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The apparent depth of water in cylindrical water tank of diameter \mathrm{2R} cm in reducing at the rate of \mathrm{x} \mathrm{cm/min} when water is being drained out at a constant rate. The amount of water drained in \mathrm{cc/min} is ( \mathrm{n_{1}=} refractive index of air, \mathrm{n_{2}=} refractive index of water)

Option: 1

\mathrm{\frac{\mathrm{x} \pi \mathrm{R}^2 \mathrm{n}_1}{\mathrm{n}_2}}


Option: 2

\mathrm{\frac{\mathrm{x} \pi \mathrm{R}^2 \mathrm{n}_2}{\mathrm{n}_1}}


Option: 3

\mathrm{\frac{2 \pi \mathrm{R}^2 \mathrm{n}_1}{\mathrm{n}_2}}


Option: 4

\mathrm{\pi R^2 x}


Answers (1)

best_answer

\mathrm{\text { As } \mathrm{y}_{\mathrm{a}}=\left(\frac{\mathrm{n}_1}{\mathrm{n}_2}\right) \mathrm{y}}
Here, \mathrm{y=} actual depth
and \mathrm{y_{a}=} apparent depth
\mathrm{\therefore \quad \mathrm{y}=\left(\frac{\mathrm{n}_2}{\mathrm{n}_1}\right) \cdot \mathrm{y}_{\mathrm{a}} \Rightarrow\left(-\frac{\mathrm{dy}}{\mathrm{dt}}\right)=\frac{\mathrm{n}_2}{\mathrm{n}_1}\left(-\frac{\mathrm{dy}_{\mathrm{a}}}{\mathrm{dt}}\right)=\frac{\mathrm{n}_2 \mathrm{x}}{\mathrm{n}_1}}
The amount of water drained in \mathrm{cc/min}
\mathrm{\therefore \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{A} \cdot\left(-\frac{\mathrm{dy}}{\mathrm{dt}}\right)=\frac{\pi \mathrm{R}^2 \mathrm{n}_2 \mathrm{x}}{\mathrm{n}_1}}

Posted by

seema garhwal

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