The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be
2Mg
\(\frac{Mg}{2}\)
Mg
\(\frac{3}{2}Mg\)
The Correct Option is B
Solution and Explanation
To solve this problem, we need to understand the forces acting on the ball as it moves through the glycerine.
The small ball, when falling through a viscous fluid like glycerine, experiences three forces:
- Gravitational Force \(F_g\): This is the force due to gravity acting downward, given by \(F_g = Mg\), where \(M\) is the mass of the ball, and \(g\) is the acceleration due to gravity.
- Buoyant Force \(F_b\): This is the upward force exerted by the fluid on the ball, given by the Archimedes' principle as \(F_b = V \cdot \rho_f \cdot g\), where \(V\) is the volume of the ball, \(\rho_f\) is the density of the fluid (glycerine, in this case), and \(g\) is the acceleration due to gravity.
- Viscous Force \(F_v\): This is the resistive force due to the viscosity of the fluid, which acts upward against the motion of the ball.
After some time, the velocity of the ball becomes constant, meaning it reaches terminal velocity. At terminal velocity, the net force acting on the ball is zero, as the forces are in equilibrium. Thus,
\(F_g = F_v + F_b\)
Substituting the expressions for the forces, we get:
\(Mg = F_v + V \cdot \rho_f \cdot g\)
The volume \(V\) of the ball can be expressed in terms of its mass \(M\) and density \(d\) as:
\(V = \frac{M}{d}\)
Plugging this back into the buoyant force equation, we have:
\(F_b = \left(\frac{M}{d}\right) \cdot \frac{d}{2} \cdot g = \frac{Mg}{2}\)
Now substitute this into the equilibrium equation:
\(Mg = F_v + \frac{Mg}{2}\)
Solve this equation for the viscous force \(F_v\):
\(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\)
Therefore, the viscous force acting on the ball is \( \frac{Mg}{2} \).
Thus, the correct answer is \(\frac{Mg}{2}\), matching Option B.
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Concepts Used:
Viscosity
Viscosity is a measure of a fluid’s resistance to flow. The SI unit of viscosity is poiseiulle (PI). Its other units are newton-second per square metre (N s m-2) or pascal-second (Pa s.) The dimensional formula of viscosity is [ML-1T-1].
Viscosity: Formula
Viscosity is measured in terms of a ratio of shearing stress to the velocity gradient in a fluid. If a sphere is dropped into a fluid, the viscosity can be determined using the following formula:
η = [2ga2(Δρ)] / 9v
Where ∆ρ is the density difference between fluid and sphere tested, a is the radius of the sphere, g is the acceleration due to gravity and v is the velocity of the sphere.
Viscosity: Types
- Dynamic viscosity: When the viscosity is measured directly by measuring force. It is defined as the ratio of shear stress to the shear strain of the motion. Dynamic viscosity is used to calculate the rate of flow in liquid.
- Kinematic viscosity: There is no force involved. It can be referred to as the ratio between the dynamic viscosity and density of the fluid. It can be computed by dividing the dynamic viscosity of the fluid with fluid mass density.
- Laminar flow: Laminar flow is the type of flow in which the fluid moves smoothly or in a regular path from one layer to the next. Laminar flow occurs in lower velocities.