#### Need solution for RD Sharma maths class 12 chapter Derivative as a Rate Measure exercise 12.1 question 7

$4\pi r^{2},16\pi \; cm^{3}/cm$

Hint:

Here the volume of ball is the volume of sphere ($V$)

$The\; volume\; o\! f\; the\; ball = the\; volume\; o\! f\; spher\! e \; V=\frac{4}{3}\pi r^{3}$

And we have to differentiate $V$ with respect to radius ($r$)

Given:

Radius of ball $=r=2cm$

Solution:

Here we have,

Radius of ball $=r=2cm$

$Volume\; o\! f\; ball = V=\frac{4}{3}\pi r^{3}$

Let’s differentiate $V$ with respect to $r$

$\therefore \frac{d V}{d r}=\frac{d}{d r}\left(\frac{4}{3}\pi r^{3}\right)$                                    $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\therefore \frac{d V}{d r}=4\pi r^{2}\; \; \; \; \; \; \; \; \; \; \; \; .....(i)$

Equation ($i$) is rate of change of volume of ball with respect to radius.

Let's put. $r=2 cm$

$\therefore \frac{d V}{d r}=4\pi 2^{2}$

$\therefore \frac{d V}{d r}=16\pi \; cm^{3}/cm$

Note:

It we take $r=2cm$

$\frac{dV}{dr}=16\pi \; cm^{3}/cm$

And we can’t write the unit of given rate cm2 because it represents volume ($V$) varying with respect to radius ($r$)