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Need solution for RD Sharma maths class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  7

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Answer: 1 \text { square/unit }

Hint: Here, we use basic formula of sphere’s area and volume

Given: Volume is changing at the same rate of it’s radius of sphere

        \text { So, } \frac{d v}{d t}=\frac{d r}{d t}(\text { Given) }

Solution:  We know the volume of sphere,

        v=\frac{4}{3} \pi r^{3}

        \begin{aligned} &\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \times \frac{d r}{d t} \\\\ &\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \times \frac{d V}{d t} \quad \text { because } \frac{d V}{d t}=\frac{d r}{d t} \quad(\text { given }) \end{aligned}

\begin{gathered} \mathrm{So}, 1=4 \pi r^{2} \\\\ r^{2}=\frac{1}{4 \pi} \end{gathered}    .......................(1)

Surface of sphere is

        \begin{aligned} &S=4 \pi r^{2} \\ &=4 \pi \times \frac{1}{4 \pi}\left(r^{2}=\frac{1}{4 \pi} \text { Frome\; quation(1) }\right) \\ &S=1 \text { square /unit } \end{aligned}

So, surface of sphere is 1 square/unit


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