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Physics MCQs for NEET — Practice Questions with Answers

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An open pipe is in resonance in its 2^nd harmonic with tuning fork of frequency f₁. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f₁ , then again a resonance is obtained with a frequency f₂. If in this case the pipe vibrates in $n^{t h}$ harmonic, then -

n = 3, $f_{2} = \frac{3}{4} f_{1}$

n = 3, $f_{2} = \frac{5}{4} f_{1}$

n = 5, $f_{2} = \frac{5}{4} f_{1}$

n = 5, $f_{2} = \frac{3}{4} f_{1}$

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Explanation

Open pipe resonance frequency $f_{1} = \frac{2 v}{2 L}$

Closed pipe resonance frequency $f_{2} = \frac{n v}{4 L}$

$f_{2} = \frac{n}{4} f_{1}$ (where n is odd and $f_{2} > f_{1}$ )

∴ n = 5

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A string of length L and mass M hangs freely from a fixed point. Then the velocity of transverse waves along the string at a distance x from the free end is

$\sqrt{g L}$

$\sqrt{g x}$

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Explanation

The velocity of transverse waves along a stretched string depends on the tension in the string. The tension at a distance x from the free end is given by T = Mgx/L. Therefore, the velocity v = sqrt(T/Î¼) = sqrt(gx), where Î¼ is the mass per unit length of the string.

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Three waves of equal frequency having amplitudes 10 μm, 4 μm and 7 μm arrive at a given point with a successive phase difference of $\frac{π}{2}$ . The amplitude of the resulting wave in μm is given by

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Explanation

Wave 1 and 3 reach out of phase. Hence resultant phase difference between them is π.

∴ Resultant amplitude of 1 and 3 = 10 – 7 = 3 $μ m$

This wave has a phase difference of $\frac{π}{2}$ with 4 $μ m$

∴ Resultant amplitude = $\sqrt{3^{2} + 4^{2}} = 5 μ m$

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An organ pipe is closed at one end has a fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is :

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Explanation

The critical hearing frequency for a person is 20,000Hz.

If a closed pipe vibrate in N^th mode, then the frequency of vibration $n = \frac{(2 N - 1) v}{4 l} = (2 N - 1) n_{1}$

(where n₁ = fundamental frequency of vibration)

Hence, 20,000 $= (2 N - 1) \times 1500$ ⇒ $N = 7 .1 \approx 7$

Also, in a closed pipe;

Number of overtones = (No. of the mode of vibration) – 1

= 7 – 1 = 6.

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An electric dipole is in unstable equilibrium in the uniform electric field. The angle between its dipole moment and the electric field is

90 $°$

120 $°$

0 $°$

180 $°$

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Explanation

U = -PE cos $θ$

For unstable equilibrium electric dipole moment opposite to the electric field vector

i.e. $θ$ = 180 $°$

U $\to$ maximum

So $θ = π$ = 180 $°$

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A particle of mass m and charge q is at rest. If it is moved with velocity $\frac{\sqrt{3} c}{2}$ where c is speed of light in vacuum, then its charge will become/remain

$\sqrt{3} q$

$\frac{q}{2}$

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Explanation

Charged be invarient quantity.

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A charge q is to be divided on two small conducting spheres. What should be the value of charges on the spheres so that when placed at a certain distance apart, the repulsion force between them is maximum

$\frac{q}{4} a n d \frac{3 q}{4}$

$\frac{q}{2} a n d \frac{q}{2}$

$\frac{q}{3} a n d \frac{q}{3}$

$\frac{q}{4} a n d \frac{q}{4}$

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Explanation

$L e t t h e c h a r g e o n o n e s p h e r e = x T h e n, c h a r g e o n o t h e r s p h e r e = q - x F o r c e b e t w e e n t h e s p h e r e s : F = \frac{k x (q - x)}{r^{2}} = \frac{k (q x - x^{2})}{r^{2}} F o r m a x i m u m f o r c e b e t w e e n t h e s p h e r e s : \frac{d F}{d x} = 0 q - 2 x = 0 x = \frac{q}{2}$

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You are traveling in a car during a thunderstorm. In order to protect yourself from lightning, you would prefer to

Touch the nearest electric pole

Take shelter under a tree

Get out and crouch on the ground

Remain in the car

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Explanation

Due to electrostatic shielding.

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If a soap bubble is given some charge, then its radius

Increases

Decreases

Remain unchanged

May increase or decrease depending upon the give charge is positive or negative

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An electric dipole of dipole moment p is placed in an electric field of intensity E such that angle between electric field and dipole moment is $θ$ . Assuming that the potential energy of the dipole is zero when $θ = 0 °$ , the potential energy of the dipole will be

-pE cos $θ$

pE(1-cos $θ$ )

pE(cos $θ$ -1)

-2pE(cos $θ$ -1)

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Explanation

$∆ U = pE ({Cosθ}_{1} - {Cosθ}_{2}) U_{f} - U_{i} = pE ({Cosθ}_{1} - {Cosθ}_{2}) θ_{1} = 0 ° \Rightarrow U_{i} = 0 (A c c o r d i n g t o q u e s t i o n) U_{f} = pE (Cos 0 ° - Cosθ) U_{f} = pE (1 - Cosθ)$

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