The altitude of a cone is 20 cm and its semivertical angle is 30^ ^...

The altitude of a cone is 20 cm and its semivertical angle is

30^{{}^{\circ}}

. If the semi-vertical angle is increasing at the rate of

2^{{}^{\circ}}

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

The altitude of a cone is 20 cm and its semivertical angle is 30^ ^ . If the semi-vertical angle is increasing at the

Options

A $30 \mathrm{ cm} / \mathrm\sec$
B $\frac{160}{3} \mathrm{ cm} / \mathrm\sec$
C $10 \mathrm{ cm} / \mathrm\sec$
D $160 \mathrm{ cm} / \mathrm\sec$

Correct answer

B. $\frac{160}{3} \mathrm{ cm} / \mathrm\sec$

Step-by-step solution

Let

\theta

be the semi-vertical angle and

r

be the radius of the cone at time

t

. the radius of the cone at time

t

. Then,

r=20 \tan \theta

\Rightarrow \quad \frac{d r}{d t}=20 \sec ^2 \theta \frac{d \theta}{d t}\Rightarrow \quad \frac{d r}{d t}=20 \sec ^2 30^{{}^{\circ}} \times 2\left[\because \theta=30^{{}^{\circ}}\right. and \left.\frac{d \theta}{d t}=2\right]\Rightarrow \quad \frac{d r}{d t}=20 \times \frac{4}{3} \times 2 \mathrm{~cm} / \mathrm{s}=\frac{160}{3} \mathrm{~cm} / \mathrm{s}

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