Category: Class 12 Physics Objective Questions

Physics Multiple Choice Questions on “Potential Energy in an External Field”.

1. Pick out the expression of electric potential energy from the following.
a) U = (frac {1}{(4pi varepsilon_o)} times [ frac {q_1q_2}{r} ])
b) U = 1 × (4πε_o) × [ (frac {q_1q_2}{r}) ]
c) U = (frac {1}{(4pi varepsilon_o)} times [ frac {q_1}{q_2}{r} ])
d) U = (frac {1}{(4pi varepsilon_o)}) × [q₁q₂]
Answer: a
Clarification: electric potential energy of a system of charges is the total amount of work done in bringing the various charges to their respective positions from infinitely large mutual separations.
The expression for electric potential energy is given by:
U = (frac {1}{(4pi varepsilon_o)} times [ frac {q_1q_2}{r} ])

2. Two isolated metallic spheres, one with a radius R and another with a radius 5R, each carries a charge ‘q’ uniformly distributed over the entire surface. Which sphere stores more electric potential energy?
a) The sphere with radius 5R
b) Both of the spheres will have the same energy
c) The sphere with radius R
d) Initially it will be the sphere with radius 5R then it will shift to the sphere with radius R
Answer: c
Clarification: The sphere with radius R stores more electric potential energy. According to the electric potential energy equation → U = (frac {1}{(4pi varepsilon_o)} times [ frac {q_1q_2}{r} ])
Potential energy is inversely proportional to radius. Therefore, the sphere with lesser radius will store more energy. So, the smaller sphere will store more energy.

3. There are two charges → Q1 = +q and charge Q2 = +2q. From the initial point (Q), Q1 is at a distance of r and Q2 is at a distance 2r. Which charge (Q1 or Q2) will have higher electrostatic potential energy?
a) Q1
b) Both will have the same energy
c) Q2
d) The information given is not enough to determine
Answer: b
Clarification: Electrostatic potential energy of Q1 → U1 = (frac {1}{(4pi varepsilon_o)}) × [Q × (frac {q}{r})] ……….1
Electrostatic potential energy of Q2 → U2 = (frac {1}{(4pi varepsilon_o)}) × [Q × (frac {2q}{2r})]
→ U2 = (frac {1}{(4pi varepsilon_o)}) × [Q x (frac {q}{r})] ………2
1 = 2
Therefore, both the charges will have the same energy.

4. Electrostatic potential energy can be negative.
a) True
b) False
Answer: a
Clarification: If the charge ‘q’ is negative, the sign should be considered in the equation. Therefore, a system consisting of negative and positive point charges will have negative potential energy. A negative potential energy means that work must be done against the electric field in order to move the charges apart.

5. Identify the dimension of electrostatic potential energy from the following.
a) ML²T^-3A^-2
b) ML³T^-2A^-1
c) M^-1L²T^-3A
d) ML²T^-3A^-1
Answer: d
Clarification: Electrostatic potential energy is a scalar quantity that means it possesses only magnitude and no direction. This quantity is denoted by U and it is measured in joules (J). The dimensional formula of electrostatic potential energy is ML²T^-3A^-1.

Physics Multiple Choice Questions on “Current Electricity – Electrical Energy & Power”.

1. The SI unit of electrical energy is ____________
a) kilojoule (KJ)
b) joules (J)
c) watt (W)
d) kilowatt (KW)
Answer: b
Clarification: Electric energy is defined as the total electric work done or energy supplied by the source of emf in maintaining the current in an electric circuit for a given time. The SI unit of electrical energy is joule (J). The commercial unit of electric energy is kilowatt-hour (kWh).

2. 1 kWh = ___________
a) 3.6 × 10⁶ J
b) 3.6 × 10⁵ J
c) 0.36 × 10⁶ J
d) 0.36 × 10⁵ J
Answer: a
Clarification: The commercial unit of electric energy is kilowatt-hour (kWh).
1 kWh = 1000 Wh = 3.6 × 10⁶J = one unit of electricity consumed. The electric energy used in factories, industries and houses are measured in kWh.

3. Calculate the number of units of electricity used if a bulb of 100 W is kept on for 5 hours.
a) 1 unit
b) 0.1 unit
c) 5 unit
d) 0.5 unit
Answer: d
Clarification: The number of units of electricity consumed is
n = (frac {(total , wattage , times , time , in , hour)}{1000})
Total wattage = 100 W      Time in hour = 5 hours
Therefore, n = (frac {100 times 5}{1000})
= 0.5 units
So, the number of units of electricity consumed is 0.5 units.

4. Electric energy is dependent on time.
a) True
b) False
Answer: a
Clarification: Yes, electric energy is dependent on time.
Electric energy = electric power × time = P × t. Electric energy is basically the energy derived from electric charge. It can be calculated by multiplying the electric power of the body with the time taken for the power emission.

5. Kabir bought 5 new light bulbs of 50 W each in addition to the 7 bulbs he already had in his house which were also 50 W each. Calculate his electricity bill, if he keeps the new bulbs on for 5 hours and the older bulbs on only for 3 hours, and the cost of one unit of electricity is Rs. 60.
a) Rs.136
b) Rs.137
c) Rs.138
d) Rs.139
Answer: c
Clarification: New bulbs = 5; Old bulbs = 7; Total wattage of new bulbs = 5 × 50;
Total wattage of old bulbs = 7 × 50; Time the new bulbs are kept on = 5 hours;
Time the old bulbs are kept on = 3 hours
The number of units of electricity consumed by new bulbs (n1) = (frac {(total , wattage , times , time , in , hours)}{1000})
= (frac {50 times 5 times 5}{1000}) = 1.25
The number of units of electricity consumed by old bulbs (n2) = (frac {50 times 7 times 3}{1000}) = 1.05 Total bill of electricity = number of units of electricity consumed × amount for one unit of electricity
= (n1 + n2) × 60
= (1.25 + 1.05) × 60
= 138
Therefore, his electricity bill will be Rs. 138.

6. Identify the correct formula of electric power.
a) Electric power = (frac {time , taken}{electric , work , done})
b) Electric power = (frac {electric , work , done}{time , taken})
c) Electric power = electric work done × time taken
d) Electric power = (frac {1}{electric , work , done})
Answer: b
Clarification: Electric power is defined as the rate at which work is done by the source of emf in maintaining the current in the electric circuit. So, the formula of electric power is:
Electric power (P) = (frac {electric , work , done}{time , taken})

7. Which one of the following is the practical unit of power?
a) Watt (W)
b) Kilowatt hour (kWh)
c) Horse power (hp)
d) Kilojoule (kJ)
Answer: c
Clarification: The practical unit of power is horse power (hp). Kilo watt is also another practical unit of power. 1 kilowatt = 1000 watt; 1 hp = 746 watt. It is usually used in reference to the output of engines or motors.

8. The power consumed by a 300 V bulb, having a resistance of 100 ohms, is 3 Watts.
a) True
b) False
Answer: b
Clarification: Electric power can also be calculated or represented in other ways, such as: Power = voltage × current; Power = current² × resistance; Power = (frac {voltage^2}{resistance}). In this case, we can use the equation – Power = (frac {voltage^2}{resistance}). Voltage = 300 V; Resistance = 100 ohms
Power = (frac {300 times 300}{100})
= 900 Watts
Therefore, the power consumed by a 300 V bulb, having a resistance of 100 ohms, is 900 Watts.

9. One watt is equal to __________
a) one kilowatt per second
b) one kilo joule per second
c) one joule per second
d) one joule per minute
Answer: c
Clarification: Watt is the SI unit of power. Power = (frac { work , done}{time}). The SI unit of work done is the same as energy, that is, joule and that of time is seconds. Therefore, one watt is equal to one joule per second.

10. An engine uses 30 A of current. The resistance offered is 15 ohms. Calculate the power consumed by the engine in horse power.
a) 18 hp
b) 19 hp
c) 17 hp
d) 13500 hp
Answer: a
Clarification: Current used = 30 A; Resistance = 15 ohms
The required equation is: Power = current² × resistance
= 30 × 30 × 15
= 13,500 Watts
We know that, 1 Watt = 746 horse power (hp). So, 13,500 Watts = 18.096 hp, which can approximately be equal to 18 hp.
Therefore, the engine consumes 18 hp power.

Physics Multiple Choice Questions on “Torque on Current Loop & Magnetic Dipole”.

1. What torque acts on a 50 turn coil of 200 cm² area carrying a current of 20 A held with its axis at right angles to a uniform magnetic field of 0.2T?
a) 8 Nm
b) 0.8 Nm
c) 4 Nm
d) 0.8 Nm
Answer: c
Clarification: Torque = NIBA sinθ.
Torque = 50 × 20 × 0.2 × 200 × 10^-4 × sin (90)
Torque = 4Nm.

2. What is the maximum torque on a rectangular coil of area 4 cm × 80 cm of 400 turns, when carrying a current of 103 A in a magnetic field of 0.2 T?
a) 1250 Nm
b) 2560 Nm
c) 3600 Nm
d) 1985 Nm
Answer: b
Clarification: Torque (Max) = NIBA.
Τ_max = 400 × 10³ × 0.2 × 320 × 10^-4
T_max = 2560 Nm.

3. Under which of these conditions, the torque acting on the current carrying loop is minimum?
a) θ = 0^o
b) θ = 90^o
c) θ = 270^o
d) θ = 450^o
Answer: a
Clarification: When θ = 0^o ➔ τ = 0.
The torque is minimum when the plane of the loop is perpendicular to the magnetic field, so the angle θ should be 0^o.

4. Identify the factor on which the torque on a planar current loop does not depend.
a) Current
b) Strength of magnetic field
c) Area of the loop
d) The shape of the loop
Answer: d
Clarification: Torque on a planar current loop depends upon current, the strength of the magnetic field, area of the loop and the orientation of the loop in the magnetic field. It is independent of the shape of the loop.

5. Give the SI unit of magnetic pole strength.
a) Tm
b) Am²
c) Am
d) Tm²
Answer: c
Clarification: The SI unit of magnetic pole strength is ampere meter (Am).
The strength of a magnetic pole is said to be one-ampere meter if it repels an equal and similar pole with a force of 10^-7 N when placed in vacuum at a distance of one meter from it.

6. The factor ((frac {eh}{4pi m_e})) is the least value of the magnetic moment. This natural unit of magnetic moment is called Planck’s constant.
a) True
b) False
Answer: b
Clarification: The factor ((frac {eh}{4pi m_e})) is the least value of the magnetic moment. This natural unit of magnetic moment is called Bohr magneton. The value of Bohr magneton is 9.27 × 10^-24 Am². So, that makes the above statement false.

7. ‘X’ is the product of the pole strength of either magnetic pole and the magnetic length of the magnetic dipole. Identify X.
a) Magnetic dipole
b) Magnetic dipole moment
c) Magnetic pole strength
d) Magnetic dipole length
Answer: b
Clarification: The product of the pole strength of either magnetic pole and the magnetic length of the magnetic dipole is called its magnetic dipole moment. The SI unit of magnetic dipole moment is Am².

8. Under which of these conditions, the torque acting on the current-carrying loop is minimum?
a) θ = 90^o
b) θ = 0^o
c) θ = 180^o
d) θ = 360^o
Answer: a
Clarification: When θ = 90^o ➔ τ = NIBA.
The torque is maximum when the plane of the loop is parallel to the magnetic field. So the angle θ has to be 90^o.

9. Calculate the torque on a 500 turn rectangular coil of length 20 cm and breadth 10 cm, carrying a current of 30 A, when placed making an angle of 60^o with a magnetic field of 0.5 T?
a) 625 Nm
b) 125 Nm
c) 75 Nm
d) 25 Nm
Answer: c
Clarification: Torque = NIBA cosθ [Angle between field and plane of coil].
Torque = 500 × 30 × 0.5 × 200 × 10^-4 × cos (60)
Torque =75 Nm.
Therefore, the torque is 75 Nm.

10. Give the SI unit of magnetic dipole moment.
a) Am
b) Am²
c) Tm²
d) Tm
Answer: b
Clarification: Magnetic dipole moment = Current × Area.
SI unit of current = Ampere (A)
SI unit of area = meter² (m²)
So, the SI unit of magnetic dipole moment = A × m² = Am²

Physics Multiple Choice Questions on “AC Generator”.

1. A 100 turn coil of area 0.1 m² rotates at half a revolution per second. It is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil?
a) 256.33 V
b) 89.12V
c) 0.314 V
d) 3.1455 V
Answer: c
Clarification: The maximum voltage generated in the coil,
e₀ = nBAω
e₀ = nBA × (2πv)
e₀ = 100 × 0.01 × 0.1 × 2π × 0.5
e₀ = 0.314 V

2. An a.c. generator consists of a coil of 1000 turns and cross-sectional area of 3m², rotating at a constant angular speed of 60 rad s^-1 in a uniform magnetic field 0.04 T. The resistance of the coil is 500Ω. Calculate the maximum current drawn from the generator.
a) 2500 A
b) 1.44 A
c) 6.25 A
d) 0.55 A
Answer: b
Clarification: e₀ = nBAω
e₀ = 100 × 0.04 × 3 × 60
e₀ = 720 V.
The maximum current drawn = (frac {e_0}{R} = frac {720}{500})
I = 1.44 A.

3. Adithya peddles a stationary bicycle the pedals of which are attached to a 500 turn coil of area 11 m². The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 20.5 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?
a) 354035 V
b) 85000V
c) 111647 V
d) 46464 V
Answer: a
Clarification: The maximum voltage generated in the coil,
e₀ = nBAω
e₀ = nBA × (2πv)
e₀ = 500 × 20.5 × 11 × 2π × 0.5
e₀ = 345035 V

4. Identify the principle behind the working of an a.c. generator.
a) Eddy currents
b) Faraday’s law
c) Lenz’s law
d) Electromagnetic induction
Answer: d
Clarification: The working of an a.c. generator is based on the principle of electromagnetic induction. When a coil is rotated about an axis perpendicular to the direction of the uniform magnetic field, an induced emf is produced across it and hence current is set up in it.

5. An armature coil consists of 30 turns of wire, each of area A = 0.05 m² and total resistance of 10 Ω. It rotates in a magnetic field of 0.15T at a constant frequency of (frac {140}{pi }) Hz. Determine the value of maximum induced emf produced in the coil.
a) 1 V
b) 500 V
c) 63 V
d) 43 V
Answer: c
Clarification: e₀ = nBAω.
e₀ = 30 × 0.15 × 0.05 × 2π × ((frac {140}{pi }))
e₀ = 63 V.

6. In an a.c. generator, electrical energy is converted to mechanical energy by electromagnetic induction.
a) True
b) False
Answer: b
Clarification: When a coil is rotated about an axis perpendicular to the direction of the uniform magnetic field, an induced emf is produced across it. In an a.c. generator, mechanical energy is converted to electrical energy by electromagnetic induction.

7. ‘X’ is a rectangular coil consisting of a large number of turns of copper wire wound over a soft iron core in an a.c. generator. Identify X.
a) Slip ring
b) Armature
c) Copper brushes
d) Field magnet
Answer: b
Clarification: A rectangular coil consisting of a large number of turns of copper-wound over a soft iron core is called the armature. The soft iron core is used to increase the magnetic flux.

8. An a.c. generator consists of a coil of 50 turns and an area 2.5 m² rotating at an angular speed of 60 rad s^-1 in a uniform magnetic field of 0.3 T between two fixed pole pieces. What is the flux through the coil, when the current is zero?
a) Maximum
b) Minimum
c) Zero
d) Independent of current
Answer: a
Clarification: The current is zero when the coil is vertical. In this position, flux through the coil is maximum. On the other hand, the current is maximum, when the coil is horizontal. So, when the current is zero, the flux through the coil is maximum.

Physics Multiple Choice Questions on “Optical Instruments”.

1. A boy uses spectacles of focal length -50 cm. Name the defect of the vision he is suffering from.
a) Astigmatism
b) Hypermetropia
c) Myopia
d) Presbyopia
Answer: c
Clarification: As the focal length is negative, the lens used is concave. When a person is prescribed a concave lens, then the person is considered to be suffering from myopic. Therefore, the boy is suffering from myopia.

2. How much intensity of the image is increased if the diameter of the objective of a telescope is doubled?
a) Two times
b) Four times
c) Eight times
d) Sixteen times
Answer: b
Clarification: On doubling the diameter, the area of the objective increases four times. Its light-gathering power increases four times. The brightness of the image also increases four times. Therefore, the intensity of the image increases by four.

3. A fly is sitting on the objective of a telescope. What will be its effect on the final image of the distant object?
a) Reduces
b) Increases
c) Remains constant
d) Indefinite
Answer: a
Clarification: The fly will not be seen in the final image. But the intensity of the final image gets reduced because light will not enter the telescope from that part of the object where the fly is sitting. Therefore, a reduction effect will be observed in the final image of the fly.

4. Two lenses of focal lengths 5 cm and 50 cm are to be used for making a telescope. Which lens will you use for the objective?
a) Both
b) Neither
c) 5 cm
d) 50 cm
Answer: d
Clarification: In a telescope, the objective should have a large focal length than the eyepiece. So the lens of 50 cm focal length will be used as the objective. So, the smaller one of the two, i.e. the 5 cm focal length lens will be used as the eyepiece.

5. Which of the following is used to increase the range of a telescope?
a) Increasing the focal length
b) Decreasing the focal length
c) Increasing the diameter of the objective
d) Decreasing the diameter of the objective
Answer: c
Clarification: The light-gathering power of the objective will increase and even faint objects will become visible. This is done by increasing the diameter of the objective. So, by increasing the diameter of the objective, the range of the telescope can be increased.

6. The upper part of the bi-focal lens is a convex lens while its lower part is a concave lens.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. The upper part of the bi-focal lens should be a concave lens, whereas the one used for distant vision as its lower part should be a convex lens. Such a bi-focal lens is used for reading purposes.

7. ‘X’ can see objects in the ultra-violet light while human beings cannot do so. Identify X.
a) Penguin
b) Bees
c) Ant
d) Tiger
Answer: b
Clarification: Bees have some retinal cones that are sensitive to ultra-violet light, so they can see objects in ultra-violet light. Humans are ultra-violet blind. Therefore, a human being cannot see UV light, whereas bees can.

8. How does the magnifying power of a telescope change on increasing the diameter of its objective?
a) Independent
b) Doubled
c) Halved
d) Becomes zero
Answer: a
Clarification: Magnifying power of a telescope is given as:
m = (frac {f_0}{f_e})
It is independent of the aperture of the object. So the magnifying power is not affected when the diameter of the objective is increased.

9. What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm?
a) + 25 cm
b) – 25 cm
c) + 50 cm
d) – 50 cm
Answer: c
Clarification: By thin lens formula,
(frac {1}{f} = frac {1}{v} – frac {1}{u})
(frac {1}{f} = frac {1}{(-50)} – frac {1}{(-25)})
f = +50 cm.

10. An object is to be seen through a simple microscope of power 10 D. Where should the object be placed to reduce maximum angular magnification? The least distance for distinct vision is 25 cm.
a) + 7.1 cm
b) – 7.1 cm
c) + 25 cm
d) – 25 cm
Answer: b
Clarification: Angular magnification is maximum when the final image is formed at the near point.
(frac {1}{u} = frac {1}{v} – frac {1}{f})
(frac {1}{u} = ( frac {-1}{25} ) – ( frac {1}{10} ) )
u = -7.1 cm.
Therefore, the object should be placed 7.1 cm before the lens to avoid angular magnification.