Question.
A submarine emits a sonar pulse, which returns from an underwater cliff in $1.02 \mathrm{~s}$. If the speed of sound in salt water is $1531 \mathrm{~ms}^{-1}$, how far away is the cliff ?
A submarine emits a sonar pulse, which returns from an underwater cliff in $1.02 \mathrm{~s}$. If the speed of sound in salt water is $1531 \mathrm{~ms}^{-1}$, how far away is the cliff ?
Solution:
Given, time taken by the sonar pulse, $\mathrm{t}=1.02 \mathrm{~s}$; speed of sound in salt water, $\mathrm{v}=1531 \mathrm{~m} \mathrm{~s}^{-}$ 1 ;
distance of cliff, $\mathrm{s}=?$
Now, $s=\frac{v}{2}=\frac{1531 \quad 1.02}{2}=780.81 \mathrm{~m}$
Given, time taken by the sonar pulse, $\mathrm{t}=1.02 \mathrm{~s}$; speed of sound in salt water, $\mathrm{v}=1531 \mathrm{~m} \mathrm{~s}^{-}$ 1 ;
distance of cliff, $\mathrm{s}=?$
Now, $s=\frac{v}{2}=\frac{1531 \quad 1.02}{2}=780.81 \mathrm{~m}$
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