The apparent depth of water in a cylindrical | vedclass

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(B) Let the apparent depth be $d_A$ and the real depth be $d_R$.The relationship between apparent depth and real depth is given by $\frac{d_R}{d_A} =

Vedclass · Accepted Answer

$\frac{x \pi R^2 n_2}{n_1}$

Vedclass · Answer

(B) Let the apparent depth be $d_A$ and the real depth be $d_R$.The relationship between apparent depth and real depth is given by $\frac{d_R}{d_A} = \frac{n_2}{n_1}$,which implies $d_A = \frac{n_1}{n_2} d_R$.Differentiating with respect to time,we get the rate of change: $\frac{d(d_A)}{dt} = \frac{n_1}{n_2} \frac{d(d_R)}{dt}$.Given that the apparent depth is reducing at a rate of $x \ cm/minute$,we have $\frac{d(d_A)}{dt} = x$.Therefore,$x = \frac{n_1}{n_2} \frac{d(d_R)}{dt} \Rightarrow \frac{d(d_R)}{dt} = \frac{n_2}{n_1} x$.The volume of water in the cylindrical tank is $V = \pi R^2 d_R$.The rate of change of volume (amount of water drained per minute) is $\frac{dV}{dt} = \pi R^2 \frac{d(d_R)}{dt}$.Substituting the value of $\frac{d(d_R)}{dt}$,we get $\frac{dV}{dt} = \pi R^2 \left( \frac{n_2}{n_1} x \right) = \frac{x \pi R^2 n_2}{n_1}$.

The apparent depth of water in a cylindrical water tank of diameter $2R \ cm$ is reducing at the rate of $x \ cm/minute$ when water is being drained out at a constant rate. The amount of water drained in $c.c./minute$ is: ($n_1 =$ refractive index of air,$n_2 =$ refractive index of water)

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