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Physics NEET Questions — Free MCQ Practice

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A long solenoid having 1000 turns per cm is carrying alternating current of one ampere peak value. A search coil of area of cross-section $1 \times 10^{- 4} m^{2}$ and of 20 turns is placed in the middle of the solenoid so that its plane is perpendicular to the axis of the solenoid. The search coil registers a peak voltage $2.5 \times 10^{- 2} V$ . The frequency of the current in the solenoid is -

A.

1.6 per second

B.

0.16 per second

C.

15.9 per second

D.

15.85 per second

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Explanation

$4 . Flux linked with the search coil ϕ = {BAN}_{s} = μ_{0} {niAN}_{s} ∴ \frac{dϕ}{dt} = μ_{0} {nAN}_{s} \frac{di}{dt} i = i_{0} \sin ωt ∴ \frac{dϕ}{dt} = μ_{0} {nAN}_{s} i_{0} ω \cos ωt E_{\max} = {(\frac{dϕ}{dt})}_{\max} = μ_{0} {nAN}_{s} i_{0} ω ∴ f = \frac{ω}{2 π} = \frac{E_{\max}}{2 {πμ}_{0} {nAN}_{s} i_{0}} f = \frac{2.5 \times 10^{- 2}}{6.28 \times 12.56 \times 10^{- 7} \times 10^{5} \times 10^{- 4} \times 20 \times 1} = 15.85 s^{- 1}$

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A coil of area $7 {cm}^{2}$ and of 50 turns is kept with its plane normal to a magnetic field B. A resistance of 30 ohm is connected to the resistance-less coil. B is 75 exp (– 200t) gauss. The current passing through the resistance at t = 5 ms will be-

A.

$0.64 mA$

B.

$1.05 mA$

C.

$1.75 mA$

D.

$14 2.60 mA$

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Explanation

$1 . E = - NA \frac{dB}{dt} ∴ i = \frac{E}{R} = \frac{NA}{R} \frac{d}{dt} (75 e^{- 200 t}) \times 10^{- 4} = \frac{NA}{R} [75 \times (- 200) e^{- 200 t}] \times 10^{- 4} = + \frac{50 \times 7 \times 10^{- 8}}{30} [15000 e^{- 1}] = \frac{175 \times 10^{- 5}}{e} = \frac{175 \times 10^{- 5}}{2.73} = 0.64 \times 10^{- 3} A i = 0.64 mA$

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The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance :

A.

0.138 H

B.

138.88 H

C.

1.389 H

D.

13.89 H

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Explanation

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A transformer has an efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A the voltage across the secondary coil and the current in the primary coil respectively are:

A.

300 V, 15 A

B.

450 V, 15 A

C.

450 V, 13.5 A

D.

600 V, 15 A

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Explanation

Current in primary coil = P/V = 3000/200 = 15A

efficiency is 90%

So output power is 90% of input

P_o = (90/100)x3000

If output voltage and current - V_oI_o

V_oI_o =(90/100)x3000

Vox6=(90/100)x3000

Vo = 450 V

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A coil of resistance 400 $Ω$ is placed in a magnetic field. If the magnetic flux $ϕ$ (Wb) linked with the coil varies with time t (sec) as $ϕ$ = 50 $t^{2}$ +4.
The current in the coil at t= 2s is:

A.

0.5 A

B.

0.1 A

C.

2 A

D.

1 A

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A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 ms^-1. The induced emf when the radius is 2cm is:

A.

2 $π μ$ V

B.

$π μ$ V

C.

$\frac{π}{2} μ V$

D.

2 $μ$ V

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Explanation

According to the formula of Magnetic flux
Magnetic flux $ϕ$ =B.A and A= $π r^{2}$
It can be also written as = B. $π r^{2}$
Therefore Induced emf can be found by using the relation magnetic flux as follows-
$|e| = \frac{d ϕ}{d t} = B (2 π r) \frac{d r}{d t}$
Now we put the given value in the equation
= 0.025 x $π$ x 2 x 2 10^-2 x 1 x 10^-3 = $π μ V$

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A circular disc of radius 0.2 m is placed in a uniform

magnetic field of induction

\frac{1}{π} (\frac{W b}{m^{2}})

in

such a way that its axis makes an angle

of

60^{0}

with

\vec{B}

. The magnetic flux linked

with the disc is:

A.

0.02 Wb

B.

0.06 Wb

C.

0.08 Wb

D.

0.01 Wb

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The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux Ï• linked with the primary coil is given by Ï•= $ϕ_{0}$ +4t, where Ï• is in weber, t is time in second and Ï•₀is a constant, the output voltage across the secondary coil is:

A.

90 V

B.

120 V

C.

220 V

D.

30 V

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Explanation

The mutual inductance between primary and secondary coils of a transformer is given by M = N1N2/R, where N1 and N2 are the number of turns in primary and secondary coils respectively, and R is a constant. The induced emf in the secondary coil is given by e = -MdÏ•/dt = -M(d/dt)(Ï•0 + 4t) = -4MN, where N is the number of turns per unit time. Substituting the given values, we get the output voltage across the secondary coil as 120 V.

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Two coils of self-inductance 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

A.

10 mH

B.

6 mH

C.

4 mH

D.

16 mH

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Explanation

Given,
Self-inductance of coil 1 = 2 mH
Self-inductance of coil 2 = 8 mH
When the total flux associated with one coil links with the other i.e., a case of maximum flux linkage, then
M₁₂ = $\frac{N_{2} ϕ_{B_{2}}}{i_{1}}$ and M₂₁ = $\frac{N_{1} ϕ B_{1}}{i_{2}}$
Similarly, L₁ = $\frac{N_{1} ϕ_{B_{1}}}{i_{1}}$ and L₂= $\frac{N_{2} ϕ_{B_{2}}}{i_{2}}$
If all the flux of coil 2 links coil 1 and vice-versa then
$ϕ_{B_{2}}$ = $ϕ_{B_{1}}$
Since, M₁₂ = M₂₁ = M, hence we have
M₁₂M₂₁=M² = $\frac{N_{1} N_{2} ϕ_{B_{1}} ϕ_{B_{2}}}{i_{1} i_{2}}$ = L₁L₂
∴ M_max = $\sqrt{L_{1} L_{2}}$
Given, L₁ = 2 mH, L₂ = 8 mH
∴ M_max = $\sqrt{2 \times 8} = \sqrt{16}$ = 4 mH

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In which of the following devices, the eddy current effect is not used?

A.

electric heater

B.

induction furnace

C.

magnetic braking in train

D.

electromagnet

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Explanation

Electric heater does not involve Eddy currents. It uses Joule's heating effect and hence (1) will be the correct answer.

Physics MCQs for NEET — Practice Questions with Answers

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