Question

A concrete mixture is pouring on ground at the rate of 8 cm³/sec to form a cone in such a way that the height of the cone is always 1/4th of the radius at the time. Find the rate of increase of the height, when the radius is 8 cm.

Answer 1

Answer:

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Answer 2

Answer:

The value of rate of increase of the height $=\frac{\mathrm{d} H}{\mathrm{d} t}=\frac{1}{8\pi }\frac{cm}{sec}$.

Step-by-step explanation:

given data in question:

 Rate of pouring a mixture =rate of change of volume =$8 \frac{cm^{3}}{sec}$

$\Rightarrow \frac{\mathrm{d} V}{\mathrm{d} t}= 8 \frac{cm^{3}}{sec}$

Also given $H=\frac{1}{4}R$

             $\Rightarrow R=4H$

Where R= radius of a cone

           H = height of a cone

As we know that the formula of volume of cone $=V=\frac{1}{3}\pi R^{2} H$

Substitue the value of R= 4H

$\Rightarrow V=\frac{1}{3}\pi \left ( 4H \right )^{2} H$

$\Rightarrow V=\frac{1}{3}\pi \times 16H^{3}$

Now,Differentiate the volume with respect to time

$\Rightarrow \frac{\mathrm{d} V}{\mathrm{d} t}=\frac{1}{3}\pi \times 16\frac{\mathrm{d} H^{3}}{\mathrm{d} t}$

 $\Rightarrow \frac{\mathrm{d} V}{\mathrm{d} t}=\frac{1}{3}\pi \times 16\times 3H^{2}\frac{\mathrm{d} H}{\mathrm{d} t}$

$\Rightarrow 8=\frac{1}{3}\pi \times 16\times 3\left ( \frac{R}{4} \right )^{2}\frac{\mathrm{d} H}{\mathrm{d} t}$

Put the value of R=8 cm

$\Rightarrow 8=\frac{1}{3}\pi \times 16\times 3\left ( \frac{8}{4} \right )^{2}\frac{\mathrm{d} H}{\mathrm{d} t}$

$\Rightarrow \frac{\mathrm{d} H}{\mathrm{d} t}=\frac{1}{8\pi }$

Therefore the value of rate of increase of the height $=\frac{\mathrm{d} H}{\mathrm{d} t}=\frac{1}{8\pi }$