Dear Student GivenTime between transmission & detection of sonar pulse = 1 2 secSpeed of sound in salt water, v = 1530 msLet, distance of cliff be "d"then distance travelled by sound = 2dNow, we know that2d = speed of sound × time=> 2d = 1530 ×1 2=> d = 1530×1 22=> d = 918 m So, no option is correct Regards

Answer the Q no. 15

Dear Student

G i v e n T i m e b e t w e e n t r a n s m i s s i o n & d e t e c t i o n o f s o n a r p u l s e = 1.2 s e c S p e e d o f s o u n d i n s a l t w a t e r, v = 1530 \frac{m}{s} L e t, d i s \tan c e o f c l i f f b e " d " t h e n d i s \tan c e t r a v e l l e d b y s o u n d = 2 d N o w, w e k n o w t h a t 2 d = s p e e d o f s o u n d \times t i m e = > 2 d = 1530 \times 1.2 = > d = \frac{1530 \times 1.2}{2} = > d = 918 m

So, no option is correct

Regards

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