\[\begin{array}{1 1}(1)23\pi&(2)33\pi\\(3)43\pi&(4)53\pi\end{array}\]

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Let V be the volume, r be the radius h be the height of the cylinder for the time 't'.

$V= \pi r^2 h$

Given $\large\frac{dr}{st}$$=-3 ;cm sec$

$r=3 \;cm$

$h=5 \;cm$

$\large\frac{dv}{dt}=?$

$V= \pi r^2 h$

$\large\frac{dV}{dt}$$=\pi \bigg[ r^2 \large\frac{dh}{dt} $$+h 2 r \large\frac{dr}{dt}\bigg]$

$\qquad= \pi [9 \times -3 +2 \times 2 \times 5 \times 3 \times 2 ]$

$\qquad =33 \pi$

Hence 2 is the correct answer

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