A small ball of mass $m$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time, the ball falls with constant velocity. The viscous force on the ball is:
- $mg \left( \frac{\rho_0}{\rho} - 1 \right)$
- $mg \left( 1 + \frac{\rho}{\rho_0} \right)$
- $mg (1 - \rho \rho_0)$
- $mg \left( 1 - \frac{\rho_0}{\rho} \right)$
The Correct Option is D
Approach Solution - 1
Applying force balance on the ball at constant velocity:
\[ mg - F_B - F_v = ma \]
Since acceleration $a = 0$ for constant velocity:
\[ \Rightarrow mg - F_B = F_v \]
The buoyant force is given by:
\[ F_B = v \rho_0 g \quad \text{where } v \text{ is the volume of the ball.} \]
Force balance equation becomes:
\[ F_v = mg - v \rho_0 g \]
Substituting $v = \frac{m}{\rho}$ (volume in terms of mass and density):
\[ \Rightarrow F_v = mg - \frac{m}{\rho} \rho_0 g \]
\[ F_v = mg \left( 1 - \frac{\rho_0}{\rho} \right) \]
Approach Solution -2
To determine the viscous force acting on a small ball moving in a viscous liquid, we need to understand the forces in play when the ball achieves its terminal velocity. At this point, the net force acting on the ball becomes zero as it falls with a constant velocity.
Let's analyze the forces:
- Gravitational Force: This force acts downward and is given by \(mg\), where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity.
- Buoyant Force: This force acts upward and is given by \(\rho_0 V g\), where \(\rho_0\) is the density of the liquid, \(V\) is the volume of the ball, and \(g\) is the acceleration due to gravity.
- Viscous Force: This is the force we need to find, and it acts upward to balance the other two forces when the ball reaches terminal velocity.
According to Archimedes' principle, the buoyant force is equal to the weight of the liquid displaced by the ball. Since the ball attains terminal velocity, the net force on the ball is zero, hence:
\(F_{\text{viscous}} + \rho_0 V g = mg\)
Solving this equation for the viscous force \(F_{\text{viscous}}\):
\(F_{\text{viscous}} = mg - \rho_0 V g\)
Since the volume \(V\) of the ball can be expressed as \(\frac{m}{\rho}\) (where \(\rho\) is the density of the ball), we substitute this into the equation:
\(F_{\text{viscous}} = mg - \rho_0 \left(\frac{m}{\rho}\right) g\)
\(F_{\text{viscous}} = mg \left( 1 - \frac{\rho_0}{\rho} \right)\)
Thus, the correct expression for the viscous force is \(mg \left( 1 - \frac{\rho_0}{\rho} \right)\).
The correct choice is:
$mg \left( 1 - \frac{\rho_0}{\rho} \right)$
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