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The radius of air bubble is increasing at the rate of 0. 25 cm/s. At what rate the volume of the bubble is increasing when the radius is 1 cm.
- a)4π cm3/s
- b)22π cm3/s
- c)2π cm3/s
- d)π cm3/s
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The radius of air bubble is increasing at the rate of 0. 25 cm/s. At w...
Given, the rate of increase of radius of the air bubble = 0.25 cm/s
We need to find the rate of increase of volume of the bubble when the radius is 1 cm.
Formula used:
Volume of a sphere = (4/3)πr^3
Differentiating both sides with respect to time t, we get:
dV/dt = 4πr^2(dr/dt)
where dV/dt is the rate of change of volume of the sphere with respect to time t and dr/dt is the rate of change of radius of the sphere with respect to time t.
Substituting the given values, we get:
dV/dt = 4π(1)^2(0.25) = π cm^3/s
Therefore, the rate of increase of volume of the bubble when the radius is 1 cm is π cm^3/s, which is the correct answer.
We need to find the rate of increase of volume of the bubble when the radius is 1 cm.
Formula used:
Volume of a sphere = (4/3)πr^3
Differentiating both sides with respect to time t, we get:
dV/dt = 4πr^2(dr/dt)
where dV/dt is the rate of change of volume of the sphere with respect to time t and dr/dt is the rate of change of radius of the sphere with respect to time t.
Substituting the given values, we get:
dV/dt = 4π(1)^2(0.25) = π cm^3/s
Therefore, the rate of increase of volume of the bubble when the radius is 1 cm is π cm^3/s, which is the correct answer.
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Community Answer
The radius of air bubble is increasing at the rate of 0. 25 cm/s. At w...
Take the air bubble as sphere:
Then volume of sphere is given by
V=4/3(πR^3)
dV/dt=d(4/3(πR^3))/dt
dV/dt=4πR^2dR/dt
where dR/dt=0.25cm/s
R=1cm
dV/dt=4π(1cm)^2(0.25cm/s)
dV/dt=πcm^3/s
Then volume of sphere is given by
V=4/3(πR^3)
dV/dt=d(4/3(πR^3))/dt
dV/dt=4πR^2dR/dt
where dR/dt=0.25cm/s
R=1cm
dV/dt=4π(1cm)^2(0.25cm/s)
dV/dt=πcm^3/s
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