# A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm

**Solution:**

Derivatives are used to find the rate of changes of a quantity with respect to the other quantity. By using the application of derivatives we can find the approximate change in one quantity with respect to the

change in the other quantity.

They are used in many situations like finding maxima or minima of a function, finding the slope of the curve, and even inflection points.

We know that V = 4/3π r^{3}

Hence,

dV/dt = dV/dt (4/3π r^{3})

= d/dr (4/3π r^{3}) dr/dt

= 4πr^{2} dr/dt

We have,

dV/dt = 900 cm^{2}/s

Therefore,

4π r^{2} dr/dt = 900

dr/dt = 900/4π r^{2}

= 225/4π r^{2}

When radius, r = 15 cm

Then,

dr/dt = 225/ π (15)^{2}

= 1/π cm/s

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.1 Question 8

## A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm

**Summary:**

Given that a balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second.The rate at which the radius of the balloon increases when the radius is 15 cm is 1/π cm/s