Question

The surface area of a ball is increasing at the rate of $2\pi sq.cm/sec$. The rate at which the radius is increasing when the surface area is $16\pi sq.cm$ is

• A. $0.125cm/sec$
• B. $0.25cm/sec$
• C. $0.5cm/sec$
• D. $1cm/sec$

Answer 1

Answer: (A)

We have, $\frac{dS}{dt}=2\pi$ sq . cm /sec         ....(i)
Now, $S=4\pi r^2$        ....(i)
Differentiating (ii) w.r.t, t, we get
    $\frac{dS}{dt}=8\pi r\frac{dr}{dy}$     $\Rightarrow 2\pi =8\pi r\frac{dr}{dt}$ [From (i)]
$\Rightarrow \frac{dr}{dt}=\frac{1}{4r}$        ...(iii)
Now, when $S=16\pi \Rightarrow 4\pi r^2=16\pi$
$\Rightarrow r^2=4\Rightarrow r=$2 cm
Hence, ${\left[\frac{dr}{dt}\right]}_{r=2}=\frac{1}{4\times 2}=\frac{1}{8}0.125$ cm /sec