Water is running into an underground right circular conical reservoir, which is 10 m deep and radius of its base is 5 m. If the rate of change in the volume of water in the reservoir is \frac{3}{2}\pi m^{3}/min.,    then the rate (in m/min) at which water rises in it, when the water level is 4 m, is :

  • Option 1)

    \frac{3}{2}

  • Option 2)

    \frac{3}{8}

  • Option 3)

    \frac{1}{8}

  • Option 4)

    \frac{1}{4}

 

Answers (1)

As we learnt in 

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}


\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)


\Rightarrow \frac{dV}{dt}(Volume)


\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)

- wherein

Where dR / dt  means Rate of change of radius.

 Move to rate measurement topic

\therefore \frac{r}{5}=\frac{h}{10}        \Rightarrow r=\frac{h}{2}

\therefore v=\frac{\pi r^{2}h}{3}=\frac{\pi }{3}.\frac{h^{3}}{4}=\frac{\pi h^{3}}{12}

\therefore\frac{dv}{dt}=\frac{3\pi }{2}=\frac{\pi }{12}\times 3h^{2}\times \frac{dh}{dt}

\frac{6}{16}=\frac{3}{8}=\frac{dh}{dt}


Option 1)

\frac{3}{2}

This option is incorrect

Option 2)

\frac{3}{8}

This option is correct

Option 3)

\frac{1}{8}

This option is incorrect

Option 4)

\frac{1}{4}

This option is incorrect

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