A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25πm3/ minute. The rate at which the surface of the oil is rising, is

A
$1 m / m i n u t e$
B
$2 m / m i n u t e$
C
$5 m / m i n u t e$
D
$1.25 m / m i n u t e$

Solution:

Let $r$ be the radius, $h$ be the height and $V$ be the

volume of the oil at time $t$ , Then,

$\begin{array}{l} V = π r^{2} h \\ \Rightarrow V = \frac{π}{4} h [∵ r = 0.5 m = \frac{1}{2} (given)] \end{array}$

$\begin{array}{l} \Rightarrow \frac{d V}{d t} = \frac{π}{4} \frac{d h}{d t} \\ \Rightarrow 0.25 π = \frac{π}{4} \times \frac{d h}{d t} [∵ \frac{d V}{d t} = 0.25 π m^{3} (given)] \end{array}$

$\Rightarrow \frac{d h}{d t} = 1 m / minute$

A cylindrical vessel of radius $0.5 m$ is filled with oil at the rate of $0.25 π m^{3}$ / minute. The rate at which the surface of the oil is rising, is

Solution:

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