#### explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 24 maths

Answer: $\frac{d h}{d t}=\frac{1}{2 \pi}cm/min$

Hint: We know, volume of the cone is $V=\frac{1}{3} \pi r^{2} h$

Given: Sand is being poured into a conical pile at the constant rate of 50 cm3/min

Solution: Let the volume be V , height be $h$  and radius be $r$  of the cone at any instant of time.

We know volume of the cone is $V=\frac{1}{3} \pi r^{2} h$ .

$r=2 h$

So, the new volume becomes,

$V=\frac{1}{3} \pi(2 h)^{2} h=\frac{4}{3} \pi h^{3}$

Differentiate the above equation with respect to time,

$\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi h^{3}\right)}{d t}=\frac{4}{3} \pi \frac{d\left(h^{3}\right)}{d t}=\frac{4}{3} \pi \times 3 h^{2} \frac{d h}{d t}$

Here, $\frac{d V}{d t}=50\: cm^{3}/min$

$50=\frac{4}{3} \pi \times 3 h^{2} \frac{d h}{d t}$

$\frac{d h}{d t}=\frac{50 \times 3}{4 \times 3 \pi h^{2}}=\frac{50 \times 3}{12 \pi(5)^{2}}=\frac{1}{2 \pi}$ (height =5 ).