Question

The slant height of a cone is fixed at $7 \,cm$. If the rate of increase of its height is $0.3\, cm/sec$, then the rate of increase of its volume when its height is $4$ cm is

• A. $\frac{\pi}{2} cc/sec$
• B. $\pi\, cc/sec$
• C. $\frac{\pi}{5} cc/sec$
• D. $\frac{\pi}{10} cc/sec$

Answer 1

Answer: (D)

We have,
Slant height of cone $(l)$ = 7 cm
$\Rightarrow \:\: l^2 = h^2 +r^2$ ....(i)
$\Rightarrow \:\: r^2 = 7^2 - 4^2$ (when $h$ = 4 cm)
$\Rightarrow \:\: r^2 = 33$
$\Rightarrow \:\: r = \sqrt{33}$ cm
Now, differentiating equation (i) w.r.t 't'
$\Rightarrow \:\: 0 = 2h \frac{dh}{dt}+2r \frac{dr}{dt}$
$\Rightarrow \:\: \frac{dr}{dt} = - \frac{h}{r} \frac{dh}{dt} $ ...(ii)
Volume of the cone, $V = \frac{1}{3} \pi r^2 h$
$\Rightarrow \:\: \frac{dV}{dt} = \frac{1}{3} \pi \bigg[ 2rh \frac{dr}{dt}+ r^2 \frac{dh}{dt}\bigg]$
$\Rightarrow \:\: \frac{dV}{dt} = \frac{1}{3} \pi \bigg[ 2rh \bigg( \frac{-h}{r} \frac{dh}{dt} \bigg) + r^2 \frac{dh}{dt}\bigg]$
$\Rightarrow \:\: \frac{dV}{dt} = \frac{1}{3} \pi \bigg[ - 2h^2 \frac{dh}{dt} + r^2 \frac{dh}{dt}\bigg]$
$\Rightarrow \:\: \frac{dV}{dt} = \frac{1}{3} \pi \frac{dh}{dt} [ -2h^2 + r^2]$
$\Rightarrow \:\: \bigg[ \frac{dV}{dt}\bigg] = \frac{1}{3} \pi (0.3) [ -2 \times 4^2 + ( \sqrt{33})^2 ]$
$ = \frac{\pi}{10} [1] = \frac{\pi}{10} cc/sec$