Question

Assume velocity of sound in air as $333.68m{s}^{-1}$. A hollow tube is placed vertically in a jar containing water. Air in the tube is vibrated by correction. Second resonance is obtained when the distance of the brass tube from the surface of water is. (Frequency of source $=388Hz$)

• A. $0,215m$
• B. $0.43m$
• C. $0.645m$
• D. $0.33m$

Answer 1

The correct option is C $0.645m$

As from the equation,

$\lambda =$ distance from surface of water

Now, as we know that $v=\nu \lambda$

where $v=$ velocity of sound, $\nu =$ frequency

So, $\lambda =\frac{v}{\nu }$

So, length from the surface of water, so that the 2nd resonance is obtained is:

$l=\frac{3}{4}\lambda$

So, $l=\frac{3}{4}\frac{v}{\nu }=0.645m$