Question

A bird in air is diving vertically downward over a tank with speed 5 $cm/s$ while base of tank is silvered. A fish in the tank is rising upward along the same line with speed 2 $cm/s$. Water level is falling at rate of 2 $cm/s$. Take ${\mu }_{water}=4/3$.

• A. Speed of image of fish as seen by bird directly is 6 $cm/s$
• B. Speed of image of fish as seen by bird directly is 3 $cm/s$
• C. Speed of image of fish as seen by bird after reflection from mirror is 3 $cm/s$
• D. Sped of image of fish as seen by bird after reflection from mirror is 6 $cm/s$

Answer 1

Answer: (A), (C)

The correct options are
A Speed of image of fish as seen by bird directly is 6 $cm/s$
C Speed of image of fish as seen by bird after reflection from mirror is 3 $cm/s$



For direct image:
$h_{app}=x+\frac{3}{4}y$ ..........(1)
By taking differentiation of equ. (1)
$\Rightarrow \frac{d{h}_{app}}{dt}=\frac{dx}{dt}+\frac{3}{4}(\frac{dy}{dt})$
$=-5+2+\frac{3}{4}(-2-2)=-6cm/s$
After reflection from mirror;
$h_{app}=x+\frac{3}{4}(y+2z)$
$\Rightarrow \frac{d{h}_{app}}{dt}=\frac{dx}{dt}+\frac{3}{4}(\frac{dy}{dt})+\frac{3}{2}\frac{dz}{dt}$
$=-5+2+\frac{3}{4}(-2-2)+\frac{3}{2}\left(2\right)=-3cm/s$