A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tub is dipped…

Q: A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tub is dipped vertically in water so that half of its length is in water. The fundamental frequency of the air column is now

(a) f /2

(b) 3 f /4

(c) f

(d) 2 f

Ans: (c)

Sol: $\large f_o = \frac{v}{2l_o}$

and , $\large f_c = \frac{v}{4l_c}$

According to question , tube is half dipped in water ,

$\large l_c = \frac{l_o}{2}$

$\large f_c = \frac{v}{4(\frac{l_o}{2})}$

$\large f_c = \frac{v}{2l_o} = f_o = f $