Two children are at opposite ends of metal rod. One strikes the end of the rod with big pebble. Find the ratio of time taken by the sound waves in air and in the metal to reach the second child :
(Speed of sound in rod $= 6420 m s^{- 1}$ )

221.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18.55

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

22.13

No worries! We‘ve got your back. Try BYJU‘S free classes today!

185.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $18.55$
$l =$ length of the rod
Time taken by the sound to travel distance $l$ in rod, $t_{1} = \frac{d i s t a n c e}{S p e e d o f s o u n d i n r o d}$
$t_{1} = \frac{l}{V_{1}}$
Time taken by sound to travel distance l in air
$t_{2} = \frac{l}{V_{2}}$

$\frac{t_{2}}{t_{1}} = \frac{l}{V_{2}} \times \frac{V_{1}}{l} = \frac{V_{1}}{V_{2}}$

$v_{1} = 6420 m / s (r o d), V_{2} = 346 m / s (a i r)$

$\frac{t_{2}}{t_{1}} = \frac{6420}{346} = 18.55$

Suggest Corrections

Similar questions

= 344 m s^{- 1}

= 5130 m s^{- 1}

m s^{- 1}

m s^{- 1} .

=

^{- 1}

=

^{- 1}

Join BYJU'S Learning Program