The volume of a sphere is increasing at the rate of 1200c.cm/sec. Find the rate of increase in its surface area when the radius is 10 cm is.

120 s q . c m / s e c

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240 s q . c m / s e c

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200 s q . c m / s e c

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100 s q . c m / s e c

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Solution

The correct option is B $240 s q . c m / s e c$
Volume of sphere= $\frac{4}{3} π r^{3}$

AS per question
$\frac{d v}{d t} = 1200 ⟹ 4 π r^{2} \frac{d r}{d t} = 1200 ⟹ \frac{d r}{d t} = \frac{3}{π} c m / s$

Rate of change of surface area= $\frac{d (4 π r^{2})}{d t} = 8 π r \frac{d r}{d t} = 8 π .10 . (\frac{3}{π})$
= $240 s q . c m / s e c$

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