Question

The velocity of sound in a gas is 4 times that in air at the same temperature. When a tuning fork is sounded in a wave of frequency 480 and wavelength ${\lambda }_1$ is produced. The same fork is sounded in the gas and of ${\lambda }_2$ is the wavelength of the wave, then ${\lambda }_2/{\lambda }_1$ is :

• A. 1
• B. 2
• C. 4
• D. 1/2

Answer 1

The correct option is C 4

We have , $v=f\lambda$ ,

when tuning fork is sounded in air ,

$v_a=f{\lambda }_1$ , ........................eq1

when tuning fork is sounded in gas ,

$v_g=f{\lambda }_2$ , .......................eq2 ,

frequency will remain same as the source of sound is same ,

dividing eq2 by eq1 ,

$v_g/v_a={\lambda }_2/{\lambda }_1$ ,....................eq3

now given , $v_g=4v_a$ ,

putting , $v_g=4v_a$ in eq3,

we get , $4v_a/v_a={\lambda }_2/{\lambda }_1$ ,

or $4={\lambda }_2/{\lambda }_1$