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Physics MCQs for NEET — Practice Questions with Answers

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A metal rod moves at a constant velocity in a direction perpendicular to its length & a constant uniform magnetic field too. Select the correct statement (s) from the following.

A.

The entire rod is at the same electrical potential

B.

There is an electric field in the rod

C.

The electric potential is highest at the centre of the rod.

D.

The electric potential is lowest at the centre of the rod.

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Explanation

When a metal rod moves in a direction perpendicular to its length and a uniform magnetic field, an electromotive force (emf) is induced in the rod. This creates an electric field within the rod. Hence, there is an electric field in the rod.

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Two co-axial solenoids are made by a pipe of cross sectional area $10 cm^2$ and length 20 cm If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is........

A. $ 4.8 \phi \times 10^{-4} H $ B. $ 4.8 \phi \times 10^{-5} H $ C. $ 2.4 \phi \times 10^{-4} H $ D. $ 4.8 \phi \times 10^{4} H $

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The self inductance of a straight conductor isâ€¦

A. Zero B. Very Large C. Very Small D. $ \infty $

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Explanation

The self-inductance of a straight conductor is zero because self-inductance depends on the flux linkage with the coil formed by the conductor. A straight conductor does not form a coil, and hence, there is no flux linkage, resulting in zero self-inductance.

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Two coils of self inductances 2 mH & 8 mH are placed so close togather that the effective flux in one coil is completely half with the other.The mutual inductance between these coils is.......

A. 4 mH B. 6 mH C. 2 mH D. 16 mH

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Explanation

$ L = k \sqrt { L_0 L_2 } , k = 1/2 $

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In circular coil. when no. of turns is doubled & resistance becomes half of the initial then inductance becomes

A. 4 times B. 2 times C. 8 times D. No change

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Explanation

$ L \alpha N^2 $

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A transformer of efficiency 90% draws an input power of 4 kW. An electrical applience connected across the secondary draws a current of 6 A. The impedence of device is.........

A. $ 60 \Omega $ B. $ 50 \Omega $ C. $ 80 \Omega $ D. $ 100 \Omega $

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The armature of dc motor has $ 20 \Omega $ resistance. It draws current of 1.5 A when run by 220 V dc suppy. The value of back induced in it will be

A. 150v B. 190v C. 170v D. 180v

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Explanation

The back EMF (Electromotive Force) in a DC motor can be calculated using Ohm's Law and the formula for the back EMF. The voltage drop across the armature resistance is given by $V_R = I imes R$, where $I$ is the current and $R$ is the resistance. Given that the armature resistance is 20 Î© and the current is 1.5 A, the voltage drop is $V_R = 1.5 imes 20 = 30V$. The back EMF $E_b$ can be found by subtracting this voltage drop from the supply voltage: $E_b = V_{supply} - V_R = 220V - 30V = 190V$.

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Two identical circular loops of metal wire are lying on a table near to each other without touching. Loop A carries a current which increasing with time. In response the loop B.......

A. Is repelled by loop A B. Is attracted by loop A C. rotates about its centre of mass D. remains stationary

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Explanation

When a current in loop A is increasing, it creates an increasing magnetic field. According to Lenz's Law, loop B will experience an induced current that opposes the change in magnetic flux. This induced current in loop B will create its own magnetic field, which will oppose the magnetic field of loop A. Therefore, loop B is repelled by loop A.

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A wire of length 2m is moving at a speed $2ms^{-1}$ keep its length perpendicular to uniform magnetic field of 0.5 T. The resistance of circuit joined with this wire is $6 \Omega $ . The rate at which work is being done to keep the wire moving at constant speed is ..........

A. $ { 1 \over 3 } \omega $ B. $ { 2 \over 3 } \omega $ C. $ { 1 \over 6 } \omega $ D. $ 2 \omega $

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Explanation

$ \rho = F.v. = {B^2 l^2 v^2 \over R } = {1 \over 6 } W $

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An inductor-resistor-battery circuit is switched on at t = 0. If the emf of battry is $ \varepsilon $ find the charge passes through the battery in one time constant $ \tau $

A. $ { i _ {max } \tau \over e } $ B. $ (\tau - 1 ) e i_{max} $ C. $ { i _{max} \over e } $ D. $ \tau i _{max} $

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Explanation

$ I = I_0 (1 - e^{-t /T}) (I_0 =I_{max}) $ $ Q = \int _0 ^ T (I_0 -I_0e ^{-t /T })dt = I_0 T/e $

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