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A small bulb is placed at the bottom of tank containing a liquid to a depth of 30 cm. The area of surface of liquid, through which light from bulb can average out is ( Refractive index of liquid is 5/4)

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$6025 {cm}^{2}$

$3256 {cm}^{2}$

$5026 {cm}^{2}$

$8044 {cm}^{2}$

answer is A.

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Detailed Solution

$r = \frac{h}{\sqrt{μ^{2} - 1}} = \frac{80}{\sqrt{\frac{5^{2}}{4^{2}} - 1}} = \frac{30 \times 4}{3} = 40 c m$

$A = π r^{2} = π \times {(40)}^{2} = 1600 \times π c m^{2} = 5026 c m^{2}$

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