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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: When a sound source moves towards observer, then frequency of sound increases. Statement – 2 : Wavelength of sound in a medium moving towards the observer decreases.
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: Newton’s equation for speed of sound was found wrong because he assumed the process to be isothermal. Statement – 2 : When sound propagates, the compressions and rarefactions happen so rapidly that there is not enough time for heat to be distributed.
Newton initially assumed that the propagation of sound in air was an isothermal process (constant temperature). However, it was later found that the process is actually adiabatic (no heat exchange due to rapid compression and rarefaction). The rapid compressions and rarefactions in sound waves do not allow enough time for heat to be distributed, thus making the process adiabatic. Therefore, both statements are true, and statement 2 correctly explains why Newton's assumption was incorrect.
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 : When pressure in a gas changes, velocity of sound in gas may change. Statement – 2 : Velocity of sound is directly proportional to square root of pressure.
The velocity of sound in a gas is given by the formula \( v = \sqrt{\frac{\gamma P}{\rho}} \), where \( v \) is the velocity, \( \gamma \) is the adiabatic index, \( P \) is the pressure, and \( \rho \) is the density. When the pressure in a gas changes, the velocity of sound may change, but this change is not directly proportional to the square root of pressure alone; it also depends on the density of the gas. Hence, statement 1 is true, statement 2 is true, but statement 2 is not the correct explanation of statement 1.
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.
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
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 $f_1 ' = 208Hz and not 192Hz$
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 ……
substituting the values of l.m and $f_1 '$ in equation 1 we get T = 27.04 N
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 ……
$ { 2 \pi \over \lambda } = 2 \Rightarrow \lambda = \pi m $ $ { 2 \pi f \over \lambda } = 3 \pi \Rightarrow \nu = { 3 \lambda \over 2 } ms^{-1} $
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 …….
Distance between two consecutive nodes = $ { \lambda \over 2 } = { \pi \over 2 } m $
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
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 $
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
$ 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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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.