Question

One end of a thin metal tube is closed by thin diaphragm of latex and the tube is lower in water with closed end downward. The tube is filled with a liquid 'x'. A plane progressive wave inside water hits the diaphragm making an angle 'q' with its normal. Assuming Snell's law to hold true for sound. Maximum angle 'q' for which sound is not transmitted through the walls of tube is (velocity of sound in liquid x = 740$\sqrt{3}$m/s and in water = 1480 m/s)

• A. $\sin^{- 1}\left( \frac{2}{3} \right)$
• B. $\sin^{- 1}\left( \frac{\sqrt{2}}{\sqrt{3}} \right)$
• C. sin^–1$\left( \frac{1}{\sqrt{3}} \right)$
• D. sin^–1$\left( \frac{1}{2} \right)$

Answer 1

Answer: (C)

sin^–1$\left( \frac{1}{\sqrt{3}} \right)$

Answer 2

Figure shows condition for just transmission of sound wave through the wall of tube.

$\frac { \sin i } { \sin r }$ = $\frac { v _ { 1 } } { v _ { 2 } }$

[v₁ = velocity of sound in water

v₂ = velocity of sound in liquid]

Ž sin i = $\frac { 1480 } { 740 \sqrt { 3 } }$ . sin(90°– q_C)

Ž i = sin^–1 $\left( \frac { 1 } { \sqrt { 3 } } \right)$