Two boats are anchored a distance 'd' apart in a sea, where seabed is at a uniform depth of D and consists of a flat rock bed. If a gun is fired at a boat, in how much time will the firing be heard at te other boat? (Given: speed of sound in air is `v_(a)`. in the water is `v_(w)` and in the rock is `v_(r)`)

Options

A. `d//v_(a)`
B. `2D//v_(w)+d//v_(r)`
C. `(dv_(w)+2Dsqrt(v_(r)^(2)-v_(w)^(2)))/(v_(r)v_(w))`
D. `(d)/(v_(r))+((2Dv_(r))/(sqrt(v_(r)^(2)-v_(w)^(2))))xx(1)/(v_(w))`

Answer

C

Explanation

method 1
`T=t_(AB)+t_(BC)+t_(CD)`
`T=(2D)/(costhetaV_(W))+((d-2Dtantheta))/(V_(R))`
`(dT)/(d theta)=0implies(2Dsecthetatantheta)/(v_(W))-(2Dsec^(2)theta)/(V_(R))`
`impliessintheta=(v_(W))/(V_(R))`
`thereforeT=(2D)/(sqrt(1-(v_(W)^(2))/(V_(R)^(2))).v_(W))+((d-2D(v_(w))/(sqrt(V_(R)^(2)-V_(W)^(2)))))/(V_(R))`
`T=(dv_(W)+2Dsqrt(v_(r)^(2)-v_(W)^(2)))/(v_(r)v_(w))`
Method 2
Assume critical angle deviation like in light `thereforesintheta(1)/(muR)=(v_(w))/(v_(R))`
Then solve `T=t_(AB)+t_(BC)+t_(CD)`.