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Physics NEET Questions — Free MCQ Practice

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A transformer has 100 turns in the primary coil and carries 8 A current. If input power is one kilowatt, the number of turns required in the secondary coil to have 500V output will be

A.

100

B.

200

C.

400

D.

300

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Explanation

$P_{P} = V_{P} i_{P} \Rightarrow 1000 = V_{P} \times 8 \Rightarrow V_{P} = \frac{1000}{8}$

$\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}} \Rightarrow \frac{(1000 / 8)}{500} = \frac{100}{N_{s}} \Rightarrow N_{s} = 400$

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A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω. The induced e.m.f. between the two ends is

A.

$\frac{1}{2} B ω l^{2}$

B.

$\frac{3}{4} B ω l^{2}$

C.

$B ω l^{2}$

D.

$2 B ω l^{2}$

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Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ >> R₂, the mutual inductance M between them will be directly proportional to

A.

R₁/R₂

B.

R₂/R₁

C.

$R_{1}^{2} / R_{2}$

D.

$R_{2}^{2} / R_{1}$

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Explanation

Mutual inductance between two coil in the same plane with their centers coinciding is given by

$M = \frac{μ_{0}}{4 π} (\frac{2 π^{2} R_{2}^{2} N_{1} N_{2}}{R_{1}})$ henry.

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A circular loop of radius R carrying current I lies in the x-y plane with its centre at the origin. The total magnetic flux through the x-y plane is

A.

Directly proportional to I

B.

Directly proportional to R

C.

Directly proportional to R²

D.

Zero

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Explanation

The circular loop behaves as a magnetic dipole whose one surface will be N-pole and another will be S-pole. Therefore magnetic lines of force emerge from N will meet at S. Hence total magnetic flux through x-y plane is zero.

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A small square loop of wire of side l is placed inside a large square loop of wire of side L (L > l). The loop are coplanar and their centre coincide. The mutual inductance of the system is proportional to

A.

l / L

B.

l² / L

C.

L/l

D.

L²/l

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A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0 so that a time-dependent current I₁(t) starts flowing through the coil. If I₂(t) is the current induced in the ring and B(t) is the magnetic field at the axis of the coil due to I₁(t), then as a function of time (t > 0), the product I₂ (t) B(t)

A.

Increases with time

B.

Decreases with time

C.

Does not vary with time

D.

Passes through a maximum

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A long solenoid of diameter 0.1m has 2 $\times 10^{4}$ turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01m is placed with its axis coinciding with the solenoid's axis. The current in the solenoid reduces at a constant rate to 0 A from 4A in 0.05s. If the resistance of the coil is $10 π^{2} Ω$ , the total charge flowing through the coil during this time is

A.

32 $πμC$

B.

16 $μ C$

C.

32 $μ C$

D.

16 $πμC$

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Explanation

(c) Current induced in the coil given by

$i = \frac{1}{R} (\frac{d ϕ}{d t})$

$\Rightarrow \frac{∆ q}{∆ t} = \frac{1}{R} (\frac{∆ ϕ}{∆ t})$

Given, the resistance of the solenoid,

R=10 $π^{2} Ω$

The radius of the second coil r= $10^{- 2}$

$∆ t = 0.05 s, ∆ i = 4 - 0 = 4 A$

The charge flowing through the coil is given by

$∆ q = (\frac{∆ ϕ}{∆ t}) \frac{1}{R} (∆ t)$

= $μ_{0} N_{1} N_{2} π r^{2} (\frac{∆ i}{∆ t}) \frac{1}{R} ∆ t$

= $4 π \times 10^{- 7} \times 2 \times 10^{4} \times 100 \times π$

$\times {(10^{- 2})}^{2} \times (\frac{4}{0.05}) \times \frac{1}{10 π^{2}} \times 0.05$

= $32 \times 10^{- 6} C = 32 μ C$

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A long solenoid has 1000 turns. When a current of 4.0 A flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{- 3}$ Wb. The self-inductance of the solenoid is-

A.

3H

B.

2H

C.

1H

D.

4H

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Explanation

(c) Given, Number of turns of solenoid, N=1000.

Current, I=4A

Magnetic flux, $ϕ_{B} = 4 \times 10^{- 3} Wb$

so, Self induction of solenoid is given by

$L = \frac{ϕ_{B} . N}{I}$ ...(1)

Substitue the given values in equation (1), we get

$L = \frac{4 \times 10^{- 3} \times 1000}{4} = 1 H$

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

A.

300V,15A

B.

450V,15A

C.

450V,13.5A

D.

600V,15A

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Explanation

Initial power=3000W

As efficiency is 90% then final power

=3000x90/100=2700W

=>V₁I₁=3000W]

V₁I₁=2700W] ...(i)

So, V₂=2700/6=900/2=450V and I₁=3000/200=15A

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A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is

A.

once per revolution

B.

twice per revolution

C.

four times per revolution

D.

six times per revolution

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Explanation

(b)

Frequency of change in direction of emf is double the frequency of rotation

Physics MCQs for NEET — Practice Questions with Answers

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