Ans. (b) decreases with the increases of temperature

Explanation: Because as temperature increases, the resistivity increases and hence the relaxation time decreases for conductors,

5. A steady current of 1 A is flowing through the conductor. The number of electrons flowing through the cross-section of the conductor in 1 sec is:

(a) 6.25 × 1015
(b) 6.25 × 1017
(c) 6.25 × 1019
(d) 6.25 × 1018

Ans. (d) 6.25 × 1018

Explanation: As we know that,

6. A metal wire is subjected to a constant potential difference. When the temperature of the metal wire increases, the drift velocity of the electron in it:

(a) increases, thermal velocity of electron increases
(b) decreases, thermal velocity of electron increases
(c) increases, thermal velocity of electron decreases
(d) decreases, thermal velocity of electron decreases

Ans. (b) decreases, thermal velocity of electron increases

Explanation: When the temperature increases, resistance increases. As the emf applied is the same, the current density decreases and the drift velocity decreases. But the rms velocity of the electron due to thermal motion is proportional to Therefore, the thermal velocity increases.

7. If N, e, τ and m are representing electron density, charge, relaxation time and mass of an electron respectively, then the resistance of wire of length l and cross-sectional area A is given by:

Explanation: If N, e, τ and m are representing electron density, charge, relaxation time and mass of an electron respectively, then the resistance of wire of length l and cross-sectional area A is,

8. Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speed of electrons in A and B is:

(a) 16 : 1
(b) 1 : 16
(c) 1 : 4
(d) 4 : 1

Ans. (a) 16 : 1

Explanation: Current flowing through the conductor, I = nevdA.

9. When a current I is set up in a wire of radius r, the drift velocity is vd. If the same current is set up through a wire of radius 2r, the drift velocity will be:

(a) 4vd
(b) 2vd
(c) vd/2
(d) vd/4

Ans. (d) vd/4

Explanation: As we know that, I = nAevd

10. A potential difference V is applied to a copper wire. If the potential difference is increased to 2V, then the drift velocity of electrons will:

(a) be double the initial velocity
(b) remain same
(c) be times the initial velocity
(d) be half the initial velocity

Ans. (a) be double the initial velocity

Explanation:
As we know that,
I = nevd
\ I ∝ vd
From Ohm’s law,
V ∝ I ∝ vd If the potential difference is doubled, drift velocity of electrons will also double.

11. A wire has a non-uniform cross-section as shown in the figure. If a steady current is flowing through it, then the drift speed of the electrons:

(a) is constant throughout the wire
(b) decreases from A to B
(c) increases from A to B
(d) varies randomly

Ans. (b) decreases from A to B

Explanation: As area increases, vd decreases.

12. A potential difference of 10 V is applied across a conductor of length 0.1 m. If the drift velocity of electrons is 2 × 10−4 m/s, the electron mobility is:

(a) 1 × 10−6
(b) 2 × 10−6
(c) 3 × 10−6
(d) 4 × 10−6

Ans. (b) 2 × 10−6

Explanation: As we know that,

13. Which of the following characteristics of electrons determines the current in a conductor? [NCERT Exemplar]

(a) Drift velocity alone
(b) Thermal velocity alone
(c) Both drift velocity and thermal velocity
(d) Neither drift nor thermal velocity

Ans. (a) Drift velocity alone

Explanation: We know that the relationship between current and drift speed is,
 = Anevd
That means, I ∝ vd
Hence, only drift velocity determines the current in a conductor.

14. The I-V characteristics shown in figure represents:

(a) ohmic conductors
(b) non-ohmic conductors
(c) insulators
(d) superconductors

Ans. (b) non-ohmic conductors

Explanation: The figure is showing I-V characteristics of non-ohmic or non-linear conductors.

15. Ohm’s law deals with the relation between:
(a) current and potential difference
(b) capacity and charge
(c) capacity and potential
(d) charge and potential difference

Ans. (a) current and potential difference

Explanation: Ohm’s law deals with the relation between current and potential difference.

4. Moving Charges and Magnetism

Multiple Choice Questions

1. Magnetic field can be produced by:

(a) a charge at rest.
(b) a changing electric field.
(c) a moving charge.
(d) both (b) and (c)

Ans. (d) both (b) and (c)

Explanation: A magnetic field is produced around a changing electric field or a moving charge.

2. A particle of mass m and charge q enters a magnetic field B perpendicularly with a velocity v. The radius of the circular path described by it will be:

3. A charged particle moving in a magnetic field experiences a resultant force:

(a) in the direction perpendicular to both the field and its velocity.
(b) in the direction of the field.
(c) in the direction opposite to that of the field.
(d) none of the above.

Ans. (a) in the direction perpendicular to both the field and its velocity.

Explanation: 

5. A charged particle of mass m and charge q travels in a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is:

Explanation: Equating magnetic force to centripetal force

Time to complete one revolution

6. Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that: [NCERT Exemplar]

(a) B ⊥ v
(b) B v
(c) it obeys inverse cube law.
(d) it is along the line joining the electron and point of observation.

Ans. (a) B ⊥ v

Explanation: According to the Biot-Savart law, the magnitude of is :

7. The Biot-Savart’s law in vector form is :

Explanation: In Biot-savart's law, magnetic field due to a tiny current element at any point is proportional to the length of the current element, the current, the sine of the angle between the current direction and line joining the current element and the point and inversely proportional to the square of the distance.

8. A charged particle moves through a magnetic field perpendicular to its direction. Then :

(a) both momentum and kinetic energy of the particle are not constant
(b) both momentum and kinetic energy of the particle are constant
(c) kinetic energy changes but the momentum is constant
(d) the momentum changes but the kinetic energy is constant

Ans. (d) the momentum changes but the kinetic energy is constant

Explanation: In circular motion, the direction of velocity changes at every point. Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by 1/2 mv2 and v 2 is the square of the velocity which does not change. If the momentum is angular then option (d) is correct.

9. If we double the radius of a coil keeping the current through it unchanged, what happens to the magnetic field on its axis at very-very far away points?

(a) Halved
(b) Doubled
(c) Becomes four times
(d) Remains unchanged

Ans. (c) Becomes four times

Explanation: At far away point,

Hence, when R is doubled. B becomes four times.

10. An electron and a proton enter a magnetic field with equal velocities. Which one of them experiences larger force?

(a) Proton
(b) Electron
(c) Both experience same force
(d) It cannot be predicted

Ans.(c) Both experience same force

Explanation: Force on a charged particle in a magnetic field is independent of mass. It is given by
Both electron and proton carry same amount of charge.

11. A charge of 1 C is moving in a magnetic field of 0.5 T with velocity of 10 m/s. Force experienced is:

(a) 0.5 N
(b) 5 N
(c) 10 N
d) 0 N

Ans. (b) 5 N

Explanation: Force on change due to magnetic field. F = qvB \ F = 1 × 0.5 × 10 = 5 N

12. Lorentz force is:

(a) the vector sum of electrostatic and magnetic force acting on a moving charged particle.
(b) the vector sum of gravitational and magnetic force acting on a moving charged particle.
(c) electrostatic force acting on a charged particle
(d) magnetic force acting on a moving charged particle.

Ans. (a) the vector sum of electrostatic and magnetic force acting on a moving charged particle.

Explanation: As Lorentz force is given by,

13. A charged particle is moving with velocity v in a magnetic field of induction B. The force on the particle will be maximum when:
(a) v and B are at an angle of 45°.
(b) v and B are perpendicular.
(c) v and B are in the same directions.
(d) v and B are in opposite directions.

Ans. (b) v and B are perpendicular.

Explanation: ... F = q(v × B) or |F| = qvB sin q F will be maximum when θ = 90°

14. An α-particle enters a magnetic field of 1 T with a velocity 106 m/s in a direction perpendicular to the field. The force on α-particle is:

(a) 1.6 × 10⁻³ N
(b) 3.2 × 10⁻¹³ N
(c) 4.8 × 10⁻¹³ N
(d) 6.4 × 10⁻¹³ N

Ans. (b) 3.2 × 10⁻¹³ N

Explanation: Force acting on a charge q moving with velocity v in magnetic field of intensity B is given by,
F = qvB sin q
= qvB (... q = 90°, ... sin q = 1)
Given that,
q = 2 × 1.6 × 10^–19 (a-particle)
v = 10⁶ m/s
B = 1 T
on substituting these values, we get
F = 2 × 1.6 × 10^–19 × 10⁶ × 1
= 3.2 × 10^–13 N

15. A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T. The force on the proton is:

(a) 2.5 × 10⁻¹⁰ N
(b) 2.5 × 10⁻¹¹ N
(c) 7.6 × 10⁻¹¹ N
(d) 7.85 × 10⁻¹² N

Ans. (d) 7.85 × 10⁻¹² N Explanation: We know that, F = qvB

5. Magnetism and Matter

Multiple Choice Questions

1. If a hole is made at the centre of a bar magnet, then its magnetic moment:

(a) does not change
(b) decreases
(c) increases
(d) vanishes

Ans. (a) does not change

Explanation: As its pole strength and length remains same.

2. The incorrect statement regarding the lines of force of the magnetic field B is :

(a) Magnetic lines of force forms a closed curve.
(b) Due to a magnet, magnetic lines of force never cut each other.
(c) Magnetic intensity is a measure of lines of force passing through unit area held normal to it.
(d) Inside a magnet, the magnetic lines of force move from north pole of a magnet towards the south pole.

Ans. (d) Inside a magnet, the magnetic lines of force move from north pole of a magnet towards the south pole.

Explanation: Inside a magnet, magnetic lines of force move from south pole to north pole.

3. Gauss’ law should be invalid if :

(a) the inverse square law were not exactly true
(b) the velocity of light were not a universal constant
(c) there were magnetic monopoles
(d) none of the above

Ans. (c) there were magnetic monopoles

Explanation: If magnetic monopoles exist, then inverse square law were not exactly true.

4. The magnetisation of a bar magnet of length 5 cm, cross-sectional area 2 cm2 and magnetic moment 1 Am2 is :

(a) 2 × 105 A/m
(b) 4 × 105 A/m
(c) 105 A/m
(d) 3 × 105 A/m

Ans. (c) 105 A/m

Explanation:
Given, l = 5 cm = 5 × 10^–2 m
A = 2 cm² = 2 × 10^–4 m²
and M = 1 Am^–2
\ Magnetisation of bar magnet is given by

5. A beam of electrons passes through crossed electric and magnetic fields of magnitudes 7.2 × 106 NC– 1 and 2.0 T respectively. The electrons which pass undeviated have velocity:

(a) 14.4 × 106 ms–1
(b) 3.6 × 106 ms–1
(c) 3.0 × 108 ms–1
(d) 0.28 × 10– 6 ms–1

Ans. (b) 3.6 × 106 ms–1

Explanation:
eE = evB

6. A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is: [NCERT Exemplar]

(a) MB q

Explanation: The rotation of loop by 30° about an axis perpendicular to its plane make no change in the angle made by axis of the loop with the direction of magnetic field, therefore, the work done to rotate the loop is zero.

7. If the current (I) flowing through a circular coil, its radius (R) and number of turns (N) in it are each doubled, magnetic flux density at its centre becomes :

(a) two times
(b) four times
(c) eight times
(d) sixteen times

Ans. (a) two times

Explanation: Magnetic flux density at the centre of a circular coil is given by

8. The north pole of a magnet is brought near a stationary negatively charged conductor. Will the pole experience any force?

(a) Yes
(b) No
(c) Depends on the magnitude of pole strength
(d) Can’t say

Ans. (b) No

Explanation: No, a stationary charge does not produces magnetic field.

9. The magnetic moment of a bar magnet is _____ to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

(a) same
(b) different
(c) unequal
(d) equal

Ans. (d) equal

Explanation: The magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

10. The magnetic dipole moment of a solenoid having N turns is given as:

(a) NIA
(b) NIA²
(c) NI²A
(d) NI²A ²

Ans. (a) NIA

Explanation: Each turn of the solenoid behaves as a small dipole having dipole moment IA.

11. In the case of bar magnet, lines of magnetic induction:

(a) run continuously through the bar magnet and outside
(b) emerge in circular paths from the middle of the bar magnet
(c) are produced only at the north pole like rays of light from a bulb
(d) start from the north pole and end at the south pole

Ans. (a) run continuously through the bar magnet and outside

Explanation: In the bar magnet, lines of magnetic induction run continuously through the bar magnet and outside.

12. A north pole of strength 50 A-m and south pole of strength 100 A-m are separated by a distance of 10 cm in air. The force between them is.

(a) 20 × 10–6 N
(b) 25 × 10–3 N
(c) 30 × 10–18 N
(d) 50 × 10–3 N

Ans. (d) 50 × 10–3 N

Explanation: Force between magnetic poles in air is given by,

13. Consider the two idealised systems: (i) a parallel plate capacitor with large plates and small separation and (ii) a long solenoid of length L >> R, radius of cross-section. In (i) E is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as below: [NCERT Exemplar]

(a) case (i) contradicts Gauss’ law for electrostatic fields.
(b) case (ii) contradicts Gauss’ law for magnetic fields.

Ans. (b) case (ii) contradicts Gauss’ law for magnetic fields.

Explanation: According to Gauss’ law in magnetism, which implies that number of magnetic field lines entering the Gaussian surface is equal to the number of magnetic field lines leaving it. Therefore, case (ii) is not possible.

14. A current carrying circular loop of radius R is placed in the x-y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y-z plane. [NCERT Exemplar]

(a) The magnitude of magnetic moment now diminishes.
(b) The magnetic moment does not change.
(c) The magnitude of B at (0, 0, z), z >> R increases.
(d) The magnitude of B at (0, 0, z), z >> R is unchanged.

Ans. (a) The magnitude of magnetic moment now diminishes.

Explanation: The magnitudes of magnetic moment of each semicircular loop of radius R lie in the x-y plane and the y-z plane is M¹ = M² = and the direction of magnetic moments are along z-direction and x-direction respectively. Their resultant,

15. The intensity of magnetisation of a bar magnet is 5.0 × 104 Am–1 . The magnetic length and the area of cross-section of the magnet are 12 cm and 1 cm2 respectively. The magnitude of magnetic moment of this bar magnet (in SI unit) is:

(a) 2.4
(b) 1.3
(c) 1.24
(d) 0.6

Ans. (d) 0.6

Explanation: Intensity of magnetisation, where, M is the magnetic moment of the bar magnet and V is the volume of the magnet.
Hence, M = IV = 5.0 × 10⁴ × 0.12 × 1 × 10^–4 = 0.6 Am²

6. Electromagnetic Induction

Multiple Choice Questions

1. A moving conductor coil in a magnetic field produces an induced emf. This is in accordance with :

(a) Lenz’s law
(b) Coulomb’s law
(c) Faraday’s law
(d) Ampere’s law

Ans. (c) Faraday’s law

Explanation: Faraday’s law of induction is a basic law of electromagnetism predicting how a magnetic field will interact with an electric circuit to produce an electromotive force.

2. Faraday’s laws are consequences of conservation of :

(a) energy and magnetic field
(b) energy
(c) magnetic field
(d) charge

Ans. (b) energy Explanation: Faraday’s laws involve conversion of mechanical energy into electrical energy. This is in accordance with the law of conservation of energy.

3. In electromagnetic induction, the induced charge is independent of :

(a) resistance of the coil
(b) change of flux
(c) time
(d) none of the above

Ans. (c) time

Explanation: Electromagnetic induction does not depend on time.

4. An induced emf is produced when a magnet is plunge into a coil. The strength of the induced emf is independent of:

(a) number of turns of coil
(b) speed with which the magnet is moved
(c) the strength of the magnet
(d) the resistivity of the wire of the coil

Ans. (d) the resistivity of the wire of the coil

Explanation: When the conductor is moved in a stationary magnetic field to procure a change in the flux linkage, the emf is statically induced.

5. The magnetic flux through a coil is inversely proportional to:

(a) magnetic field
(b) number of turns
(c) area
(d) none of these

Ans. (d) none of these

Explanation: The magnetic flux through some surface is proportional to the number of field lines passing through that surface.

6. Which of the following phenomena makes use of electromagnetic induction?
(a) Magnetising an iron piece with a bar magnet
(b) Generation of hydroelectricity
(c) Magnetising a soft iron piece by placing inside a current carrying solenoid
(d) Charging a storage battery Ans. (b) Generation of hydroelectricity

Explanation: Hydroelectric plant uses mechanical energy of water to move a magnetic field passes coils of wire to generate voltage.

7. The magnetic flux (f) linked with a coil due to its own magnetic field is related to the number of turns (N) of the coil as:

(a) f ∝N 2
(b) f ∝ N –1
(c) f ∝N
(d) f ∝ N –2

Ans. (a) f ∝N 2

Explanation:

8. Faraday's law of induction states that an electromotive force is induced by a __________.

(a) resistance of the coil
(b) change of flux
(c) time
(d) none of the above

Ans. (b) change of flux

Explanation: We know that,

where f = NBA

9. To induce an emf in a coil, the linking magnetic flux:

(a) must remain constant
(b) can either increase or decrease
(c) must increase
(d) must decrease

Ans. (b) can either increase or decrease

Explanation: Emf is induced when there is change in magnetic flux in the coil.

10. The magnetic flux through a circuit of resistance R changes by an amount Df in a time Dt. Then the total quantity of electric charge Q that passes through any point in the circuit during the time Dt is represented by:

11. A coil has 200 turns and area of 70 cm² . The magnetic field perpendicular to the plane of the coil is 0.3 Wb/m² and take 0.1 s to rotate through 180°. The value of the induced emf will be:

(a) 84 V
(b) 42 V
(c) 8.4 V
(d) 4.2 V

Ans. (c) 8.4 V

Explanation: As we know that, Change in flux, f = 2BAN

12. The expression for the induced emf contains a negative sign . What is the significance of the negative sign?

(a) The induced emf is opposite to the direction of the flux.
(b) The induced emf is produced only, when the magnetic flux decreases.
(c) The induced emf opposes the changes in the magnetic flux
(d) None of the above

Ans. (c) The induced emf opposes the changes in the magnetic flux

Explanation: The negative sign gives the direction of the induced emf.

13. Lenz’s law gives:

(a) the direction of the induced current
(b) the magnitude of the induced emf
(c) the magnitude of the induced current
(d) both the magnitude and direction of the induced current

Ans. (a) the direction of the induced current

Explanation: Lenz’s law states that an induced electric current flows in a direction such that the current opposes the change that induced it.

14. Lenz’s and Faraday’s law is expressed by the following formula (here e = induced emf, f = magnetic flux in one turn and N = number of turns):

15. A rectangular coil of 100 turns and size 0.1 m × 0.05 m is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 s, the magnitude of the emf induced in the coil is:

(a) 0.5
(b) 3
(c) 2
(d) 6

Ans. (a) 0.5

Explanation: Hence, area ofcoil
A = 0.1 m × 0.05 m
= 5 × 10^–3 m²
N = 100
Initial flux linked with the coil,
ϕ1 = BA cos θ
= 0.1 × 5 × 10^–3 cos 0°
 5 × 10^–4 Wb
Final flux linked with the coil,
ϕ2 = 0.05 × 5 × 10^–3 cos 0°
= 25 × 10–5 Wb
= 2.5 × 10–4 Wb
The magnitude of induced emf in the coil is,

7. Alternating Current

Multiple Choice Questions

1. Alternating current cannot be measured by DC ammeter because:

(a) AC is virtual
(b) AC changes its direction
(c) AC cannot pass through DC ammeter
(d) Average value of complete cycle is zero

Ans. (d) Average value of complete cycle is zero

Explanation: In DC ammeter, a coil is free to rotate in the magnetic field of a fixed magnet. If an alternating current is passed through such a coil, the torque will reverse its direction each time the current changes direction and the average value of the torque will be zero.

2. When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. This means: [NCERT Exemplar]

(a) Input voltage cannot be AC voltage, but a DC voltage.
(b) maximum input voltage is 220 V

3. The alternating current of equivalent value of is:

(a) r.m.s. current
(b) DC current
(c) Peak current
(d) All of these

Ans. (a) r.m.s. current

Explanation: We know that,

4. In an AC circuit I = 100 sin 200pt. The time required for the current to achieve its peak value will be:

5. The ratio of mean value over half cycle to r.m.s. value of AC is:

Explanation:
We know that,

7. The frequency of an alternating voltage is 50 cps and its amplitude is 120 V. Then the rms value of voltage is:

(a) 56.5 V
(b) 70.7 V
(c) 101.3 V
(d) 84.8 V

Ans. (d) 84.8 V

Explanation: We know that,

8. A 40 Ω electric heater is connected to a 200 V, 50 Hz mains supply. The peak value of electric current flowing in the circuit is approximately:

(a) 10 A
(b) 5 A
(c) 7 A
(d) 2.5 A

Ans. (c) 7 A

Explanation:

9. A resistance of 20 Ω is connected to a source of an alternating potential V = 220 sin (100pt). The time taken by the current to change from its peak value to rms value is:

10. The rms value of an AC of 50 Hz is 10 A. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

(a) 1 × 10−2 s and 7.07 A
(b) 2 × 10−2 s and 14.14 A
(c) 5 × 10−3 s and 14.14 A
(d) 5 × 10−3 s and 7.07 A

Ans.(c) 5 × 10−3 s and 14.14 A

Explanation: Time taken by the current to reach the maximum value.

11. Determine the rms value of the emf given by, E (in V) = 8 sin (wt) + 6 sin (2wt)

12. A generator produces a voltage that is given by V = 240 sin 120t V, where t is in seconds. The frequency and rms voltage are nearly:

(a) 19 Hz and 120 V
(b) 19 Hz and 170 V
(c) 60 Hz and 240 V
(d) 754 Hz and 170 V

Ans. (b) 19 Hz and 170 V

Explanation: Given, V = 240 sin 120t V Comparing with V = V0 sin ωt
V0 = 240 V ω
= 120 rad/s

13. The instantaneous voltage through a device of impedance 20 Ω is e = 80 sin 100πt. The effective value of the current is:

(a) 1.732 A
(b) 2.828 A
(c) 3 A
(d) 4 A

Ans. (b) 2.828 A

Explanation: Given, e = 80 sin 100πt …(i)
Standard equation of instantaneous voltage is given by
e = em sin ωt …(ii)
Compare (i) and (ii), we get
em = 80 V where em is the voltage amplitude.
Current amplitude,

14. In an AC circuit an alternating voltage V = 200 sin 100t is connected to a capacitor of capacity 1 μF. The rms value of the current in the circuit is:

(a) 10 mA
(b) 20 mA
(c) 100 mA
(d) 200 mA

Ans. (b) 20 mA

Explanation: We know that,

15. If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 s after its value becomes zero is: [NCERT Exemplar]

Explanation: According to the question,
f = 50 Hz

8. Electromagnetic Waves

Multiple Choice Questions

1. Fundamental particle in an electromagnetic wave is :

(a) proton
(b) electron
(c) phonon
(d) photon

Ans. (d) photon

Explanation: Photon is a fundamental particle in electromagnetic waves.

2. Which of the following is not transported by the electromagnetic waves ?

(a) Charge
(b) Energy
(c) Information
(d) Momentum

Ans. (a) Charge

Explanation: Electromagnetic waves carry energy (light), momentum (can be seen by photoelectric experiment) and information (like radio waves) but doesn’t carry any charge.

3. Electromagnetic waves carry :

(a) negative charge
(b) positive charge
(c) no charge
(d) both positive and negative charge

Ans. (c) no charge

Explanation: There is no electric charge in electromagnetic waves i.e., exactly zero. Electromagnetic waves may be generated by moving charges, but they carry no charge themselves. They are not deflected from their path.

4. Electromagnetic waves are produced by:

(a) accelerated charged particle
(b) decelerated charged particle
(c) charge in uniform motion
(d) none of the above

Ans. (a) accelerated charged particle

Explanation: An accelerating charged particle produces an electromagnetic wave.

5. The E and B fields in electromagnetic waves are oriented:

(a) parallel to the wave’s direction of propagation, as well as to each other.
(b) parallel to the wave’s direction of propagation and perpendicular to each other.
(c) perpendicular to the wave’s direction of propagation and parallel to each other.
(d) perpendicular to the wave’s direction of propagation and also to each other.

Ans. (d) perpendicular to the wave’s direction of propagation and also to each other.

Explanation: For electromagnetic waves, E and B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E × B.

6. The wavelength of electromagnetic waves of frequency 6 × 10¹² Hz in free space is:

(a) 5 × 10–5 m
(b) 6 × 10-5 m
(c) 3 × 10–5 m
(d) 8 × 10–5 m

Ans. (a) 5 × 10–5 m

Explanation:

7. In electromagnetic waves the phase difference between electric and magnetic field vectors are:

(a) 2π
(b) 2/π
(c) π
(d) Zero

Ans. (d) Zero

Explanation: Electric and magnetic field vectors always remain in the same phase.

8. What is the velocity of electromagnetic waves in free space?

Explanation: As the constant m0 and e0 do not depend on frequency or wavelength of electromagnetic. waves so the value of c remains same. Hence, the velocity of electromagnetic waves in free space is :

9. In electromagnetic wave if ue and um are mean electric and magnetic energy densities respectively, then:

(a) um = ue
(b) um < ue
(c) um > ue
(d) None of these

Ans. (a) um = ue

Explanation: Energy is equally distributed among electric field and magnetic field in e.m. waves.

10. A frequency of oscillation of the electric field vector of a certain electromagnetic wave is 5 × 10¹⁴ Hz. What is the frequency of oscillation of the corresponding magnetic field and to which part of the electromagnetic spectrum does it belongs?

(a) 10 × 10¹⁴ Hz; Visible region
(b) 5 × 10¹⁴ Hz; UV region
(c) 5 × 10¹⁴ Hz; Visible region
(d) 7.5 × 10¹⁴ Hz; UV region

Ans. (c) 5 × 10¹⁴ Hz; Visible region

Explanation: Frequency of oscillation of magnetic field is same as that of electric field vector i.e., 5 × 10¹⁴ Hz and it lies in visible region.

11. The ratio of frequencies of UV rays and infrared rays in glass is:

(a) More than 1
(b) Less than 1
(c) Equal to 1
(d) None of the above

Ans. (a) More than 1

Explanation: As frequency of UV rays > frequency of infrared rays, so the ratio is greater or more than 1.

12. Which of the following can act as a source of electromagnetic waves?

(a) A charge at rest
(b) A charge moving with a constant velocity
(c) A charge moving in a circular orbit
(d) None of the above

Ans. (c) A charge moving in a circular orbit

Explanation: A charge moving in a circular orbit can act as a source of electromagnetic waves as the electromagnetic. waves are produced by accelerated charged particles only. A charge moving in a circular orbit has an accelerated motion

13. The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is B0 = 500 nT. What is the amplitude of the electric field part of the wave?

(a) 200 N/C
(b) 350 N/C
(c) 1000 N/C
(d) 150 N/C

Ans. (d) 150 N/C

Explanation:

14. The wavelength of electromagnetic radiation is doubled, what will happen to the energy of the photon?

(a) Double
(b) Halved
(c) Remains same
(d) None of these

Ans. (b) Halved

Explanation:

9. Ray Optics and Optical Instruments

Multiple Choice Questions

1. A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is :

(a) blue
(b) green
(c) violet
(d) red

Ans. (d) red

Explanation: Since v ∝ l, the light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerges first.

2. If the critical angle is 49°, and a fish is a little away below the surface of a lake,the fish could see things above the water surface within an angular range of θ°, where:

(a) θ = 49°
(b) θ = 98°
(c) θ = 90°
(b) 24.5°

Ans. (b) θ = 98°

Explanation: Since, critical angle qc = 49° and q° = 2qc = 2 × 49° = 98°

3. The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in figure, the path shown is correct?

(a) 1
(b) 2
(c) 3
(d) 4

Ans. (b) 2

Explanation: Here, light ray goes from (optically) rarer medium air to optically denser turpentine, then it bends towards the normal i.e., i > r. Whereas when it goes from optically denser medium turpentine to rarer medium water, then it bends away the normal i.e., i < r.

4. There are certain material developed in laboratories which have a negative refractive index number. A ray incident from air (Medium 1) into such a medium (Medium 2) shall follow a path given by :

Explanation: The negative refractive index materials are those in which incident ray from air (Medium 1) to them refract or bend differently to that of positive refractive index medium.

5. In the ray diagram for the refraction given below. The maximum value of angle q for which the light suffers total internal reflection at the vertical surface, is:

6. A piece of glass immersed in a colourless liquid is not visible. It is so because :

(a) glass and liquid are both colourless
(b) both are of same density
(c) both are of same refractive index
(d) glass reflects the light transmitted by the liquid.

Ans. (c) both are of same refractive index

Explanation: We won’t be able to recognize the glass plate as different from the liquid, if the refractive index of the glass plate is same as that of the liquid.

7. A fish which is at a depth of 12 cm in water viewed by an observer on the bank of a lake. Its apparent depth as observed by the observer is:

(a) 3 cm
(b) 9 cm
(c) 12 cm
(d) 16 cm

Ans. (b) 9 cm

Explanation: D = Real depth, d = apparent depth

8. Refractive index of a transparent material is :

(a) same for all colours
(b) maximum for violet colour
(c) minimum for violet colour
(d) maximum for red colour

Ans. (b) maximum for violet colour

Explanation: The index of refraction varies with frequency, it doesn’t changes as light travels from one medium to another. As violet colour has the shortest wavelength and so the refractive index is maximum for it.

9. A physical quantity in optics without unit :

(a) Power
(b) Refractive Index
(c) Wavelength
(d) Frequency

Ans. (b) Refractive index

Explanation: Refractive index of a material is a dimensionless number.

10. Can refractive index of a medium be less than unity?

(a) Yes
(b) No
(c) Data insufficient
(d) None of these

Ans. (a) Yes

Explanation:

11. A ray of monochromatic light passes from medium 1 to medium 2. If the angle of incidence in medium 1 is θ and the corresponding angle of refraction in medium 2 is θ/2, which of the media is optically denser ?

(a) Medium 1
(b) Medium 2
(c) Both Medium 1 and 2
(d) None of the above

Ans. (b) Medium 2

Explanation:

12. You are given four sources of light each one providing a light of a single colour - red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?

(a) The beam of red light would undergo total internal reflection
(b) The beam of red light would bend towards normal while it gets refracted through the second medium
(c) The beam of blue light would undergo total internal reflection
(d) The beam of green light would bend away from the normal as it gets refracted through the second medium

Ans. (c) The beam of blue light would undergo total internal reflection

Explanation: According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength. According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle. So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

13. The phenomena involved in the reflection of radio waves by ionosphere is similar to :

(a) reflection of light by a plane mirror
(b) total internal reflection of light in air during a mirage
(c) dispersion of light by water molecules during the formation of a rainbow
(d) scattering of light by the particles of air

Ans. (b) total internal reflection of light in air during a mirage

Explanation: The phenomenon involved in the reflection of radio waves by ionosphere is similar to total internal reflection of light in air during a mirage i.e., angle of incidence is greater than critical angle.

14. Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because

(a) the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge
(b) the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air
(c) some of the points of the object far away from the edge may not be visible because of total internal reflection
(d) water in a trough acts as a lens and magnifies the object

Ans. (a, b, c)

Explanation: When immersed object is seen from close to the edge of the trough the object looks distorted because the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge. The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air and some of the points of the object far away from the edge may not be visible because of total internal reflection.

15. A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (figure). When observed from the face AD, the pin shall :

(a) appear to be near A
(b) appear to be near D
(c) appear to be at the centre of AD
(d) not be seen at all

Ans. (d) not be seen at all

Explanation: For m = 1.6, the critical angle, m = 1/sin C, we have C = 38.7°, when viewed from AD, as long as angle of incidence on AD of the ray emanating from pin is greater than the critical angle the light suffers from total internal reflection and cannot be seen through AD.

10. Wave Optics

Multiple Choice Questions

1. For light diverging from a point source,

(a) the wavefront is spherical
(b) the intensity decreases in proportion to the distance squared
(c) the wavefront is parabolic
(d) the intensity at the wavefront does not depend on the distance

Ans. (a, b) Explanation: Consider the diagram in which light diverges from a point source (O).

Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram. If power of the source is P, then intensity of the source will be

where, r is radius of the wavefront at any time.

2. If a light source is placed at the focus of a concave mirror, the reflected wave front obtained will be:

(a) Spherical wavefront
(b) Plane wavefront
(c) Elliptical wavefront
(d) None of the above

Ans. (b) Plane wavefront

Explanation: We know that the light rays coming from focus becomes parallel to the principal axis, after reflection through concave mirror. So the wavefront obtained will be a plane wavefront.

3. If plane wavefront is parallel through a convex lens then the refractive wavefront will converge at:

(a) Optical centre
(b) Centre of curvature
(c) Infinity
(d) Focus of lens

Ans. (d) Focus of lens

Explanation: The plane wavefront after refraction through convex lens always converges at focus of the lens.

4. During reflection of a plane wavefront from plane reflecting surface, the secondary wavelets are drawn with :

(a) decreasing radii
(b) increasing radii
(c) same radii
(d) any radii

Ans. (a) decreasing radii

Explanation: The secondary wavelets are drawn with decreasing radii during reflection of a plane wavefront from a plane reflecting surface.

5. What will be the value of 1n2 for the given figure:

6. Which relation is correct for the given figure:

(a) n1 > n2
(b) n1 > n2
(c) n1 = n2
(d) None of these

Ans. (c) n1 = n2

Explanation: A wavefront does not deviate from its path only when the refractive index of both the medium is same.

7. The electric field vector for a circularly polarised light varies with time as:

Explanation: We know for circularly polarised light. E₁ = a sin wt E₂ = a cos wt

8. When light suffers reflection at the interface between water and glass, the change of phase in the reflected wave is :

(a) Zero
(b) p
(c) 2p

Ans. (b) p

Explanation: On reflection from denser medium, phase reversal occurs i.e. phase change of π takes place, because glass is much denser than water.

9. When a ray of light passes from glass to air then which of the following is true :

(a) v1 > v2
(b) v2 > v1
(c) v1 = v2
(d) None of these

Ans. (b) v2 > v1

Explanation: Velocity of light is low in denser medium and high in rarer medium.

10. Find the value of O from given figure:

(a) 45°
(b) 35°
(c) 30°
(d) 25°

Ans. (c) 30°

Explanation: ∠i = 90° – 60° = 30°
From figure, ∠i = ∠AOC (Angle of reflection)
\ ∠i = ∠q = 30°.

11. Two waves of equal amplitude and wavelength but differing in phase are superimposed. The amplitude of the resultant wave is maximum when phase difference is :

(a) zero

Ans. (a) zero

Explanation: Two waves of equal amplitude and wavelength but differing in phase are superimposed. The amplitude of the resultant wave is maximum when phase difference is zero.

12. When light passes from air to glass then :

(a) the wavelength increases
(b) the wavelength decreases
(c) the frequency increases
(d) nothing is changed

Ans. (b) the wavelength decreases

Explanation: Glass is denser than air, so a light ray on passing from air to glass slows down and hence its wavelength decreases.

13. When a wave undergoes reflection at an interface from rarer to denser medium, adhoc change in its phases is :

Ans. (c) p

Explanation: When a wave is reflected into rarer medium from the surface of a denser medium, it undergoes a phase change of π radian.

14. The refractive index of glass is 1.5 for light waves of l = 6000 Å in vacuum. Its wavelength in glass is :

(a) 2000 Å
(b) 4000 Å
(c) 1000 Å
(d) 3000 Å

Ans. (b) 4000 Å

Explanation:

15. Monochromatic light is refracted from air into a medium of refractive index n. The ratio of the wavelengths of the incident and the refracted waves is :

(a) 1 : 1
(b) 1 : n
(c) n : 1
(d) n2 : 1

Ans. (c) n : 1

Explanation:

11. Dual Nature of Radiation and Matter

Multiple Choice Questions

1. ________ is the type of nature of electromagnetic waves.

(a) Particle nature
(b) Wave nature
(c) Photon nature
(d) Dual nature

Ans.(d) Dual nature

Explanation: Electromagnetic radiations have a wave nature as well as properties alike to those of particles. Therefore, electromagnetic radiations are emissions with a dual nature, i.e. it has both wave and particle aspects.

2. Highest energy photoelectrons will be produced by :

(a) visible light
(b) X-rays
(c) ultraviolet light
(d) g-rays

Ans. (d) g-rays Explanation: A collaboration of Chinese and Japanese astrophysicists has reported the highest energy photons ever seen are gamma rays with energies up to 450 trillion electron volts (TeV).

3. Which of the following proves particle nature of light ?

(a) Refraction
(b) Interference
(c) Polarisation
(d) Photoelectric effect

Ans. (d) Photoelectric effect

Explanation: Photoelectric effect states that light travels in the form of bundles or packets of energy, called photons. This effect is explained on the basis of quantum nature of light. So, it clearly explains the particle’s nature of light.

4. Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency n is used to convert 1 kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion :

(a) decreases with increasing n, with n fixed
(b) decreases with n fixed, v increasing
(c) remains constant with n and n changing such that nn = constant
(d) all of the above

Ans. (d) all of the above

Explanation: Energy spent to convert ice into water
= mass × latent heat = m × L
= (1000 g) × (80 cal/g)
= 80000 cal
Energy of photons used = nT × E = nT × hn

5. There are _______ types of electron emissions.

(a) One
(b) Two
(c) Three
(d) Four

Ans. (d) Four

Explanation: There are four types of electron emissions, namely, thermionic emission, photoelectric emission, secondary emission and field emission. These are the different methods of producing electron emissions.

6. Which of the following is incorrect?

(a) Vacuum tubes are thermionic devices
(b) An electron gun is to create electrons and then accelerate them to a very high speed
(c) Thermionic emissions and photoelectric emissions are the same
(d) The kinetic energy of photoelectrons vary

Ans. (c) Thermionic emissions and photoelectric emissions are the same

Explanation: No, thermionic emissions and photoelectric emissions are not the same. During thermionic emissions, the electrons are emitted from the metal surface by providing heat energy, whereas, during photoelectric emission light energy is emitted when electrons are emitted from the surface of the metal. So, both emissions are the opposite in operation. 

7. Who performed photoelectric effect first?

(a) Heinrich Rudolf Hertz
(b) Albert Einstein
(c) de-Broglie
(d) None of the above

Ans. (a) Heinrich Rudolf Hertz

Explanation: German physicist Heinrich Rudolf Hertz discovered the photoelectric effect in 1887 when he was working in connection with his work on radio waves.

8. According to Einstein’s photoelectric equation, the plot of kinetic energy of the emitted photoelectrons from a metal vs frequency of the incident radiation, is shown in figure.

Choose the correct option depicted by the given graph.

(a) Slope of the graph depends on the nature of the metal
(b) Slope of the graph depends upon the intensity of radiation
(c) Slope of the graph is same for all metals
(d) All of the above

Ans. (c) Slope of the graph is same for all metals

Explanation: According to Einstein’s equation Kinetic energy = hn – f where, kinetic energy and n are variables. Now, compare the given equation with, y = mx + c \ Slope of the line = h where, h is planck’s constant. Hence, slope is same for all metals and independent of the nature of the metal and intensity of radiation.

9. A point source is placed at a distance 0·8 m from a metallic sphere of radius 8·0 × 10^{– 3} m, as shown in figure.

Point source is of 3·2 × 10– 3 W power emitting photons of 5·0 eV energy. Work function for metallic sphere is 3·0 eV. Assuming 100% absorption, calculate the number of photons incident on the sphere per second.

(a) 10¹⁹
(b) 10¹⁸
(c) 10¹¹
(d) 10²²

Ans.
(c) 1011

Explanation: Energy of emitted photons = 5 eV
= 8 × 10^{– 19} J
Power of source = 3·2 × 10^{– 3} W
\ Number of photons emitted by source per second,

= 4 × 10¹⁵ photons per second Number of photons reaching the metallic sphere,

= 5 × 10¹⁴ m^{– 2}s ^{– 1} Effective area of the sphere over which photons fall, A = 4pr ² = 4p(8 × 10^{– 3}) ² = 8·02 × 10^{– 4} m² \ Number of photons incident on the sphere in 1 s, = 5 × 10¹⁴ × 8·02 × 10^{– 4} = 4 × 10¹¹ s ^{– 1} ~ 10¹¹ s^{– 1}

10. Calculate the work function of Cesium from the observations of the photoelectric experiment plotted on the graph shown in figure.

(a) 1·2 eV
(b) 2 eV
(c) 1·75 eV
(d) 2·5 eV

Ans. (b) 2 eV

Explanation: Work function, f0
= hn0 = 6·6 × 10^{– 34} × 0·49 × 10¹⁵ J

11. Planck’s constant :

(a) depends upon medium
(b) depends upon wavelength of light
(c) depends upon frequency of light
(d) is an universal constant

Ans. (d) is an universal constant

Explanation: The value of Planck’s constant is 6.6 × 10^–34 J-s and this value is constant all over the universe Hence, it is an universal constant.

12. Einstein’s theory of photoelectric effect based on the equation :

(a) E = mc2
(b) E = hn

13. Einstein’s photoelectric equation is :

(a) Emax = hl – f0

14. Dimensions of Planck’s constant are same as of:

(a) force × time
(b) energy × distance
(c) energy / time
(d) energy × time

Ans. (d) energy × time

Explanation: The dimensions of Planck’s constant is the same as the dimensions of the angular momentum.

15. The work function of a surface of photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the :

(a) UV region
(b) Visible region
(c) Infra-red region
(d) X-ray region

Ans. (a) UV region

Explanation:

\ It lies in UV region

12. Atoms

Multiple Choice Questions

1. Rutherford’s experiment on scattering of a-particles proved that :

(a) atom contains positrons
(b) number of positive charges is not equal to the number of negative charges
(c) positive charge is uniformly distributed in the atom
(d) atom is mostly empty

Ans. (d) atom is mostly empty

Explanation: Rutherford’s experiment on scattering of α-particles proved that atom is mostly empty.

2. In Geiger-Marsden scattering experiment, the trajectory traced by an a-particle depends on :

(a) impact parameter
(b) number of scattered a-particles
(c) number of collisions
(d) none of the above

Ans. (a) impact parameter

Explanation: Trajectory of α-particle depends on impact parameter which is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus.

3. Scattering of an a-particle by an inverse square field (like produced by a charged nucleus in Rutherford’s model) is shown in figure. If charge of nucleus Z is 79 and kinetic energy of a-particle is 10 MeV, then find the impact parameter.

(a) 2·1 × 10– 14 m
(b) 1·1 × 10– 16 m
(c) 1·1 × 10– 14 m
(d) 2·2 × 10– 16 m

Ans. (c) 1·1 × 10– 14 m

Explanation: In Rutherford's model, the impact parameter is given by

4. In the a-particle scattering experiment, the shape of the trajectory of the scattered a-particles depend upon :

(a) only on impact parameter.
(b) only on the source of a-particles.
(c) both impact parameter and source of a-particles.
(d) impact parameter and the screen material of the detector.

Ans. (a) only on impact parameter.

Explanation: In the α-particle scattering experiment, the shape of the trajectory of the scattered α-particles only depends upon impact parameters.

5. The number of alpha particles scattered at 60o is ٧٠ per minute in an alpha particle scattering experiment. The number of alpha particles scattered per minute at ٦٠o are:

(a) 45
(b) 0
(c) 25
(d) 70

Ans. (d) 70

Explanation:

6. The kinetic energy of a-particle incident on gold foil is doubled. How does the distance of closest approach changes?

(a) Double
(b) Halved
(c) Remains same
(d) None of the above

Ans. (b) Halved

Explanation: The distance of closest approach is given by

7. The impact parameter for scattering of alpha particle by 180o is :

\(a) 1
(b) 0
(c) 2
(d) 4

Ans. (b) 0

Explanation: The impact parameter for scattering of alpha particle by 180o is zero (0).

8. The first model of atom in 1898 was proposed by :

(a) Niels Bohr
(b) J.J. Thomson
(c) Albert Einstein
(d) Ernst Rutherford

Ans. (b) J.J. Thomson
Explanation: In 1898, J.J. Thomson proposed the first atomic model or plum pudding model of the atom.

9. The mass of a neutron is:

(a) 1.866 u
(b) 1.00866 u
(c) 0.1866 u
(d) 1.0866 u

Ans. (b) 1.00866 u

Explanation: The mass of a neutron is 1.00866 u.

10. The principle that a quantum orbital cannot be occupied by more than two electrons was given by:

(a) Millikan
(b) Hund
(c) Becquerel
(d) None of these

Ans. (d) None of these

Explanation: The principle that a quantum orbital cannot be occupied by more than two electrons was given by Pauli.

11. Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr's model, will be about [NCERT Exemplar]

(a) 53 pm
(b) 27 pm
(c) 18 pm
(d) 13 pm

Ans. (c) 18 pm

Explanation: The atomic number of lithium is 3, therefore, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr's model, will be about times to that of Bohr radius. Therefore, the radius of lithium ion is nealry 18 pm.

12. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be  
This last expression is not correct, because

(a) n would not be integral
(b) Bohr's quantisation applies only two electron
(c) the frame in which the electron is at rest is not inertial
(d) the motion of the proton would not be in circular orbits, even approximately. 

Ans. (c) the frame in which the electron is at rest is not inertial

Explanation: When one decides to work in a frame of reference where the electron is at rest, the given expression is not true as it forms the non-inertial frame of reference.

13. The simple Bohr's model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because [NCERT Exemplar]

(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no longer be given by Coulomb's law

Ans. (a) of the electrons not being subject to a central force

Explanation: The simple Bohr's model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.

14. For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr's model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true, [NCERT Exemplar]

(a) because Bohr's model gives incorrect values of angular momentum
(b) because only one of these would have a minimum energy
(c) angular momentum must be in the direction of spin of electron
(d) because electrons go around only in horizontal orbits

Ans. (a) because Bohr's model gives incorrect values of angular momentum

Explanation: In the simple Bohr's model, only the magnitude of angular momentum is kept equal to some integral multiple of where, h is Planck's constant and thus, the Bohr model gives incorrect values of angular momentum.

15. An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state, [NCERT Exemplar]

(a) the electron would not move in circular orbits
(b) the energy would be (2)⁴ times that of a H-atom
(c) the electrons orbit would go around the protons
(d) the molecule will soon decay in a proton and a H-atom

Ans. (a, c)

Explanation: The protons are separated by a small distance of the order of angstrom. In the ground state the electron would not move in circular orbits and the electrons orbit would go around the protons.

13. Nuclei

Multiple Choice Questions

1. The size of the nucleus was first measured by:

(a) Einstein
(b) Rutherford
(c) Bohr
(d) None of these

Ans. (b) Rutherford

Explanation: Rutherford was first to measure the size of nucleus of an atom. The size of nucleus is smaller than 4 × 10^–14 m.

2. The neutron was discovered by :

(a) James Chadwick
(b) Rutherford
(c) Pierre Curie
(d) None of the above

Ans. (a) James Chadwick

Explanation: Neutrons are chargeless sub-atomic particles. It was discovered by James Chadwick.

3. The ratio of the nuclear radii of elements with mass number 216 and 125 is :

Ans. (c) 6 : 5

Explanation: Since, R = R0A ^1/3

4. Two nuclei have mass numbers in the ratio 1 : 8. The ratio of their nuclear radii is:

(a) 1 : 2
(b) 1 : 4
(c) 2 : 3
(d) 3 : 1

Ans. (a) 1 : 2

Explanation:

5. The stability of a nucleus is decided by its :

(a) number of protons
(b) number of nucleons
(c) binding energy per nucleon
(d) binding energy

Ans. (b) number of nucleons

Explanation: The two main factors that determine nuclear stability are the neutron/proton ratio and the total number of nucleons in the nucleus.

6. The amount of 1 mg of matter converted into energy gives :

(a) 9 J
(b) 9 × 10³ J
(c) 9 × 10⁵ J
(d) 9 × 10¹⁰J

Ans. (d) 9 × 10¹⁰J

Explanation: Here, m = 1 mg = 1 × 10^–3 g = 1 × 10^–6 kg
According to Einstein mass-energy equivalence.
E = mc²
= 1 × 10^–6 × (3 × 10⁸ ) ²= 9 × 10¹⁰ J

7. An atom bomb weighing 1 kg explodes releasing 9 × 10¹² J of energy. What percentage of mass is converted into energy?

(a) 0.1%
(b) 1%
(c) 2%
(d) 10%

Ans. (a) 0.1%

Explanation: E = mc²
9 × 10¹³ = m × (3 × 10⁸ ) ²


= 0.1%

8. Heavy stable nuclei have more neutrons than protons. This is because of the fact that : [NCERT Exemplar]

a) neutrons are heavier than protons
(b) electrostatic force between protons are repulsive
(c) neutrons decay into protons through beta decay
(d) nuclear forces between neutrons are weaker than that between protons

Ans. (b) electrostatic force between protons are repulsive

Explanation: Stable heavy nuclei have more neutrons than protons. This is because electrostatic force between protons is repulsive, which may reduce stability.

9. MP and MN are masses of proton and neutron, respectively, at rest. If they combine to form deuterium nucleus. The mass of the nucleus will be :

(a) less than MP
(b) less than (MP + MN)
(c) less than (MP + 2MN)
(d) greater than (MP + 2MN) Ans. (b) less than (MP + MN)

Explanation: We know that whenever there is fusion or fission of nucleoids and nuclei, some mass is lost (mass defect) which converts into energy. So net mass of products is slightly less than that of substance.

10. Consider a plot of Neutron number (N) versus proton number (Z) for the different nuclei. Let three nucleoids A, B and C are at the positions as shown in the figure. The order of their stability may be:

(a) A > B > C
(b) A < B < C
(c) B > A > C
(d) C < A < B

Ans. (c) B > A > C
Explanation: Nucleus with more neutron number (N) is slightly more stable than nucleus which have more atomic number (Z). So, order of stability is B > A > C.

11. If 200 MeV energy is released in the fission of a single nucleus of 23592U, how many fission must occur per second to produce a power of 1 kW?

(a) 3.12 × 10¹³
(b) 4.12 × 10³
(c) 4.1 × 10¹⁷
(d) 5.12 × 10¹⁹

Ans. (a) 3.12 × 10¹³

Explanation:

12. The energy released in nuclear fission is due to:

(a) the disappearance of few neutrons
(b) the total binding energy of fragments is more than the BE of parental element
(c) the total BE of fragments is less than BE of parental element
(d) the total BE of fragments is equal to the BE of parental element

Ans. (b) the total binding energy of fragments is more than the BE of parental element

Explanation: The difference in mass appears as the energy released.

13. If power of nuclear reactor is 100 W then rate of nuclear fission is :

(a) (3.6 × 10⁶ )s ^–1
(b) (3.2 × 10¹²)s ^–1
(c) (1.8 × 10⁶ )s ^–1
(d) (1.8 × 10¹²)s ^–1

Ans. (b) (3.2 × 10¹²)s ^–1
Explanation: Power of nuclear reactor = 100 W

= 3.2 × 10¹²/s^–1

14. The operation of a nuclear reactor is said to be critical if the multiplication factor (k) has a value :

(a) 1
(b) 1.5
(c) 2.1
(d) 2.5

Ans. (a) 1

Explanation: If multiplication factor k = 1, the operator of nuclear reactor is said to be critical. If it is less than one, reaction can be sustained and if its more than 1, reaction become uncontrolled and it is an explosion.

15. Boron is used in atomic reactor for :

(a) absorption of neutrons
(b) absorption of a-particles
(c) speed up the reaction
(d) change the reaction

Ans. (a) absorption of neutrons

Explanation: Boron is used as neutron absorber material thereby creating the possibility of controlling a nuclear reactor by changing the neutron multiplication factor.

14. Semiconductor Electronics: Materials, Devices And Simple Circuits

Multiple Choice Questions

1. Band structure of a particular semiconductor is shown in figure. If lattice constant of this semiconductor is decreased, then choose the correct option from the following:

(a) EC decreases
(b) EV decreases
(c) Eg increases
(d) All of these

Ans. (d) All of these

Explanation: When we decreases the lattice parameter of given semiconductor, the electrons are more tightly bound to the atom in this case, and hence require more energy to excite. Thus, width of both valence and conduction band decreases, which led to the increase in band gap between the valence and conduction band.

2. The forbidden energy band gap in conductors, semiconductors and insulators are Eg1, Eg2, and Eg3 respectively. The relation among them is :

(a) Eg1 = Eg2 = Eg3
(b) Eg1 < Eg2 < Eg3
(c) Eg1 > Eg2 > Eg3
(d) Eg1 < Eg2 > Eg3

Ans. (b) Eg1 < Eg2 < Eg3

Explanation: Band gap of the insulator is largest as it restricts the flow of electrons through it. So, Eg1 < Eg2 < Eg3.

3. The distinction between conductors, insulators and semiconductors is largely connected with :

(a) the type of crystal lattice
(b) binding energy of their electrons
(c) relative width of their energy gap
(d) None of the above

Ans. (c) relative width of their energy gap

Explanation: According to band theory, distinction between conductors, insulators and semiconductors is based on relative width of energy gaps between valence band and conduction band.

4. The energy gaps in energy band diagrams of a conductor, semiconductor and insulator are E₁, E₂ and E₃. Arrange them in increasing order.

(a) E1 < E₂ < E₃
(b) E₁ > E2 > E₃
(c) E₂ < E1 < E₃
(d) E₃ < E₂ < E₁

Ans. (a) E₁ < E₂ < E₃

Explanation: Increasing order is: E₁ < E^₂ < E₃

5. A semiconductor has equal electron and hole concentration of 6 × 10⁴ m^{–3 .} On doping with a certain impurity, electron concentration increases to 8 × 10¹² m^–3. Identify the type of semiconductor.

(a) n-type
(b) p-type
(c) Both (a) and (b)
(d) None of these

Ans. (a) n-type

Explanation: n-type, as ne >> nh

6. The _________, a property of materials C, Si and Ge depends upon the energy gap between their conduction and valence bands.

Ans. Conductivity.

7. The conductivity of a semiconductor increases with increase in temperature, because : [NCERT Exemplar]

(a) number density of free current carriers increases
(b) relaxation time increases
(c) both number density of carriers and relaxation time increase
(d) number density of carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density

Ans. (d) number density of carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density

Explanation: The conductivity of a semiconductor increases with increase in temperature, because the number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation is much less than increase in number density.

8. When an electric field is applied across a semiconductor : [NCERT Exemplar]

(a) electrons move from lower energy level to higher energy level in the conduction band
(b) electrons move from higher energy level to lower energy level in the conduction band
(c) holes in the valence band move from higher energy level to lower energy level
(d) holes in the valence band move from lower energy level to higher energy level

Ans. (a, c)

Explanation: When electric field is applied across a semiconductor, the electrons in the conduction band get accelerated and acquire energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

9. In a semiconductor :

(a) there are no free electrons at 0 K
(b) there are no free electrons at any temperature
(c) the number of free electrons increases with pressure
(d) the number of free electrons is more than that in a conductor

Ans. (a) there are no free electrons at 0 K Explanation: In semiconductors, the valence band is full at 0 K, but the conduction band is empty. So, no free electron is available for conduction at 0 K.

10. The resistivity of a semiconductor at room temperature is in between :

(a) 10– 2 to 10– 5 W cm
(b) 10– 3 to 106 W cm
(c) 106 to 108 W cm
(d) 1010 to 1012 W cm

Ans. (b) 10– 3 to 106 W cm

Explanation: The resistivity of a semiconductor at room temperature is in between 10–3 to 106 Ω-cm.

11. When an impurity is doped into an intrinsic semiconductor, the resistance of the semiconductor :

(a) increases
(b) decreases
(c) remains unchanged
(d) becomes zero

Ans. (b) decreases

Explanation: When an impurity is doped into an intrinsic semiconductor, the resistance of the semiconductor decreases. i.e. it becomes more conductive.

12. How does the conductivity of a semiconductor change with the rise in its temperature?

(a) Decreases
(b) Increases
(c) Remains same
(d) None of these

Ans. (b) Increases Explanation: Conductivity of a semiconductor increases with increase in temperature.

13. In figure given below V0 is the potential barrier across a p-n junction, when no battery is connected across the junction : [NCERT Exemplar]

(a) 1 and 3 both correspond to forward bias of junction
(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction
(c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction
(d) 3 and 1 both correspond to reverse bias of junction

Ans. (b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction

Explanation: When p-n junction is forward biased, it opposes the potential barrier junction, when p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction.

14. In figure given below, assuming the diodes to be ideal: [NCERT Exemplar]

(a) D₁ is forward biased and D₂ is reverse biased and hence current flows from A to B
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice-versa
(c) D₁ and D₂ are both forward biased and hence current flows from A to B
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice-versa

Ans. (b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice-versa

Explanation: In the given circuit p-side of p-n junction D₁ is connected to lower voltage and n-side of D₁ to higher voltage. Thus, D₁ is reverse biased. The p-side of p-n junction D₂ is at higher potential and n-side of D₂ is at lower potential. Therefore, D₂ is in forward biased. Hence, no current flows through the junction B to A

15. Hole is: [NCERT Exemplar]

(a) an anti-particle of electron
(b) a vacancy created when an electron leaves a covalent bond
(c) absence of free electrons
(d) an artificially created particle

Ans. (b) a vacancy created when an electron leaves a covalent bond

Explanation: The concept of hole describes the lack of an electron at a position where one could exist in an atom or atomic lattice. If an electron is excited into a higher state, it leaves a hole in its old state. Thus, hole can be defined as a vacancy created when an electron leaves a covalent bond.

15. Communication System

Multiple Choice Questions

1. Out of the following which is an essential element of a communication system?

(a) Transistor
(b) Transmitter
(c) Computer
(d) Both (a) and (b)

Ans. (d) Both (a) and (b)

Explanation: The communication system is a system that describes the information exchange between two points. The process of transmission and reception of information is called communication. Components of Communication: The three essential components of a communication system are: Transmitter, Transistor, Receiver. Option (d) is correct.

2. The distortion in the transmission and processing of signals is called:

(a) Interference
(b) Diffraction
(c) Noise
(d) Scattering

Ans. (c) Noise

Explanation: The distortion in the transmission and processing of signals is called noise. In any communication system, during the transmission of the signal, or while receiving the signal, some unwanted signal gets introduced into the communication, making it unpleasant for the receiver, questioning the quality of the communication. Such a disturbance is called Noise.

3. If the maximum frequency of an audio signal be f, then what would be the bandwidth of an Amplitude Modulated (AM) wave?

Explanation: Bandwidth can be predicted using B.W. = 2 fm where fm = the maximum modulating frequency.

4. If the frequencies of carrier waves for AM and FM be fA and fF respectively, then:

(a) fA ≈ fF
(b) fA < fF
(c) fA > fF
(d) fA ≥ fF

Ans. (b) fA < fF

Explanation: Amplitude modulation is a technique where the amplitude of carrier wave varies depending on the information signal. If the frequency range between 535 to 1705 Hz. Whereas the frequency modulation is the encoding of of information in a carrier wave by varying the instantaneous frequency of the wave. The frequency range of FM is 88 to 108 MHz in the higher spectrum. So fa < ff

5. If modulation frequency of an FM wave is f, then the modulation index will be directly proportional to :

Explanation: In FM (Frequency Modulation), the modulation index is defined as the ratio of frequency deviation to the modulating frequency. We observe that the modulation index for FM signal is inversely proportional to the modulating frequency f.

6. In amplitude modulation modulation index of a transmitted carrier wave is b. The increase in power dissipation would be directly proportional to

(a) b
(b) b ²
(c) 1 + b ²

Ans. (b) b 2

Explanation: If the modulation index β=1,
then the power of AM wave is equal to 1.5 times the carrier power. So, the power required for transmitting an AM wave is 1.5 times the carrier power for a perfect modulation.
Pt = Pc (1+β2/2)
Hence it is directly proportional to β2

7. Rate of energy dissipation in a carrier wave transmission is 10kW. What would be the rate of energy dissipated if the wave is frequency modulated to 10% level?

(a) 10 kW
(b) 10.05kW
(c) 10.1kW
(d) 10.5 kW

Ans. (a) 10 kW

Explanation: Rate of energy dissipation in a carrier wave transmission is 10kW.
Pt = Pc (1+β² /2)
β = 10%
β = 10/100
β = 0.1 Pt = 10 KW
10 × 10³ =
Pc = (1 + (0.1)² /2)
Pc = 9.95 KW Pc = 10 KW(approx)
Hence option (a) is correct.

8. A MODEM is used:

(a) To superimpose a data signal on a carrier wave
(b) To retrieve a data signal from its mixture with a carrier wave
(c) To amplify a data signal (
d) To convert an analogue signal to a digital signal and vice-versa

Ans. (d) To convert an analogue signal to a digital signal and vice-versa

Explanation: A modem is a device that connects your home, usually through a coax cable connection, to your Internet service provider (ISP), like Xfinity. The modem takes signals from your ISP and translates them into signals your local devices can use, and vice versa. or we can say that to convert an analogue signal to a digital signal and vice-versa

9. The process in which the amplitude of the carrier wave is made praportional to the instantaneous amplitude of the signal wave is called:

(a) Amplitude modulation
(b) Demodulation
(c) Rectification
(d) Amplification

Ans. (a) Amplitude modulation

Explanation: The process in which the amplitude of the carrier wave is made proportional to the instantaneous amplitude of the signal wave is called amplitude modulation. In amplitude modulation, the amplitude of the modulating signal is superimposed on the amplitude of the carrier wave and due to this the amplitude of the carrier wave varies in accordance with the amplitude of the modulating signal.

10. Radio waves of low frequencies cannot be transmitted to long distances. So these are superposed on a high frequency carrier signal. This process is known as:

(a) Amplification
(b) Rectification
(c) Modulation
(d) Demodulation

Ans. (c) Modulation

Explanation: The main advantage of high frequency signals is that they can be transmitted over very long distances by dissipating very small power. Thus a low frequency audio signals must be sent along with the high frequency signals for communication. This can be done by superimposing electrical audio signals on a high frequency wave called the carrier wave. The carrier wave is generated from radio-frequency oscillators and is undamped in nature. Thus, when the audio-frequency signal is superimposed on a carrier wave, the resulting wave gets all the characteristics of the audio signal. The method of superimposing an audio signal over the carrier wave is called modulation.

11. Long distance transmission is not possible using ground waves due to which characteristics of electromagnetic waves?

(a) Scattering
(b) Interference
(c) Diffraction
(d) Polarisation

Ans. (d) Polarisation

Explanation: Ground wave propagation is a form of signal transmission in which the signal travels along the ground's surface, and it's used to provide regional coverage on the long and medium wave bands. The type of antenna and its polarization has a major effect on ground wave propagation. Vertical polarization is subject to considerably less attenuation than horizontally polarized signals.

12. The example of employing electromagnetic wave, with a range of frequency from 30 MHz to very high 300 MHz, as carrier wave is:

(a) Radio transmission of short distance
(b) Transmission of FM radio
(c) Television transmission
(d) Radar system

Ans. (b) Transmission of FM radio

Explanation: The radio waves themselves span upon a huge range of frequencies, starting as low as 30Hz up to as high as 300GHz. 30MHz to 300MHz belongs to the very high-frequency range. The radio wave communication signals travel through the air in a straight line and are mainly used for broadcasting in radio and televisions. A few very common uses of the very high frequency band of the radio waves are, the digital audio broadcasting (DAB), the FM broadcasting, low range data communication with radio modems, etc.

13. In distant communication, the data signal is not transmitted directly without the help of carrier wave. The reason is:

(a) Extremely large transmitting antenna is necessary
(b) The transmitting antenna is to be installed at very high attitude
(c) The chance of mixing noise in data signal increased
(d) Due to overlapping of many data signals, the data at receiving end becomes very unclear

Ans. (a) Extremely large transmitting antenna is necessary

Explanation: For the transmission of audio signals at distant places, the high frequency carrier waves are used, because these frequency carrier waves travel through space or medium with the speed of light and they are not obstructed by earth's atmosphere. Extremely large transmitting antenna is necessary.

14. If properly modulated carrier wave is transmitted through antenna, then rate of energy dissipated at the antenna

(a) Remains same for AM wave
(b) Increases for AM wave
(c) Remains same for FM wave
(d) Increases for FM wave

Ans. (c) Remains same for FM wave

Explanation: If properly modulated carrier wave is transmitted through antenna, then rate of energy dissipated at the antenna remain same for FM.

15. Which of the following is not the element of the communication system?

(a) Transmitter
(b) Channel
(c) Receiver
(d) Rectifier

Ans. (d) Rectifier

Explanation: Transmitter - Which converts the signal into an electrical signal and then sends it. Channel - Which carries the signal. Receiver - Which receives and converts the signal in original form. These three are generalized parts of a communication system. Rectifier - Which converts AC to DC, hence not a part of a communication system.