On dropping a stone in stationary water,circu | vedclass

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Question

(D) Let $r$ be the radius of the circular ripple and $A$ be its area.Given that the rate of change of the radius is $\frac{dr}{dt} = 6 	ext{ cm/sec}$

Vedclass · Accepted Answer

$120 \pi 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(D) Let $r$ be the radius of the circular ripple and $A$ be its area.Given that the rate of change of the radius is $\frac{dr}{dt} = 6 	ext{ cm/sec}$.The area of a circle is given by $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.Substitute the given values $r = 10 	ext{ cm}$ and $\frac{dr}{dt} = 6 	ext{ cm/sec}$:$\frac{dA}{dt} = 2 \pi 	imes 10 	imes 6 = 120 \pi 	ext{ cm}^2/	ext{sec}$.Thus,the rate of increase in the area is $120 \pi 	ext{ cm}^2/	ext{sec}$.

On dropping a stone in stationary water,circular ripples are observed. The rate of flow of ripples is $6 \text{ cm/sec}$. When the radius of the circle is $10 \text{ cm}$,the rate of increase in its area is:

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