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  • Explain solution RD Sharma class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question 8 maths

Explain solution RD Sharma class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  8 maths

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Answer: \begin{aligned} &\\\ &r=\frac{1}{2 \sqrt{\pi}} \end{aligned} \; \; \; units

Hint: Here we use basic rate of change concept with respect to bodies and quantities

Given: The ratio of change of volume of sphere is equal to the rate of change of its radius

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

Solution:  Here V be the volume of sphere

        \begin{aligned} &v=\frac{4}{3} \pi r^{3} \\\\ &\frac{d v}{d t}=\frac{4}{3} \times 3 r^{2} \times \pi \frac{d r}{d t} \end{aligned}    [derivative with respect to time t]

        \begin{aligned} &\frac{d v}{d t}=\frac{4}{3} \times 3 r^{2} \times \pi \frac{d v}{d t}\left(\frac{d v}{d t}=\frac{d r}{d t}\right) \\\\ &1=4 \pi r^{2} \\\\ &r^{2}=\frac{1}{4 \pi} \quad----------(1) \end{aligned}

From eq (1)

So,

        \begin{aligned} &r=\sqrt{\frac{1}{4 \pi}} \\\\ &r=\frac{1}{2 \sqrt{\pi}} \end{aligned}

 

 

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