A spherical balloon is filled with 4500 πcubic meters of helium gas. If a leak in the balloon ca

Name: Problem 50, Chapter 14: A Complete Resource Book in Mathematics for JEE Main
Uploaded: 2021-10-05T11:38:05-08:00
Duration: 4 min 33 s
Channel: Anurag Kumar
Description: A spherical balloon is filled with 4500 πcubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 πcubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 40 minutes after the leakage began is (A) 9 / 7 (B) 7 / 9 (C) 2 / 9 (D) 9 / 2

step 1 In this question given a spherical balloon is filled with 4 ,500 times pi cubic meters of helium

Question

A spherical balloon is filled with $4500 \pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72 \pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 40 minutes after the leakage began is (A) $9 / 7$ (B) $7 / 9$ (C) $2 / 9$ (D) $9 / 2$

   A spherical balloon is filled with $4500 \pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72 \pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 40 minutes after the leakage began is
(A) $9 / 7$
(B) $7 / 9$
(C) $2 / 9$
(D) $9 / 2$

A Complete Resource Book in Mathematics for JEE Main

Dinesh Khattar 2019 Edition

Chapter 14, Problem 50

Instant Answer

Solved by Expert Anurag Kumar

10/05/2021

Step 1

The rate at which the gas is escaping is given by $\frac{dV}{dt} = -72\pi$ cubic meters per minute (the negative sign indicates that the volume is decreasing). Show more…

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A spherical balloon is filled with $4500 \pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72 \pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases 40 minutes after the leakage began is (A) $9 / 7$ (B) $7 / 9$ (C) $2 / 9$ (D) $9 / 2$

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Key Concepts

Related Rates

Related rates problems involve finding the rate at which one quantity changes in relation to another quantity whose rate of change is known. These problems typically require setting up an equation that relates the two quantities, differentiating with respect to time, and then solving for the desired rate.

Volume of a Sphere

The formula for the volume of a sphere, V = (4/3)?r³, is fundamental when dealing with spherical objects. It relates the volume directly to the cube of the radius, and it plays a pivotal role in problems where the volume changes over time.

Chain Rule

The chain rule in differentiation is used to compute the derivative of a composite function. In related rates problems, it allows us to differentiate an equation that links two variables (such as volume and radius) with respect to time, yielding a relation between their rates of change.

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