Question

Water is dropping out at the steady rate of $2cc/sec$ through a tiny hole at the vertex of a conical whose axis is vertical. When the slant height of the water is $4cm$ then the rate of decrease height of water is [Given that the vertex angle of the funnel is ${120}^{\circ }$]

• A. $\frac{32}{27\pi }cm/sec$
• B. $\frac{2}{3\pi }cm/sec$
• C. $\frac{4}{3\pi }cm/sec$
• D. $\frac{1}{3\pi }cm/sec$

Answer 1

The correct option is A $\frac{32}{27\pi }cm/sec$

Rate = 2 cc / s

Slant height $=4$ cm

Vertex angle$={120}^0$ .

To find the rate of dripping of water from conical funnel,

By formula,

$dv/dt=4$

$Volume=1/3\pi r^{2}h$

$sin60=r/l$

$cos60=h/l$

Where,

$l-$ slant height

$r=\surd 32.l$

$h=l/2$

Substituting,

We get,

$Volume=\left(1/3\right)\ast \pi \ast 3/4\ast l^2\ast l/2$

$( 1 / 8 ) * π * l^344

Differentiating with respect to t,

$dvolume/dt=\left(1/8\right)\ast \pi \ast 3\left(l^2\right)\left(dl/dt\right)$

$dvolume/dt=4$

$dl/dt=(4\ast 8)/27\pi$

$32/\left(27\pi \right)cm/s$

Hence, the option $A$ is the correct answer.