Category: Class 12 Physics Objective Questions

Physics Written Test Questions for Schools on “Magnetic Field on the Axis of a Circular Current Loop”.

1. Pick out the expression for magnetic field strength at any point at the center of a circular loop from the following?
a) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡90
b) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡45
c) B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡30
d) B = [μ_o × 4πr]( int_0^{2pi r} )dl sin⁡90
Answer: a
Clarification: The magnetic field strength at any point at center of circular loop carrying current I and radius r is given as:
B = ( [ frac {mu_o I}{4 pi r} ] int_0^{2pi r} )dl sin⁡90
It can also be expressed as → B = ( [ frac {mu_o I}{4 pi r} ] ) × 2πr or B = ( [ frac {mu_o I}{2 r} ] ) direction.
It is inwards if the current is flowing in the clockwise direction and it is onwards if the current is flowing in the anticlockwise direction.

2. Identify the expression for the magnetic field on the axis of circular loop.
a) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {1}{2}} big ] } ) × 2πr
b) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } )
c) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } ) × 2πr
d) B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {1}{2}} big ] } ) × 4πr
Answer: c
Clarification: The expression for magnetic field on the axis of circular loop is given as:
B=( frac {mu_o Ir}{4 pi big ( (r^2 , + , x^2)^{frac {3}{2}} big ] } ) × 2πr
It is towards the loop if current in it is in clockwise direction and it is away from the loop if current in it is in anticlockwise direction.

3. The magnetic field due to a current carrying circular loop of radius 4 cm at a point on the axis at a distance of 7 cm from the center is 48 μT. What will be the value at the center of the loop?
a) 390 μT
b) 393 μT
c) 395 μT
d) 397 μT
Answer: b
Clarification: Field along axis of coil → B =( frac {mu_o IR^2}{2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] } )
At the coil of the coil → B₁ = ( frac {mu_o I}{2R})
(frac {B_1}{B} = frac {mu_o I}{2R} , times , frac {2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] }{mu_o IR^2} = frac {2 big ( (R^2 , + , x^2)^{frac {3}{2}} big ] }{R^3})
B1 = (frac {48 [4^2 + 7^2]^{frac {3}{2}}}{4^3})
B1 = (frac {48 times (65)^{frac {3}{2}}}{4^3})
B1 = 131.01 × 3
B1 = 393.03 = 393 μT

4. When an arc of a circle of radius R subtends an angle of (frac {pi }{4}) at the center, and carries a current I, the magnetic field at the center is B = (frac {mu_o I}{16R}).
a) True
b) False
Answer: a
Clarification: Magnetic field at the center of a circular arc of radius R, carries a current I and makes an angle θ at the center is given by
B = (frac {mu_o I theta}{4 pi R})
In this case θ = (frac {pi }{4})
B = (frac {mu_o I (frac {pi }{4})}{4 pi R})
B = (frac {mu_o I}{16 R})

5. A horizontal overhead power line carries a current of 100A in east to west direction. What is the magnitude and direction of the magnetic field due to the current, 2 m below the line?
a) 3 x 10^-5 T
b) 2 x 10^-5 T
c) 1 x 10^-5 T
d) 4 x 10^-5 T
Answer: c
Clarification: Given: I = 100A; r = 2 m; (frac {mu_o}{4 pi }) = 10^-7
The required equation → B = ((frac {mu_o}{4 pi }) , times , frac {2I}{r})
B = 10^-7 × 2 × (frac {100}{2})
B = 10^-7 × 100
B = 1 × 10^-5 T
Therefore, the magnitude of the magnetic field is 1 × 10^-5 T and the direction of the magnetic field is south.

Physics Written Test Questions for Schools,

Physics Multiple Choice Questions on “Electromagnetic Induction – Motional Electromotive Force”.

1. Identify the expression for the motional electromotive force from the following?
a) E = -vLB
b) E = vLB
c) E = (frac {v}{LB})
d) E = (frac {LB}{v})
Answer: a
Clarification: Motional electromotive is the emf induced by the motion of the conductor across the magnetic field. The expression for motional electromotive force is given by:
E = -vLB
This equation is true as long as the velocity, magnetic field, and length are mutually perpendicular to each other. The negative sign is associated with Lenz’s law.

2. A bar of length 0.7 m slides along metal rails at a speed of 1 m/s. The bar and rails are in a magnetic field of 20 T, pointing out into the page. Calculate the motional emf.
a) 0.7 V
b) 7 V
c) 14 V
d) 1.4 V
Answer: c
Clarification: Length (L) = 0.7 m; Speed (v) = 1 m/s; Magnetic field (B) = 20 T
The required equation E = -vLB (The negative sign only applies to the direction)
E = 1 × 0.7 × 20
E = 0.7 × 20
E = 14 V
Therefore, the motional emf in the bar and rails is 14 V.

3. A bar of length 0.15 m slides along metal rails at a speed of 5 m/s. The bar and rails are in a magnetic field of 40 T, pointing out into the page. The resistance of two resistors in parallel is both 20 Ω, and the resistance of the bar is 5 Ω. What is the current in the bar?
a) 1 A
b) 2 A
c) 3 A
d) 5 A
Answer: b
Clarification: L = 0.15 m; B = 40 T; v = 5 m/s; R1 = 10 Ω; R2 = 10 Ω; R3 = 5 Ω
Emf (E) = vLB = 5 × 0.15 × 40
E = 5 × 6
E = 30 V
R1 and R2 are in parallel ➔ (frac {1}{R} = frac {1}{R1} + frac {1}{R2} = frac {1}{20} + frac {1}{20} = frac {2}{20} = frac {1}{10}) ➔ R = 10 Ω
R_TOT = R + R3 = 10 + 5 = 15 Ω
Therefore, current (I) = (frac {E}{R})
I = (frac {30}{15})
I = 2 A

4. Induced emf and motional emf are exactly the same.
a) True
b) False
Answer: b
Clarification: No. they are not exactly the same. According to Faraday’s Law, an induced emf is created whenever there’s a changing magnetic flux through a loop. If the changing emf is due to some kind motion of a conductor in a magnetic field, then it would be called as motional emf.

5. A bar of length 2m is said to fall freely in a magnetic field of magnitude 50 T. What is the motional emf in the bar when it has fallen 40 meters?
a) 700 V
b) 2100 V
c) 2800 V
d) 1400 V
Answer: c
Clarification: Given: L = 2 m; B = 50 T;
The bar is said to be falling freely, so ➔ u = 0
According to Newton’s third equation ➔ v² – u² = 2gh
v² = 2gh; h = 40 m
v = (sqrt{(2 × 9.8 × 40)})
v = 28 m/s
Motional emf (E) = vLB
E = 28 × 2 × 50
E = 2800 V

6. A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end across a magnetic field (B) of 0.5 T, pointing out of the page. The rod is of length 45 cm and the speed of the rod is 70 cm/s. The rod has a resistance of 10 Ω and the resistance of the rails and connector is negligible. What is the rate at which energy is being transferred to thermal energy?
a) 0.225 W
b) 22.55 W
c) 2.25 × 10^-4 W
d) 2.25 × 10^-3 W
Answer: d
Clarification: Given: B = 0.5 T; v = 70 cm/s = 70 × 10^-2 m/s; L = 45 cm = 45 × 10^-2 m; R = 10 Ω
Motional emf (E) = vLB = 70 × 10^-2 × 45 × 10^-2 × 0.5
E = 0.15 V
Current (I) = (frac {E}{R})
I = (frac {0.15}{10})
I = 0.015 A
Rate of energy or power (P) = I²R
P = 0.015² × 10
P = 2.25 × 10^-3 W
Therefore, the rate of energy transfer is 2.25 × 10^-3 W.

7. Find the true statement.
a) Motional emf is inversely proportional to speed of electric conductor
b) Motional emf is inversely proportion to the length of the conductor
c) Motional emf is directly proportional to the magnetic field
d) Motional emf is inversely proportional to the magnetic field
Answer: c
Clarification: Motional emf is directly proportional to the magnetic field across which the electric conductor moves. It is also directly proportional to the (frac {velocity}{speed}) of the electric conductor as well as to the length of the conductor. All the other statements are not valid.

Physics Multiple Choice Questions on “Ray Optics – Total Internal Reflection”.

1. What causes haloes (rings) around the sun or the moon?
a) Total internal reflection
b) Refraction of light
c) Reflection of light
d) Dispersion
Answer: c
Clarification: When the sun or the moon is seen through a thin veil of high clouds, haloes, i.e. rings, are seen. These are formed due to the reflection of light by the icy crystals present in the atmosphere. So, this is the cause of rings around the sun or the moon.

2. Why is the sequence of colors in the secondary rainbow reverse of that in the primary rainbow?
a) Refraction of light
b) Two internal reflection
c) Reflection of light
d) Dispersion
Answer: b
Clarification: The sequence of colors in the secondary rainbow is the reverse of that in the primary rainbow because a secondary rainbow is formed by two internal reflections of light in water droplets while a primary rainbow is formed by just one total internal reflection.

3. Which of the following is a necessary condition for total internal reflection?
a) The angle of incidence in the denser medium must be greater than the critical angle for the two media
b) The angle of incidence in the rarer medium must be greater than the critical angle for the two media
c) The angle of incidence in the denser medium must be lesser than the critical angle for the two media
d) The angle of reflection in the denser medium must be greater than the critical angle for the two media
Answer: a
Clarification: The necessary conditions for total internal reflection are ➔ the light must travel from an optically denser to an optically rarer medium and the angle of incidence in the denser medium must be greater than the critical angle for the two media.

4. Which of the following color is the ocean?
a) Black
b) Orange
c) Blue
d) Red
Answer: c
Clarification: The Ocean is blue because water absorbs colors in the red part of the light spectrum. Like a filter, this leaves behind colors in the blue part of the light spectrum for us to see. So, the blue color of the ocean is due to the preferential scattering of light by water molecules.

5. The secondary rainbow is brighter than the primary rainbow.
a) True
b) False
Answer: b
Clarification: Light suffers two total internal reflections in the formation of the secondary rainbow. So more light intensity is absorbed. Therefore, the secondary rainbow will be fainter compared to the primary rainbow and the latter will be brighter.

6. ‘X’ is an optical illusion observed in deserts or over hot extended surfaces like a coal-tarred road, due to which a traveler sees a shimmering pond of water some distance ahead of him and in which the surrounding objects like trees appear inverted. Identify X.
a) Mileage
b) Mirage
c) Optical activity
d) Total internal reflection
Answer: b
Clarification: Mirage is an optical illusion observed in deserts or over hot extended surfaces like a coal-tarred road, due to which a traveler sees a shimmering pond of water some distance ahead of him and in which the surrounding objects like trees appear inverted.

7. Identify the principle behind the sparkling of diamonds.
a) Total internal reflection
b) Refraction
c) Reflection
d) Optical activity
Answer: a
Clarification: The faces of a diamond are cut suitably so that light entering it suffers total internal reflections repeatedly and gets collected inside but it comes out through only a few faces. Diamond sparkles when seen in the direction of emerging light.

Physics Assessment Questions for IIT JEE Exam on “Wave Nature of Matter”.

1. What type of nature do electromagnetic waves have?
a) Dual nature
b) Wave nature
c) Particle nature
d) Photon nature
Answer: a
Clarification: 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. The magnitude of which of the following is proportional to the frequency of the wave?
a) Electrons
b) Neutrons
c) Photons
d) Protons
Answer: c
Clarification: The energy conveyed by an electromagnetic wave is always carried in packets whose magnitude is proportional to the frequency of the wave. These packets of energy are known as photons. The energy of a photon is given as:
E = hv
Where h is the Planck’s constant and v is the frequency of the wave.

3. Identify the de – Broglie expression from the following.
a) λ=h×p
b) λ=(frac {h}{p})
c) λ=h+p
d) λ=h-p
Answer: b
Clarification: de – Broglie equation states that matter can act as waves as well as particles. So, the de Broglie equation helps us understand the concept of matter having a wavelength. The expression for de – Broglie wavelength is given as:
λ=(frac {h}{p} = frac {h}{mv})
Where h is the Planck’s constant; p is the momentum; m is the mass of the particle and v is the velocity of the particle.

4. When the wavelength of an electron increases, the velocity of the electron will also increase.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. According to the de – Broglie equation, the velocity of the particle and the de – Broglie are inversely proportional to each other. Therefore, when the wavelength of an electron increases, the velocity of the electron decreases.

5. The sun gives light at the rate of 1500 W/m² of area perpendicular to the direction of light. Assume the wavelength of light as 5000Å. Calculate the number of photons/s arriving at 1 m² area at that part of the earth.
a) 4.770 × 10²¹
b) 3.770 × 10¹¹
c) 3.770 × 10²¹
d) 3.770 × 10²⁰
Answer: c
Clarification: Given: I = 1500 W/m²; Wavelength = 5000Å
Required equation ➔ E=hv=(frac {hc}{lambda })
Speed of light (c) = 3 × 10⁸ m/s
Number of photons/s received = n = (frac {IA}{E} = frac {(1500 times 1) times (5000 times 10^{-10})}{6.63 times 10^{-34} times 3 times 10^8} )
n = 3.770 × 10²¹
Therefore, the number of electrons received per second is 3.770 × 10²¹.

6. What is the de – Broglie wavelength associated with an electron, accelerated through a potential difference of 200 volts?
a) 1 nm
b) 0.5 nm
c) 0.0056 nm
d) 0.086 nm
Answer: d
Clarification: Given: Potential difference (V) = 200 V
The de – Broglie wavelength is given as:
λ=(frac {h}{p} = frac {h}{mv} = frac {1.227}{sqrt {V}}) nm
λ=(frac {1.227}{sqrt {200}})
λ=0.086 nm

7. What is the de – Broglie wavelength of a proton accelerated through a potential difference of 2 kV?
a) 0.65 × 10^-13 m
b) 0.65 × 10^-10 m
c) 0.65 × 10^-11 m
d) 0.65 × 10^-20 m
Answer: b
Clarification: Given: charge of proton = 1.6 × 10^-19; mass of proton = 1.6 × 10^-27; V = 2 kV
λ=(frac {h}{p} = frac {h}{sqrt {2mE}} = frac {h}{sqrt {2mqV}})
λ=(frac {6.6 times 10^{-34}}{sqrt {2times (1.6 times 10^{-27})times (1.6 times 10^{-19}) times 2000}})
λ=0.65 × 10^-12 m
Therefore, the wavelength is 0.65 × 10^-12 m.

8. While comparing the alpha particle, neutron, and beta particle, the alpha particle has the lowest de – Broglie wavelength.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. In comparison with beta particle and neutron, the alpha particle has a higher mass, followed by neutron and then beta particle. According to de – Broglie wavelength equation, the wavelength is inversely proportional to the mass. Hence, the alpha particle has the lowest wavelength since it has the highest mass.

9. What is the de – Broglie wavelength of a ball of mass 150 g moving at a speed of 50 m/s?
a) 8.8 × 10^-34 m
b) 8.8 × 10^-30 m
c) 8.8 × 10^-25 m
d) 8.8 × 10^-35 m
Answer: d
Clarification: Given: m = 150 g; v = 50 m/s
The required equation ➔ λ=(frac {h}{p} = frac {h}{mv})
λ=(frac {6.6 times 10^{-34}}{150 times 10^{-3} times 50})
λ=8.8 × 10^-35 m

10. What will be the de – Broglie wavelength when the kinetic energy of the electron increases by 5 times?
a) √5
b) 5
c) (frac {1}{sqrt {5}})
d) (frac {1}{5})
Answer: c
Clarification: The required equation ➔ λ=(frac {h}{mv} = frac {h}{sqrt {2mK}} )
Where h is the Planck’s constant, m is the mass of the electron and K is the kinetic energy of the electron.
Since the mass of the electron remain unchanged, the wavelength will be inversely proportioned to the kinetic energy, so,
(frac {lambda }{lambda^{‘}} = sqrt {frac {K^{‘}}{K} } = sqrt { frac {5K}{K} } ) = √5
Therefore, λ’=(frac {lambda }{sqrt {5}})
Hence, the wavelength is reduced by a factor of √5.

Physics Assessment Questions for IIT JEE Exam,

Physics Multiple Choice Questions on “Semiconductor Electronics – Special Purpose p-n Junction Diodes”.

1. Which of the following is operated in forward bias?
a) LED
b) Zener diode
c) Photodiode
d) Solar cell
Answer: a
Clarification: A light-emitting diode (LED) converts electric energy into light energy. A LED is a heavily doped p-n junction which under forward bias emits spontaneous radiation. The semiconductor used for the fabrication of visible LEDs must at least have a bandgap of 1.8 eV.

2. Which of the following converts light energy to electric current?
a) LED
b) Zener diode
c) Photodiode
d) Solar cell
Answer: c
Clarification: A photodiode is a semiconductor device, which converts light energy to electric energy. It is a special type p-n junction diode fabricated with a transparent window to allow light to fall on the diode. Photodiodes are operated in reverse bias.

3. Identify the one on which no external bias is applied.
a) Zener diode
b) Solar cell
c) Photodiode
d) Light-emitting diode
Answer: b
Clarification: Solar cells convert solar energy into electric energy. A solar cell is a p-n junction that generates emf when solar radiation falls on the p-n junction. It works on the same principle as the photodiode, except that no external bias is applied and the junction area is kept large.

4. The I-V characteristics of a LED are similar to that of Si junction diode.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. The I-V characteristics of a LED are indeed similar to that of Si junction diode. But the threshold voltages are much higher and slightly different for each color. The reverse breakdown voltages of LEDs are very low, typically around 5 V.

5. Which of the following is used as a voltage regulator?
a) Photodiode
b) Solar cell
c) LED
d) Zener diode
Answer: b
Clarification: Zener diode was invented by Clarence Melvin Zener. It is designed to operate under reverse bias in the breakdown region and is used as a voltage regulator. It is designed to allow current to flow backward.