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  2. Stuck here, help me understand: - Oscillations and Waves - JEE Main-6
# Engineering

The temperature at which the speed of the sound in air becomes double of its value at 0\degree C is

  • Option 1)

    273K

  • Option 2)

    546K

  • Option 3)

    1092K

  • Option 4)

    0K

 

Answers (1)
D Divya Saini

v \alpha \sqrt T \Rightarrow \sqrt{\frac{T_{2}}{T_{1}}} = \frac{v_{2}}{v_{1}} \Rightarrow T_{2} = T_{1}\left [\left ( \frac{v_{2}}{v_{1}} \right )^{2} \right ]\Rightarrow T_{2} = 273 X 4 = 1092 K

 

Effect of temperature on speed of sound -

\because \frac{P}{\rho }= \frac{RT}{M}

\therefore V= \sqrt{\frac{\gamma RT}{M}}

 

- wherein

V \alpha \sqrt{T}

T is in Kelvin

So velocity increses with increase in temperature.

 

 

 

 


Option 1)

273K

This is incorrect.

Option 2)

546K

This is incorrect.

Option 3)

1092K

This is correct.

Option 4)

0K

This is incorrect.

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