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need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 23

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Answer: \frac{d R}{d t}=0.25\; cm/sec

Hint: The volume of the hallow sphere is V=\frac{4}{3} \pi\left(R^{3}-r^{3}\right)

Given: radius =1 cm/sec and outer radius are 4 cm and 8 cm

Solution: Differentiating the volume with respect to time,

\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi\left(R^{3}-r^{3}\right)\right)}{d t}

This is the rate of the volume of the hollow sphere

0=\frac{4}{3} \pi \frac{d\left(R^{3}-r^{3}\right)}{d t}=\frac{d\left(R^{3}\right)}{d t}-\frac{d\left(r^{3}\right)}{d t}=0

\frac{d r}{d t}=1\: cm/sec

 \begin{aligned} &3 R^{2} \frac{d R}{d t}-3 r^{2}(1)=0 \\\\ &3 R^{2} \frac{d R}{d t}=3 r^{2} \end{aligned}

Here radius are 4 cm and 8 cm

\begin{aligned} &3(8)^{2} \frac{d R}{d t}=3(4)^{2} \\\\ &\frac{d R}{d t}=\frac{3(8)^{2}}{3(4)^{2}} \\\\ &=0.25 \: \mathrm{cm} / \mathrm{sec} \end{aligned}

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