The radius of a sphere increases at the rate | vedclass

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(D) Let $r$ be the radius of the sphere. Given,the rate of change in radius is $\frac{dr}{dt} = 0.04 	ext{ cm/sec}$. The volume of the sphere is $V =

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) Let $r$ be the radius of the sphere. Given,the rate of change in radius is $\frac{dr}{dt} = 0.04 	ext{ cm/sec}$. The volume of the sphere is $V = \frac{4}{3} \pi r^3$. Differentiating with respect to $t$,we get: $\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \cdot \frac{dr}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \quad \dots(i)$ The surface area of the sphere is $S = 4 \pi r^2$. Differentiating with respect to $t$,we get: $\frac{dS}{dt} = 4 \pi (2r) \cdot \frac{dr}{dt} = 8 \pi r \cdot \frac{dr}{dt} \quad \dots(ii)$ Dividing equation $(i)$ by equation $(ii)$,we get: $\frac{dV/dt}{dS/dt} = \frac{dV}{dS} = \frac{4 \pi r^2 \cdot \frac{dr}{dt}}{8 \pi r \cdot \frac{dr}{dt}}$ $\frac{dV}{dS} = \frac{r}{2}$ Given $r = 10 	ext{ cm}$,we substitute the value: $\frac{dV}{dS} = \frac{10}{2} = 5 	ext{ cm}$. Thus,the rate of increase in volume with respect to surface area is $5 	ext{ cm}$. Hence,option $(D)$ is correct.

The radius of a sphere increases at the rate of $0.04 \text{ cm/sec}$. The rate of increase in the volume of that sphere with respect to its surface area,when its radius is $10 \text{ cm}$ is

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