If by dropping a stone in a quiet lake a wave | vedclass

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(A) Given that the rate of increase of the radius is $\frac{dr}{dt} = 3.5 \, cm/sec$ and the radius is $r = 10 \, cm$.The area of the circular region

Vedclass · Accepted Answer

$220$

Vedclass · Answer

(A) Given that the rate of increase of the radius is $\frac{dr}{dt} = 3.5 \, cm/sec$ and the radius is $r = 10 \, cm$.The area of the circular region is $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.Substituting the given values: $\frac{dA}{dt} = 2 	imes \left( \frac{22}{7} ight) 	imes 10 	imes 3.5$.Since $3.5 = \frac{7}{2}$,we have $\frac{dA}{dt} = 2 	imes \left( \frac{22}{7} ight) 	imes 10 	imes \left( \frac{7}{2} ight)$.Simplifying the expression: $\frac{dA}{dt} = 22 	imes 10 = 220 \, cm^2/sec$.

If by dropping a stone in a quiet lake a wave moves in a circle at a speed of $3.5 \, cm/sec$,then the rate of increase of the enclosed circular region when the radius of the circular wave is $10 \, cm$ is ......... $cm^2/sec$. $\left( \pi = \frac{22}{7} \right)$

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