9. A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Answers (1)

We need to find the value of \frac{dV}{dr}  at r =10 cm
The volume of the sphere (V) = \frac{4}{3}\pi r^{3}
\frac{dV}{dr} = \frac{d(\frac{4}{3}\pi r^{3})}{dr} = \frac{4}{3}\pi \times 3r^{2} = 4\pi r^{2} = 4\pi (10)^{2} = 4\pi \times 100 = 400\pi \ cm^{3}/s
Hence,  the rate at which its volume is increasing with the radius when the later is 10 cm is 400\pi \ cm^{3}/s

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 39999/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions