$A$ spherical snowball is forming such that its volume is increasing at the rate of $8 \text{ cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2 \text{ cm}$.

  • A
    $\pi \text{ cm/sec}$
  • B
    $\frac{1}{8\pi} \text{ cm/sec}$
  • C
    $2\pi \text{ cm/sec}$
  • D
    $\frac{1}{2\pi} \text{ cm/sec}$

Explore More

Similar Questions

If displacement $s = 5 \sin(2t)$,then the velocity at the end of $t = \frac{\pi}{3} \text{ s}$ is

The rate of change of the volume of a sphere with respect to its surface area,when the radius is $5 \ m$ is

The volume of a spherical balloon is increasing at the rate of $2 \ cm^3/sec$. When its radius is $4 \ cm$, the rate of change of its surface area (in $cm^2/sec$) is

The displacement of a particle in time $t$ is given by $s = 2t^2 - 3t + 1$. The acceleration is

The rate of change of the area of a circle with respect to its radius at $r = 3 \text{ cm}$ is . . . . . . $\text{cm}^2/\text{cm}$. (in $\pi$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo