Question

A point source $\text{S}$ is placed at a height $h$ from the bottom of a vessel of height $H(<h)$. The vessel is polished at the base. Water is gradually filled in the vessel at a constant rate $\alpha {\text{m}}^3/\text{s}$. The distance $d$ of image of the source after reflection from the mirror, from the bottom of the vessel will be varies with time $t$ as :

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let the base area of the vessel be $A$.

Given,

Rate of increase of level of water $A\frac{dy}{dt}=\alpha \Rightarrow y=\frac{\alpha t}{A}$

From the diagram, we can write that,

Apparent distance of the source from the mirror :

$d=y+(h-y)\mu$

$d=h\mu -y(\mu -1)$

$d=h\mu -\frac{\alpha t}{A}(\mu -1)$

We know that, the image formed by a plane mirror is at the same distance as the distance of the object as seen from the mirror. Therefore, the image distance from the bottom of the mirror is also $d$.

Comparing this with $y=mx+c$ we get straight line graph with positive y - intercept and negative slope.

Since, water will get filled only upto $H$ after that, the water level will be constant. we can say that, $d=$ constant with time.

Hence, option (b) is the correct answer.