Q. 205.0( 1 Vote )

# The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when the radius is 5 cm. (CBSE 2017)

Given: the volume of a spherical balloon is increasing at the rate of 25 cm3/sec.

To find the rate of change of its surface area at the instant when the radius is 5 cm

Let the radius of the given spherical balloon be r cm, and V be its volume at any instant time.

Then according to the given criteria,

The rate of the volume of the spherical balloon is increasing is,

But volume of the spherical balloon is,

Applying derivative with respect to time on both sides we get,

Substituting the value from equation (i) in above equation, we get

Now the surface area of the spherical balloon at any time t will be

S = 4r2 cm2.

Applying derivative with respect to time on both sides we get,

Substituting the value from equation (ii), we get

So when the radius is 5cm, the rate of surface area will become,

Hence the rate of change of its surface area at the instant when the radius is 5 cm is 10cm2/sec

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
How to find Maxima & Minima?43 mins
Tangent & Normal To A Curve53 mins
Test your knowledge of Tangents & Normals (Quiz)52 mins
Interactive quizz on tangent and normal, maxima and minima43 mins
Interactive quiz on maxima and minima48 mins
Tangents & Normals (Concept Builder Class)55 mins
Application of Biotechnology48 mins
Application of Biotechnology | Concepts - 0256 mins
Application of Biotechnology | Concepts - 0160 mins
Application of Biotechnology Part 229 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses