Problem 72 - Master Resource Book in JEE Main Physics

Question

A spherical solid ball of volume \(V\) is made of a material of density \(\rho_{1}\). It is falling through a liquid of denaity \(\rho_{2}\left(\rho_{2}>\rho_{1}\right)\). Assume that the liquid applied a viscous force on the ball that is proportional to the square of its speeds \(v\), ie., \(F_{\text {vasous }}=-k v^{2}(k>0)\). The terminal speed of the ball is (a) \(\frac{V g\left(\rho_{1}-p_{2}\right)}{k}\) (b) \(\sqrt{\frac{V_{g\left(\rho_{1}-\beta_{2}\right)}}{k}}\) (c) \(\frac{V g \rho_{1}}{k}\) (d) \(\sqrt{\frac{V g \rho_{1}}{k}}\)

Step-by-Step Solution

Answer

Option (b) \( \sqrt{\frac{Vg(\rho_1-\rho_2)}{k}} \).

Step 1: Understand the Forces

First, identify the forces acting on the falling spherical ball. The ball experiences gravitational force, buoyant force, and viscous force. The gravitational force is given by:\[ F_g = V \rho_1 g \]where \( V \) is the volume of the ball, \( \rho_1 \) is the density of the material, and \( g \) is the acceleration due to gravity. The buoyant force is given by:\[ F_b = V \rho_2 g \]where \( \rho_2 \) is the density of the liquid. The viscous force is:\[ F_v = -k v^2 \]where \( k \) is a positive constant, and \( v \) is the speed of the ball.

Step 2: Applying Newton's Second Law

To find the terminal speed, apply Newton's second law. At terminal speed, the forces should balance, meaning net force is zero. Therefore, the sum of the forces must equal zero. Set up the equation:\[ F_g - F_b - k v_t^2 = 0 \]where \( v_t \) is the terminal speed.

Step 3: Solve the Equation

Substitute the expressions for the forces into the equation:\[ V \rho_1 g - V \rho_2 g - k v_t^2 = 0 \]Simplify by factoring out \( Vg \):\[ Vg (\rho_1 - \rho_2) = k v_t^2 \]

Step 4: Solve for the Terminal Speed

To find the terminal speed \( v_t \), divide both sides of the equation by \( k \) and solve for \( v_t \):\[ v_t^2 = \frac{Vg(\rho_1 - \rho_2)}{k} \]Taking the square root of both sides, we find:\[ v_t = \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}} \]

Step 5: Choose the Correct Answer

Compare the expression \( \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}} \) with the given options. The correct answer is option (b) \( \sqrt{\frac{Vg(\rho_1 - \rho_2)}{k}} \).

Key Concepts

Buoyancy

Buoyancy is the upward force exerted by a fluid that opposes the weight of an object immersed in it. This force is why objects feel lighter in water and can even float. The principle behind buoyancy was discovered by Archimedes, who stated that the buoyant force on an object is equal to the weight of the fluid it displaces.
In the context of our exercise, the ball falls through a liquid, and the buoyant force acting on it can be expressed as \( F_b = V \rho_2 g \), where \( V \) is the volume of the ball, \( \rho_2 \) is the density of the liquid, and \( g \) is the acceleration due to gravity.
This force acts upwards, opposing the gravitational force pulling the ball down. Although this force doesn't change with the speed of the ball, it plays a critical role in determining the net force acting on the ball and thus its terminal velocity.

Viscous Force

Viscous force is a type of resistive force that acts against the motion of objects through a fluid. Unlike buoyancy, this force depends directly on the speed of the object. More specifically, the viscous force in this problem is described by \( F_v = -k v^2 \), a negative sign indicating that the force opposes the direction of velocity.
Here, \( k \) is a constant that represents the fluid's viscosity and how effectively it resists motion, and \( v \) is the velocity of the ball. Because this force is proportional to the square of the velocity, it increases rapidly with faster speeds.
At lower speeds, viscous force may play a smaller role, but as the ball speeds up, the quadratic increase in viscous force becomes significant. This force continues to build up until it balances out the other forces, leading to the terminal velocity of the ball.

Newton's Second Law

Newton's Second Law states that the acceleration of an object is dependent upon the net force acting upon it and the object's mass. Formally, it's expressed as \( F = m \cdot a \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration.
In our exercise about terminal velocity, this law is applied to determine when the forces acting on the falling ball balance out. When the ball reaches terminal velocity, the net force is zero because the upward buoyant and viscous forces perfectly counteract the downward gravitational force.
The equation showing this balance is \( F_g - F_b - k v_t^2 = 0 \), leading to no acceleration or change in speed at terminal velocity. This crucial concept ensures the ball continues to fall at a constant speed, highlighting the interaction of different forces in a fluid dynamic environment.