The apparent depth of water in a cylindrical | vedclass

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(B) The apparent depth $h'$ is related to the real depth $h$ by the formula $h' = h / \mu$,where $\mu = n_2/n_1$ is the refractive index of water with

Vedclass · Accepted Answer

$x \pi R^2 n_2/n_1$

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(B) The apparent depth $h'$ is related to the real depth $h$ by the formula $h' = h / \mu$,where $\mu = n_2/n_1$ is the refractive index of water with respect to air.Given that the apparent depth is reducing at a rate of $x \, cm/min$,we have $\frac{dh'}{dt} = -x$.Since $h' = h \cdot (n_1/n_2)$,differentiating with respect to time $t$ gives $\frac{dh'}{dt} = \frac{n_1}{n_2} \frac{dh}{dt}$.Substituting the given rate: $-x = \frac{n_1}{n_2} \frac{dh}{dt}$,which implies $\frac{dh}{dt} = -x \cdot (n_2/n_1)$. Thus,the real depth is decreasing at a rate of $x \cdot (n_2/n_1) \, cm/min$.The volume of water in the cylindrical tank is $V = \pi R^2 h$.The rate of change of volume is $\frac{dV}{dt} = \pi R^2 \frac{dh}{dt}$.Substituting the magnitude of $\frac{dh}{dt}$,the volume of water drained per minute is $\frac{dV}{dt} = \pi R^2 \cdot x \cdot (n_2/n_1) = x \pi R^2 (n_2/n_1)$.

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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