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

Answers (1)


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


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)


Radius of ball =r=2cm


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


It we take r=2cm

The correct answer would be

\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)

Posted by

Gurleen Kaur

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