Question

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

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

Answer 1

Answer: (A), (C)

Speed of the image of the fish as seen by the bird after reflection from the mirror is $3\text{cm/s}$.


For direct image, apparent depth:
$h_{app}=x+\frac{3}{4}y$
$\Rightarrow \frac{d{h}_{app}}{dt}=\frac{dx}{dt}+\frac{3}{4}\left(\frac{dy}{dt}\right)$
Here, $\frac{dx}{dt}=-v_{\text{bird}}+v_{\text{water surface}}$
and $\frac{dy}{dt}=-v_{\text{fish}}-v_{\text{water surface}}$

$\Rightarrow \frac{d{h}_{app}}{dt}=-5+2+\frac{3}{4}(-2-2)$
$=-6\text{cm/s}$

After reflection from mirror,
$h_{app}=x+\frac{3}{4}(y+2z)$
$\Rightarrow \frac{d{h}_{app}}{dt}=\frac{dx}{dt}+\frac{3}{4}\left(\frac{dy}{dt}\right)+\frac{3}{2}\frac{dz}{dt}$
$=-5+2+\frac{3}{4}(-2-2)+\frac{3}{2}\left(2\right)$
$=-3\text{cm/s}$

Option (a) and (c) are correct.