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Q.

A spherical balloon is being inflated at the rate of $35 c m^{3} / m i n .$ The rate of increase in the surface area (in $c m^{3} / m i n$ ) of the balloon when its diameter is 14 cm is ______

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answer is 10.

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Detailed Solution

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$\frac{d v}{d t} = 35 c m^{3} / \min, 2 r = 14, r = 7$

$\frac{d v}{d t} = \frac{r}{2} \frac{d s}{d t}$ $35 = \frac{7}{2} \frac{d s}{d t}$

$\frac{d s}{d t} = 35 \times \frac{2}{7} = 10$

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A spherical balloon is being inflated at the rate of 35 cm3/min. The rate of increase in the surface area (in cm3/min ) of the balloon when its diameter is 14 cm is ______