Question

A container contains water up to a height of $20\text{cm}$ and there is a point at the centre of the bottom of the container. A rubber ring of radius $R$ floats centrally on the water. The ceiling of the room is $2\text{m}$ above the water surface. Then, choose the correct option(s):

• A. Maximum value of $R$ of the ring for which the ring forms a shadow on the ceiling is $22.6\text{cm}$.
• B. If $R=15\text{m}$, radius of the shadow of the ring on the ceiling is $2.8\text{m}$.
• C. Maximum value of $R$ for which the shadow of the ring forms on the ceiling is $11.3\text{cm}$.
• D. If $R=15\text{m}$, radius of the shadow of the ring on the ceiling is $1.4\text{m}$.

Answer 1

Answer: (A), (B)

If $R=15\text{m}$, radius of the shadow of the ring on the ceiling is $2.8\text{m}$.

Given:
$h=20\text{cm};H=2\text{m}$


From above figure,

$\tan i=\frac{R}{20\text{cm}}$

$\Rightarrow \tan i=\frac{15}{20}$ $(\text{For}R=15\text{cm})$

$\Rightarrow i={37}^{\circ }$

At water - air interface,

${\mu }_1\sin i={\mu }_2\sin r$
$\Rightarrow \frac{4}{3}\times \sin {37}^{\circ }=1\times \sin r$
$\Rightarrow \sin r=\frac{4}{5}\Rightarrow r={53}^{\circ }$

Again from geometry,
$\tan {53}^{\circ }=\frac{4}{3}=\frac{R^{\prime }}{2\text{m}}$

$\Rightarrow R^{\prime }=2.67\text{m}$

So, radius of the shadow for $R=15\text{m}$

$=R+R^{\prime }=2.67+0.15=2.82\approx 2.8\text{m}$

For the maximum value of radius of ring, the boundary of the ring should overlap the circle of illuminance. Circle of illuminance is the circle formed by collection of all the rays coming out of the water.


For outer most rays on the circle of illuminance,
$\sin {\theta }_c=\left(\frac{{\mu }_a}{{\mu }_w}\right)$

$\Rightarrow \sin {\theta }_c=\left(\frac{1}{4/3}\right)$

$\Rightarrow \sin {\theta }_c=\left(\frac{3}{4}\right)$

From geometry,

$\tan {\theta }_c=\frac{R_i}{h}$

where , $R_i$ is the radius of circle of illuminance.

$\Rightarrow \frac{\sin {\theta }_c}{\sqrt{1-{\sin }^2{\theta }_c}}=\frac{R_i}{20\text{cm}}$

$\Rightarrow \frac{R_i}{20}=\frac{\frac{3}{4}}{\sqrt{1-\frac{9}{16}}}=\frac{3}{\sqrt{7}}$

$\Rightarrow R_i=22.67\text{cm}$

Hence, options (a) and (b) are correct alternatives.