Category: Class 12 Physics Objective Questions

Physics Quiz for Class 12 on “Ampere and Forces between Two Parallel Currents”.

1. What is the expression for the force experienced by two parallel current carrying conductors of unequal length?
a) F=(frac {mu_o}{4pi } times frac {2I_1 I_2}{r}) × l
b) F=(frac {mu_o}{4pi } times frac {I_1 I_2}{r}) × l
c) F=(frac {mu_o}{4pi } times frac {frac {2I_1 I_2}{r}}{l})
d) F=(frac {mu_o}{4pi } times frac {2I_1 I_2}{4r}) × l
Answer: a
Clarification: If two linear current carrying conductors of unequal length are held parallel to each other, then the force on a long conductor is due to magnetic field interaction due to currents of short conductor and long conductor. The force on long conductor is force to that on the short conductor given by:
F=(frac {mu_o}{4pi } times frac {2I_1 I_2}{r}) × l

2. Two parallel, long wires carry currents I₁ and I₂ where I₁ > I₂. When the currents are in the same direction, the magnetic field at a point midway between the wires is 50 μT. If the direction of I₂ is revered, the field becomes 100 μT. What is the value of (frac {I_1}{I_2})?
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: At the midpoint ➔ CASE 1:
B1=(frac {mu_o I_1}{2pi d} – frac {mu_o I_2}{2pi d})
B1=(frac {mu_o (I_1 – I_2)}{2pi d})=50 μT
At the midpoint ➔ CASE 2:
B2=(frac {mu_o I_1}{2pi d} + frac {mu_o I_2}{2pi d})
B1=(frac {mu_o (I_1 + I_2)}{2pi d})=100 μT
(frac {B1}{B2}=frac {I_1-I_2}{I_1+I_2}=frac {50 mu T}{100 mu T})
(frac {I_1-I_2}{I_1+I_2}=frac {1}{2})
Using componendo and Dividendo rule:
(frac {(I_1+I_2) , + , (I_1-I_2)}{(I_1-I_2) , – , (I_1+I_2)}=frac {1+2}{1-2})
(frac {I_1}{-I_2}=frac {3}{-1})
Therefore, (frac {I_1}{I_2})=3

3. There are two conductors X and Y carrying a current I and moving in the same direction. p and q are two electron beams also moving in the same direction. Will there be attraction or repulsion between the 2 conductors and between the two electron beams separately?
a) The electron beams will repel each other and conductors attract each other
b) The electron beams will attract each other and the conductors also attract each other
c) The electron beams will attract each other and the conductors repel each other
d) The electron beams will repel each other and the conductors also repel each other
Answer: b
Clarification: Since, both the current carrying conductors are moving in the same direction, they will attract each other. Moving electron beams is equivalent to an electric current in opposite direction. Therefore, there will attraction between the conductors X and Y as well as between the electron beams p and q.

4. When infinitely long parallel wires carry equal currents in the same direction, the magnetic field at the midpoint in between the two wires is zero.
a) True
b) False
Answer: a
Clarification: Yes, when infinitely long parallel wires carry equal currents in the same direction, the magnetic field at the midpoint in between the two wires is zero. This is because, when the wires carry equal currents in the same direction, the magnetic field created at the midpoint will cancel out each other and thus, the field is zero.

5. Two long conductors, separated by a distance r carry current I₁ and I₂ in the same direction. They exert a force F on each other. Now, the current in one them is increased to 3 times and the direction is reversed. The distance is also increased to 5r. What is the new value of the force between them?
a) (frac {3F}{5})
b) (frac {-3F}{5})
c) (frac {5F}{3})
d) (frac {-5F}{3})
Answer: b
Clarification: First case:
F=(frac {mu_o}{2pi } big [ frac {I_1 I_2}{r} big ] )l
Second case:
F^‘=(frac {mu_o}{2pi } big [ frac {(-3I_1)(I_2)}{5r} big ] )l
(frac {F^{‘}}{F} = frac {frac {mu_o}{2pi } big [ frac {(-3I_1)(I_2)}{5r} big ] l}{frac {mu_o}{2pi } big [ frac {I_1 I_2}{r} big ] l})
(frac {F^{‘}}{F}= -frac {3}{5})
Therefore, F^‘=((-frac {3}{5}))F=(underline{frac {-3F}{5}})

6. There are 2 long parallel conductors AB and CD. AB carries 4A current and CD carries 3A current. The magnetic field at the midpoint of these 2 conductors is B. If 4A current is switched off, then what is the magnetic field at the midpoint now?
a) (frac {B}{3})
b) (frac {2B}{3})
c) B
d) 3B
Answer: d
Clarification: Firstly, the magnetic field at the midpoint due to current in AB:
B_AB=μ_o×(frac {2 times 4}{4pi d}=frac {8mu_o}{4pi d}) ………………..1
Magnetic field at midpoint due to current in CD:
B_CD=μ_o×(frac {2 times 3}{4pi d}=frac {6mu_o}{4pi d}) …………………..2
The net magnetic field = B_AB – B_CD ➔ 1 – 2
B = (frac {2 mu_o}{4pi d}) ……………………….3
Now, when 4A is switched off, then the magnetic field at the midpoint will be due to the current in CD, i.e. due to 3A current
B^‘ = μ_o×(frac {2 times 3}{4pi d}=frac {6mu_o}{4pi d}) ……………..4
Comparing 3 and 4:
Therefore, B^‘ = 3B

7. Pick out the expression for force per unit length between two parallel carrying conductors if the current passing through both of them are the same?
a) (frac {mu_o 2I^2}{4πb})
b) (frac {mu_o 2I^3}{4πb})
c) (frac {mu_o I^2}{4πb})
d) (frac {mu_o 4I^2}{4πb})
Answer: b
Clarification: The expression for force between 2 parallel current carrying conductors is given by:
f=(frac {mu_o}{4pi } big [ frac {I_1 I_2}{b} big ] )
In this case, the current passing through both of them are the same, i.e. I₁=I₂
Therefore, the force per unit length is given as:
f=(frac {mu_o}{4pi } big [ frac {I^2}{b} big ] )

Physics Quiz for Class 12,

Physics Multiple Choice Questions on “Electromagnetic Inductance”.

1. Give the SI unit of self-inductance.
a) Farad
b) Ampere
c) Henry
d) Maxwell
Answer: c
Clarification: The self-inductance of a coil is said to be one henry if an induced emf of one volt is set up in it when the current in it changes at the rate of one ampere per second. Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.

2. Write the dimensions of self-inductance.
a) [M³ L² T^-2 A^-2]
b) [M L² T^-2 A^-2]
c) [M² L² T^-2 A^-2]
d) [M L² T² A^-2]
Answer: b
Clarification: Self-inductance = (frac {Magnetic , flux}{Current})
(frac {[ M L^2 T^{-2} ]}{[ A^2 ]})
Self-inductance = [M L² T^-2 A^-2]

3. Identify the factor on which self-inductance does not depend.
a) Number of turns
b) Area of cross-section
c) Permeability of the core material
d) The permittivity of the core material
Answer: d
Clarification: Self-inductance is directly proportional to the number of turns and the area of cross-section. The self-inductance of a solenoid increases μ_r times if it is wound over an iron core of relative permeability μ_r.

4. Which of the following is the unit of mutual inductance?
a) VsA^-2
b) V³sA²
c) V²s
d) VsA^-1
Answer: d
Clarification: Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil. Unit of Mutual inductance = (frac {V}{As^{-1}}).
SI unit = VsA^-1.

5. Identify the factor on which mutual inductance does not depend.
a) Relative separation
b) The relative orientation of the two coils
c) Reciprocity
d) Permeability of the core material
Answer: c
Clarification: The mutual inductance of two coils is the property of their combination. It does not matter which one of them functions as the primary or the secondary coil. Hence, mutual inductance does not depend on reciprocity.

6. Mutual inductance is called the inertia of electricity.
a) True
b) False
Answer: b
Clarification: Self-induction of a coil is that the property by which it tends to take care of the magnetic flux linked with it and opposes any change within the flux by inducing a current in it. This is the reason why self-induction is named inertia of electricity.

7. Calculate the mutual inductance between two coils if a current 10 A in the primary coil changes the flux by 500 Wb per turn in the secondary coil of 200 turns.
a) 10 H
b) 10⁴ H
c) 1000 H
d) 100 H
Answer: b
Clarification: NΦ = MI
200 × 500 = M × 10
M = 10⁴ H

8. What is the relative permeability of an air-core inductor, if its self-inductance increases from 0.5 mH to 50 mH due to the introduction of an iron core into it?
a) 100
b) 0.1
c) 10000
d) 0.0001
Answer: a
Clarification: μ_r = (frac {mu }{mu_0})
μ_r = (frac {L}{L_0})
μ_r = (frac {50}{0.5})
μ_r = 100

9. What is the self-inductance of the coil, if the magnetic flux of 10 microwebers is linked with a coil when a current of 5 mA flows through it?
a) 20 mH
b) 5 mH
c) 2 mH
d) 250 mH
Answer: c
Clarification: Self-inductance = (frac {Magnetic , flux}{Current})
Self inductance = (frac {10 times 10^{-6}}{5 times 10^{-3}})
Self inductance = 2 × 10^-3 H
Self inductance = 2mH

10. Determine the self-inductance of a coil, which has a magnetic flux of 50 milliwebers that is produced when a current of 5 A flows through it?
a) 1 × 10^-2 Wb
b) 1 × 10^-3 Wb
c) 100 Wb
d) 1 × 10³ Wb
Answer: a
Clarification: Self-inductance = (frac {Magnetic , flux}{Current})
Self-inductance = (frac {50 times 10^{-3}}{5})
Self-inductance = 1 × 10^-2 Wb

Physics Exam Questions for IIT JEE Exam on “Ray Optics – Natural Phenomena due to Sunlight”.

1. Which among the following relates scattering and wavelength of the particle?
a) Rayleigh scattering
b) Photon scattering
c) Wave theory
d) Cauchy’s formula
Answer: a
Clarification: Light of shorter wavelengths is scattered much more than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering.

2. Which one of the following is not an illustration of the scattering of light?
a) The blue color of the sky
b) The white color of the clouds
c) Rainbow
d) The red color of danger signals
Answer: c
Clarification: A rainbow is an example of a dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to the combined effect of dispersion, refraction, and reflection of sunlight by spherical water droplets of rain. Others all are common illustrations of scattering of light.

3. How many kinds of rainbows are there, generally?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Generally, there are two kinds of rainbows, namely primary rainbow and secondary rainbow. The primary rainbow occurs due to one internal reflection and two refractions, whereas secondary rainbow occurs due to two internal reflections and two refractions, from the water drops suspended in the air.

4. The secondary rainbow is fainter than the primary rainbow.
a) True
b) False
Answer: a
Clarification: Yes, the secondary rainbow is fainter than the primary rainbow. As the secondary rainbow is formed by one more reflection than the primary rainbow, it is much fainter than the primary rainbow. The colors are also reversed in the secondary rainbow. In the primary rainbow, the violet color is on the inner edge and red color is on the outer edge. It is the other way round for the secondary rainbow.

5. Find the wrong statement.
a) Clouds appear white for most of the day because the sun radiates all visible wavelengths of light
b) The rays have to travel a shorter distance than usual at sunrise and sunset
c) Danger signals are red because red color gets scattered the least
d) The sky is blue because blue light is scattered more than other colors since it travels as shorter waves
Answer: b
Clarification: During sunrise and sunset, the rays have to travel a large part of the atmosphere since the rays are close to the horizon at that time. That is why, red being the least scattered enters our eyes, as all the other colors get scattered away.

Physics Exam Questions for IIT JEE Exam,

Physics Multiple Choice Questions on “Atomic Spectra”.

1. Which source is associated with a line emission spectrum?
a) Electric fire
b) Neon street sign
c) Red traffic light
d) Sun
Answer: b
Clarification: Neon street sign gives a line emission spectrum. When neon atoms gain enough energy to become excited, light is produced. Atom releases a photon when it returns to a lower energy state. Therefore, the source associated with a line emission spectrum is the neon street sign.

2. According to the uncertainty principle for an electron, time measurement will become uncertain if which of the following is measured with high certainty?
a) Energy
b) Momentum
c) Location
d) Velocity
Answer: a
Clarification: According to the uncertainty principle,
ΔE.Δt >= (frac {h}{2pi }).
Thus the time measured will become uncertain if ΔE is measured with high certainty.

3. Find the true statement.
a) An electron will not lose energy when jumping from the 1^st orbit to the 3^rd orbit
b) An electron will not give energy when jumping from the 1^st orbit to the 3^rd orbit
c) An electron will release energy when jumping from the 1^st orbit to the 3^rd orbit
d) An electron will absorb energy when jumping from the 1^st orbit to the 3^rd orbit
Answer: d
Clarification: An electron will absorb energy when jumping from the 1^st orbit to the 3^rd orbit. Only by absorbing energy, an electron will be able to jump from the first orbit to the third orbit in the atomic spectrum.

4. The size of the atom is proportional to which of the following?
a) A
b) A^1/3
c) A^2/3
d) A^-1/3
Answer: b
Clarification: The size of the atom is proportional to A^1/3.
The electron affinity of the atom is inversely proportional to the atomic size. As the number of energy levels increases, the atomic size must increase.

5. Calculate the ratio of the kinetic energy for the n = 2 electron for the Li atom to that of Be⁺ ion?
a) (frac {9}{16})
b) (frac {3}{4})
c) 1
d) (frac {1}{2})
Answer: a
Clarification: (frac {KE_{Li}}{KE_{Be}} = big [ frac {(frac {Z_{Li}}{2} )}{(frac {Z_{Be}}{2})} big ]^2 )
(frac {KE_{Li}}{KE_{Be}} = big [ frac {(frac {3}{2} )}{(frac {4}{2})} big ]^2 )
(frac {KE_{Li}}{KE_{Be}} = frac {9}{16}).

6. The Bohr model of atoms uses Einstein’s photoelectric equation.
a) True
b) False
Answer: b
Clarification: Bohr model assumes that the angular momentum of electrons is quantized. Therefore, the Bohr model of the atoms involved is independent of Einstein’s photoelectric equation.

7. What is the ratio of minimum to maximum wavelength in the Balmer series?
a) 5:9
b) 5:36
c) 1:4
d) 3:4
Answer: a
Clarification: For a wavelength of Balmer series,
(frac {1}{lambda }) = R( [ frac {1}{4} – frac {1}{9} ] )
(frac {1}{lambda } = frac {5R}{30}).
(frac {lambda_{min}}{lambda_{max}} = frac {5R}{36} times frac {4}{R} )
(frac {lambda_{min}}{lambda_{max}} = frac {5}{9})

8. The energy of characteristic X-ray is a consequence of which of the following?
a) The energy of the projectile electron
b) The thermal energy of the target
c) Transition in target atoms
d) Temperature
Answer: c
Clarification: The energy of characteristic X-ray is a consequence of transition in target atoms. They cause emission or absorption of electromagnetic radiation. The other options are not responsible for the energy of characteristic X rays.

9. Find out the minimum energy required to take out the only one electron from the ground state of Li⁺?
a) 13.6 eV
b) 122.4 eV
c) 25.3 eV
d) 67.9 eV
Answer: b
Clarification: Ionization energy is given as:
E = 13.6 Z² eV
For Li⁺, Z = 3
E = 13.6 × 9
E = 122.4 eV

10. What is the energy required to ionize an H-atom from the third excited state, if ground state ionization energy of H-atom is 13.6 eV?
a) 1.5 eV
b) 3.4 eV
c) 13.6 eV
d) 12.1 eV
Answer: a
Clarification: From third excited state, E = (frac {-13.6}{16})
E = -0.85 eV
Energy required to ionize H-atom from second excited state = 0 – (-0.85)
E = +0.85 eV

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.