Question

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

पर्याय

• 30 cm/sec

• \[\frac{160}{3} cm/\sec\]

• 10 cm/sec
  
  

• 160 cm/sec

Answer 1

 \[\frac{160}{3} cm/\sec\]

\[\text {Let r be the radius, h be the height and } \alpha\text {  be the semi - vertical angle of the cone} .\]

\[\text { Then }, \]
\[\tan \alpha = \frac{r}{h}\]
\[ \Rightarrow se c^2 \alpha\left( \frac{d\alpha}{dt} \right) = \frac{dr}{h dt}\]
\[ \Rightarrow \frac{dr}{dt} = h \times se c^2 \alpha\left( \frac{d\alpha}{dt} \right)\]
\[ \Rightarrow \frac{dr}{dt} = 20 \times se c^2 30 \times 2 \left[ \because h = 20 cm, \alpha = 30^\circ \text { and } \frac{d\alpha}{dt} = 2^\circ \text { per second } \right]\]
\[ \Rightarrow \frac{dr}{dt} = 40 \times \left( \frac{2}{\sqrt{3}} \right)^2 \]
\[ \Rightarrow \frac{dr}{dt} = \frac{160}{3} cm/\sec\]