The volume of a sphere is increasing at the r | vedclass

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(B) Let $V$ be the volume,$S$ be the surface area,and $r$ be the radius of the sphere.It is given that $\frac{dV}{dt} = 1200 	ext{ cm}^3/	ext{s}$ an

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$240$

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(B) Let $V$ be the volume,$S$ be the surface area,and $r$ be the radius of the sphere.It is given that $\frac{dV}{dt} = 1200 	ext{ cm}^3/	ext{s}$ and $r = 10 	ext{ cm}$.The volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting the given values: $1200 = 4 \pi (10)^2 \frac{dr}{dt} \implies 1200 = 400 \pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{3}{\pi} 	ext{ cm/s}$.The surface area of a sphere is $S = 4 \pi r^2$.Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.Substituting $r = 10 	ext{ cm}$ and $\frac{dr}{dt} = \frac{3}{\pi} 	ext{ cm/s}$:$\frac{dS}{dt} = 8 \pi (10) \left( \frac{3}{\pi} ight) = 80 	imes 3 = 240 	ext{ cm}^2/	ext{s}$.

The volume of a sphere is increasing at the rate of $1200 \text{ cm}^3/\text{s}$. The rate of increase in its surface area when the radius is $10 \text{ cm}$ is: (in $\text{ cm}^2/\text{s}$)

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