Question

A fish rising vertically up towards the surface of water at a speed of $3m{s}^{-1}$ observes a bird diving vertically downwards towards it at a speed $9m{s}^{-1}$. The real speed of bird will be:
(Given ${\mu }_w=\frac{4}{3}$)

• A. $5m{s}^{-1}$
• B. $\frac{9}{2}m{s}^{-1}$
• C. $3m{s}^{-1}$
• D. $\frac{7}{2}m{s}^{-1}$

Answer 1

The correct option is B $\frac{9}{2}m{s}^{-1}$

Let at a given instant the fish is at distance 'y' from water surface and bird is at a real distance of 'x' from water-air interface.

Since the observer fish is in denser medium, bird will appear to it at a further distance.
$\Rightarrow d_{app}=y+{\mu }_{w}x$
on differentiating both sides w.r.t time:
$\frac{d}{dt}\left(d_{app}\right)=\left(\frac{dy}{dt}\right)+{\mu }_w\left(\frac{dx}{dt}\right)$
$\Rightarrow {\left(v_{app}\right)}_B={\left(v_{real}\right)}_F+{\mu }_w{\left(v_{real}\right)}_B$
$[\because {\left(v_{app}\right)}_B=9m{s}^{-1},{\left(v_{real}\right)}_F=3m{s}^{-1}]$
$\Rightarrow 9=3+\frac{4}{3}{\left(v_{real}\right)}_B$
$\therefore {\left(v_{real}\right)}_B=\frac{18}{4}=\frac{9}{2}m{s}^{-1}$

Why this question? It challenges you to use the tool of differentiation in order to estimate ${}^{\prime }{v}_{real}^{\prime }$ from concept of apparent depth.