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  • Explain solution RD Sharma class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 24 maths

Explain solution RD Sharma class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 24 maths

Answers (1)

Answer:

C. Surface area times the rate of change of radius

Hint:

The volume and surface area of a sphere with radius r, is defined as,

Given:

V(r)=\frac{4}{3} \pi r^{3} \quad \ldots \ldots \text { (i) } \quad \text { and } \quad A(r)=4 \pi r^{2}

Solution:

\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t}, \text { we get }\\ &V(r)=\frac{4}{3} \pi r^{2} \frac{d r}{d t}=A \times \frac{d r}{d t} \end{aligned}

(Surface area of sphere) X (rate of change of radius)

Posted by

Gurleen Kaur

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