The altitude of a cone is 20 cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of

$\frac{160}{3}$ cm/sec

Let $\theta$ be the semi-vertical angle and r be the radius of the cone at time t. Then,

$r=20 \tan \theta$

$\Rightarrow \frac{d r}{d t}=20 \sec ^2 \theta \frac{d \theta}{d t}$

$\Rightarrow \frac{d r}{d t}=20 \sec ^2 30^{\circ} \times 2$         $\left[∵ \theta=30^{\circ} \text { and } \frac{d \theta}{d t}=2 \text { (given)}\right]$

$\Rightarrow \frac{d r}{d t}=20 \times \frac{4}{3} \times 2 \mathrm{~cm} / \mathrm{sec}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$