Question

if the speed of sound in air at given pressure"V" then doubling the pressure the new speed become equal equal to
a)2V
b)0.5V
c)V
D)4V

Answer 1

Answer:

4v

Explanation:

Because the speed of sound increases with pressure.

Answer 2

Answer:

The new speed remains the same.

Explanation:

• Speed of sound in an ideal gas is given by

       $v=\sqrt{\dfrac{\gamma RT}{M} }$

• where M is the molecular mass of gas, R is the universal gas constant, T is the temperature and $\gamma$ is the adiabatic constant.
• Assuming the air as an ideal gas, from ideal gas law, we have

       $PV=nRT$                                ...(1)

• where P is the pressure, V is the volume and n is the number of moles.
• And also the density of gas is given by

        $\rho =\dfrac{M}{V}$                                       ...(2)

• Using equation (1) and (2), for n = 1, the speed of sound in air is given by
• $v=\sqrt{\dfrac{\gamma P}{\rho} }$

Given the speed of sound in air at given pressure is $v$.

The pressure is doubled, thus the new pressure is 2P.

Since the medium air is same, $\gamma$ is constant.

As we know,

$P\propto\dfrac{1}{V}~~\mbox{and}~~\rho =\dfrac{M}{V}$  

equating both, we can say for unit mass, $P\propto\rho$

So, if pressure is doubled, then density also gets doubled.

Then, the new speed is

$v=\sqrt{\dfrac{\gamma 2P}{2\rho} }=\sqrt{\dfrac{\gamma P}{\rho} }$

Therefore, the new speed remains the same.

So, the correct answer is option C.

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