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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?
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A spherical balloon is filled with 4500 π cubic meters of helium ...
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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.
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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?
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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?.
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