
A small sphere of mass M is dropped from a great height, after it has fallen 100m, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100m of fall is (g = acceleration due to gravity)
A. Greater than the work done by the friction in the second 100 m
B. Less than the work done by the friction in the second 100 m
C. Equal to 100 Mg
D. Greater than 100 Mg
Answer
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Hint: Before solving this problem let us understand Stoke's Law. It states that the retarding force of a sphere that is moving through a viscous fluid is directly proportional to the velocity, radius of the sphere, and viscosity of the fluid.
Formula Used:
According to Stoke’s law, the viscous force is,
\[F = 6\pi \eta Rv\]
Here, R is radius of the ball, V is terminal velocity of the ball and \[\eta \] is coefficient of viscosity.
Complete step by step solution:
Consider a small sphere of mass M that is dropped from a great height, after it has fallen 100m, it will attain its terminal velocity and continue to fall at the same speed. We need to find the work done by air friction against the sphere during the first 100m of fall.
When the spherical ball is allowed to move in a fluid, the viscous force that will experience is,
\[F = 6\pi \eta Rv\]
We know that,
\[\text{Workdone} = \text{Force} \times \text{displacement}\]
Here, according to the question, the displacement is the same that is 100 m, and the work done depends upon the force, force here we have to consider the viscous force.
Here, in the first 100 m the body starts from rest and its velocity goes on increasing and after 100 m it acquires maximum velocity which is terminal velocity. Further, because of the air friction that is the viscous force which is proportional to velocity is low in the beginning and maximum at \[v = {v_r}\].Hence work done against air friction in the first 100 m is less than the work done in second 100m. Therefore, as the force is lower in the first 100 m the work done would also be lower compared to the next (second) 100 m.
Hence, option B is the correct answer.
Note: Terminal velocity is the highest velocity attained by an object falling through a fluid. It is observed that when the sum of drag force and buoyancy is equal to the downward gravitational force acting on the object.
Formula Used:
According to Stoke’s law, the viscous force is,
\[F = 6\pi \eta Rv\]
Here, R is radius of the ball, V is terminal velocity of the ball and \[\eta \] is coefficient of viscosity.
Complete step by step solution:
Consider a small sphere of mass M that is dropped from a great height, after it has fallen 100m, it will attain its terminal velocity and continue to fall at the same speed. We need to find the work done by air friction against the sphere during the first 100m of fall.
When the spherical ball is allowed to move in a fluid, the viscous force that will experience is,
\[F = 6\pi \eta Rv\]
We know that,
\[\text{Workdone} = \text{Force} \times \text{displacement}\]
Here, according to the question, the displacement is the same that is 100 m, and the work done depends upon the force, force here we have to consider the viscous force.
Here, in the first 100 m the body starts from rest and its velocity goes on increasing and after 100 m it acquires maximum velocity which is terminal velocity. Further, because of the air friction that is the viscous force which is proportional to velocity is low in the beginning and maximum at \[v = {v_r}\].Hence work done against air friction in the first 100 m is less than the work done in second 100m. Therefore, as the force is lower in the first 100 m the work done would also be lower compared to the next (second) 100 m.
Hence, option B is the correct answer.
Note: Terminal velocity is the highest velocity attained by an object falling through a fluid. It is observed that when the sum of drag force and buoyancy is equal to the downward gravitational force acting on the object.
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