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A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball is greater than the weight of the ball by a factor of
A) 2
B) 3
C) 4
D) 6

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Last updated date: 13th Jun 2024
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Answer
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Hint:If the ball is dipped inside a liquid then the liquid applies a force of buoyancy on the body in the upward direction. The weight of the body will continuously act in the downward direction and the viscous force due to the liquid will act in the opposite direction of the motion of the ball.

Step by step solution:
As it is given that the ball is rising up with a constant velocity. If the ball is moving with constant velocity then the value of net force is zero as there is no acceleration to the body.
Let us write the equation of the forces acting on the ball.
As the ball is rising upwards which means that the weight of the body is acting in the downwards direction the viscous force will act in the direction opposite to the direction of the motion of the ball and the buoyancy force will act on the ball upwards. The buoyancy force always acts in the upwards direction, it tries to lift the body up.
The relation of the force is given by,
$ \Rightarrow {F_{net}} = 0$
And the net force is given by,
$ \Rightarrow {\text{weight}} + {\text{viscous force}} = {\text{buoyant force}}$
The weight of the body is equal to,
$ \Rightarrow w = \rho gV$………eq. (1)
The buoyant force is given by,
$ \Rightarrow B = V\left( {4\rho } \right)g$………eq. (2)
Let the viscous force be represented by the term${F_v}$.
The equation will be,
$ \Rightarrow {\text{weight}} + {\text{viscous force}} = {\text{buoyant force}}$
$ \Rightarrow {\text{w}} + {F_v} = B$………eq. (3)
Replace the value of weight and buoyant force in equation (3) from equation (1) and equation (2).
$ \Rightarrow \rho gV + {F_v} = V\left( {4\rho } \right)g$
$ \Rightarrow {F_v} = V\left( {4\rho } \right)g - \rho gV$
$ \Rightarrow {F_v} = 3\rho gV$………eq. (4)
The viscous force is given by${F_v} = 3\rho gV$.
The ratio of viscous force to the weight of the body is given by,
$ \Rightarrow \dfrac{{{F_v}}}{w}$
Replace the value of viscous force and weight from equation (1) and equation (4).
$ \Rightarrow \dfrac{{{F_v}}}{w} = \dfrac{{3\rho gV}}{{\rho gV}}$
After solving the above relation we get,
$ \Rightarrow \dfrac{{{F_v}}}{w} = 3$
So the value of the ratio of viscous force to the weight is equal to 3. So the correct answer

for this problem is option is option B.

Note:The buoyant force is always in the direction upward direction the weight is always in the downward direction but the viscous force will act in the opposite direction to the motion of the ball as the viscous force will try to stop the motion of the ball