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Physics MCQs for NEET — Practice Questions with Answers

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Assertion – Reason type questions : For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. Statement – 1 : If wave enters from one medium to another medium then sum of amplitudes of reflected wave and transmitted wave is equal to the amplitude of incident wave. Statement – 2 : If wave enters from one medium to another medium some part of energy is transmitted and rest of the energy is reflected back.

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A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second is heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is $320 ms– 1$ The frequency of the string vibrating in its 1st overtone is …… hz

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Explanation

Since the beat frequency is 8, the frequency of the string vibrating in its first Overtone is 192 Hz or 208 Hz Where for 1st Overtone frequency $ f_1 = { 1 \over l } \sqrt { T \over m} ....(1) $ It is given that the beat frequency decreases if the tension in the string is decreased. $ \therefore f_1 ' \gt f_1$ Hence $f_1 ' = 208Hz and not 192Hz$

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A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second is heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is $320 ms– 1$ The tension in the string is very nearly equal to ……

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Explanation

substituting the values of l.m and $f_1 '$ in equation 1 we get T = 27.04 N

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Standing waves are produced by the superposition of two waves
$y_1 = 0.05Sin(3pt – 2x) and y_2 = 0.05Sin(3pt + 2x) $where x and y are in meters and t is in seconds. The speed $( in ms^{– 1} ) $ of each wave is ……

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Explanation

$ { 2 \pi \over \lambda } = 2 \Rightarrow \lambda = \pi m $ $ { 2 \pi f \over \lambda } = 3 \pi \Rightarrow \nu = { 3 \lambda \over 2 } ms^{-1} $

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Standing waves are produced by the superposition of two waves
$y_1 = 0.05Sin(3pt – 2x) and y_2 = 0.05Sin(3pt + 2x) $where x and y are in meters and t is in seconds. The distance ( in meters ) between two consecutive nodes is …….

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Explanation

Distance between two consecutive nodes = $ { \lambda \over 2 } = { \pi \over 2 } m $

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Standing waves are produced by the superposition of two waves
$y_1 = 0.05Sin(3pt – 2x) and y_2 = 0.05Sin(3pt + 2x) $where x and y are in meters and t is in seconds. The amplitude of a particle at x = 0.5 m is

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Explanation

The resultant displacement is given by, $ y = 0.1 cos 2x sin3 \pi t Or y = A sin 3 \pi t$ Where Ais the Amplitude of standing waves given by 0.1 cos 2x $ At x = 0.5m, cos 2x = cos (1rad) = cos \left ( {\pi \over 3.14} \right ) = cos 57.3 ^\circ = 0.054 m $ $ Amplitude A at (x = 0.5 m ) = 0.1 \times 0.54 =0.54 m $

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Standing waves are produced by the superposition of two waves
$y_1 = 0.05Sin(3pt – 2x) and y_2 = 0.05Sin(3pt + 2x) $where x and y are in meters and t is in seconds. The velocity $( in ms^{– 1} )$ of a particle at x = 0.25 m at t = 0.5 s is

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Explanation

$ Particle velocity \nu = { dy \over dt} = { d \over dt } ( 0.1 cos 2x sin 3 \pi t ) = 0.1 \times 3 \pi cos 2 x sin 3 \pi t $ $ at x = 0.25 m and t =0.5 s , v =0 $

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When two sound waves travel in the same direction in a medium, the displacement of a particle located at x at time t is given by $y_1 = 0.05Cos(0.50px - 100pt ) & y_2 = 0.05Cos ( 0.46px - 92pt )$, where $y_1, y_2$ and x are in meter and t is in seconds. What is the speed of sound in the medium?

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Explanation

The two displacements can be written as $y_1 = A cos (k_1x - \omega_1t)$ and $ y_2 = A cos (k_2x - \omega_ 2 t)$ compare this equation with given equation and get solution.

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When two sound waves travel in the same direction in a medium, the displacement of a particle located at x at time t is given by $y_1 = 0.05Cos(0.50px - 100pt ) & y_2 = 0.05Cos ( 0.46px - 92pt )$, where $y_1, y_2$ and x are in meter and t is in seconds. How many times per second does an observer hear the sound of maximum intensity?

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Explanation

$ Beat frequency f_1 - f_2 = { \omega_1 \over 2 \pi } - { \omega_2 \over 2 \pi } $

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When two sound waves travel in the same direction in a medium, the displacement of a particle located at x at time t is given by $y_1 = 0.05Cos(0.50px - 100pt ) & y_2 = 0.05Cos ( 0.46px - 92pt )$, where $y_1, y_2$ and x are in meter and t is in seconds. At x = 0, how many times between t = 0 and t = 1 s does the resultant displacement become zero?

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Explanation

The resultant displacement is given by $y = y_1 + y_2$ $ = A cos ( k_1x - \omega_1t) + A cos (k_2x - \omega_2t) $ For x = 0 we have $ y =A cos \omega_1 t + A cos \omega_2 t $ $ \therefore y = 0.10 cos (96 \pi t) cos (4 \pi t) $ Between t = 0 and t = 1 s, $Cos 96 \pi t $ becomes zero 96times and $cos 4 \pi t$ becomes zero 4 times. Hence the resultant displacement Y at x = 0 becomes zero 100 times between t = 0 and t = 15.

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