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A spherical snowball is melting in such a way that its volume is decreasing at a rate of $1 cm^{3}/min.$ The rate at which the diameter is decreasing when the diameter is $10 cms$ is

\[\begin{array}{1 1}(1)\frac{-1}{50\pi}cm/min&(2)\frac{1}{50\pi}cm/min\\(3)\frac{-11}{75\pi}cm/min&(4)\frac{-2}{75\pi}cm/min\end{array}\]

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Let V be the volume of the spherical snow all and r be the radius at time 't'
$V= \large\frac{4}{3}$$\pi r^3$
We know diameter $d=2r$
$V= \large\frac{8}{6} \pi r^3=\large\frac{\pi}{6} $$(2r)^3$
$V= \large\frac{\pi}{6}$$.d^3$
$\large\frac{dV}{dt} =\frac{\pi}{6} $$ \times 3d^2 \times \large\frac{d}{dt}$$(d)$
$I=\large\frac{\pi}{2} $$ \times 10^2 \times \large\frac{d}{dt}$$(d)$
$I=\large\frac{\pi}{2} $$ \times 10 \times 10 \times \large\frac{d}{dt}$$(d)$
$\qquad= \large\frac{1}{50 \pi}$$cm/min$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
 

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