JEE Exam > JEE Questions > A spherical balloon is filled with 4500 π...
Start Learning for Free
A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is
- a)2/9
- b)9/2
- c)9/7
- d)7/9
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A spherical balloon is filled with 4500 π cubic meters of helium ...
Most Upvoted Answer
A spherical balloon is filled with 4500 π cubic meters of helium ...
Given:
Volume of the balloon, V = 4500 cubic meters
Rate of leakage of gas, r = 75 cubic meters per minute
We need to find the rate at which the radius of the balloon decreases after 49 minutes.
Let's assume the radius of the balloon at any time t is r(t) meters.
The volume of a sphere is given by the formula V = (4/3)πr^3.
1. Find the initial radius of the balloon
The initial volume of the balloon is given as V = 4500 cubic meters.
Therefore, (4/3)πr^3 = 4500.
Simplifying, r^3 = (3/4)(4500/π).
Taking the cube root on both sides, we get r = [(3/4)(4500/π)]^(1/3).
2. Find the rate of change of volume with respect to time
Differentiating the volume formula with respect to time t, we get dV/dt = 4πr^2(dr/dt).
Since the volume is constant (no additional gas is being filled), dV/dt = 0.
Therefore, 4πr^2(dr/dt) = 0.
Simplifying, dr/dt = 0 / (4πr^2) = 0.
3. Find the rate of change of radius with respect to time
Since dr/dt = 0, it means the radius is not changing with time when there is no leakage.
4. Find the rate of change of radius after 49 minutes of leakage
After 49 minutes, the volume of the balloon would have decreased by (75 * 49) cubic meters.
The new volume of the balloon is V_new = V - (75 * 49).
The new volume can be expressed as (4/3)π(r_new)^3.
Therefore, (4/3)π(r_new)^3 = V - (75 * 49).
Simplifying, (4/3)π(r_new)^3 = 4500 - 3675 = 825.
(r_new)^3 = (3/4)(825/π).
Taking the cube root on both sides, we get r_new = [(3/4)(825/π)]^(1/3).
5. Find the rate of change of radius after 49 minutes
The rate of change of radius after 49 minutes is given by dr/dt = (r_new - r) / 49.
Substituting the values of r_new and r, we get dr/dt = {[(3/4)(825/π)]^(1/3) - [(3/4)(4500/π)]^(1/3)} / 49.
Calculating the expression, we find that dr/dt ≈ 2/9.
Therefore, the rate at which the radius of the balloon decreases 49 minutes after the leakage began is approximately 2/9 meters per minute.
Volume of the balloon, V = 4500 cubic meters
Rate of leakage of gas, r = 75 cubic meters per minute
We need to find the rate at which the radius of the balloon decreases after 49 minutes.
Let's assume the radius of the balloon at any time t is r(t) meters.
The volume of a sphere is given by the formula V = (4/3)πr^3.
1. Find the initial radius of the balloon
The initial volume of the balloon is given as V = 4500 cubic meters.
Therefore, (4/3)πr^3 = 4500.
Simplifying, r^3 = (3/4)(4500/π).
Taking the cube root on both sides, we get r = [(3/4)(4500/π)]^(1/3).
2. Find the rate of change of volume with respect to time
Differentiating the volume formula with respect to time t, we get dV/dt = 4πr^2(dr/dt).
Since the volume is constant (no additional gas is being filled), dV/dt = 0.
Therefore, 4πr^2(dr/dt) = 0.
Simplifying, dr/dt = 0 / (4πr^2) = 0.
3. Find the rate of change of radius with respect to time
Since dr/dt = 0, it means the radius is not changing with time when there is no leakage.
4. Find the rate of change of radius after 49 minutes of leakage
After 49 minutes, the volume of the balloon would have decreased by (75 * 49) cubic meters.
The new volume of the balloon is V_new = V - (75 * 49).
The new volume can be expressed as (4/3)π(r_new)^3.
Therefore, (4/3)π(r_new)^3 = V - (75 * 49).
Simplifying, (4/3)π(r_new)^3 = 4500 - 3675 = 825.
(r_new)^3 = (3/4)(825/π).
Taking the cube root on both sides, we get r_new = [(3/4)(825/π)]^(1/3).
5. Find the rate of change of radius after 49 minutes
The rate of change of radius after 49 minutes is given by dr/dt = (r_new - r) / 49.
Substituting the values of r_new and r, we get dr/dt = {[(3/4)(825/π)]^(1/3) - [(3/4)(4500/π)]^(1/3)} / 49.
Calculating the expression, we find that dr/dt ≈ 2/9.
Therefore, the rate at which the radius of the balloon decreases 49 minutes after the leakage began is approximately 2/9 meters per minute.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer?
Question Description
A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer?.
A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer?.
Solutions for A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer?, a detailed solution for A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A spherical balloon is filled with 4500 π cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 75 π cubic meters per minute, then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began isa)2/9b)9/2c)9/7d)7/9Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup