Problem 79 - Master Resource Book in JEE Main Physics

Question

A metallic sphere of mass \(M\) falls through glycerine with a terminal velcity \(v\). If we drop a ball of mass \(8 M\) of same metal into a column of glycerine, the terminal velocity of the ball will be (a) \(2 \underline{\underline{v}}\) (b) \(4 \underline{\mathrm{v}}\) (c) \(8 v\) (d) \(16 \mathrm{v}\)

Step-by-Step Solution

Answer

The terminal velocity of the ball will be (c) 8v.

Step 1: Understand Terminal Velocity

Terminal velocity is achieved when the gravitational force on the falling object is balanced by the drag force and the buoyant force. For a sphere, we usually use the Stokes' law to describe the drag force in a fluid like glycerine.

Step 2: Determine Forces Involved

The force due to gravity on the sphere can be expressed as \(F_g = Mg\). The drag force opposing this motion can be stated as \(F_d = kv\) based on Stokes' law, where \(k\) is a constant of the fluid and object's properties.

Step 3: Equation of Motion at Terminal Velocity

At terminal velocity, the sum of forces equals zero: \(Mg = kv + F_b\). For simplification, assume buoyant force \(F_b\) remains constant and can be omitted for calculation of proportionality.

Step 4: Construct Equation for First Sphere

Rewriting the equation for the first sphere given it reaches terminal velocity \(v\): \(kv = Mg\).

Step 5: Derive the Equation for Increased Mass Sphere

For the sphere with increased mass \(8M\), the new equation using the terminal velocity \(v'\) is: \(k v' = 8Mg\).

Step 6: Relate Velocities of Different Masses

Since \(kv = Mg\) from Step 4, we can relate \(k v' = 8Mg\), thus substituting gives \(k v' = 8kv\). Removing the constant \(k\) gives, \(v' = 8v\).

Step 7: Conclusion and Choice

Compare the result \(v' = 8v\) with the options provided. The correct choice is (c) \(8v\).

Key Concepts

Metallic Sphere Motion

When we study a metallic sphere moving through a fluid, it's fascinating to consider the forces acting upon it. These forces include gravity pulling the sphere downward, drag force opposing this motion, and buoyancy that slightly counters gravity.
The motion of a metallic sphere, particularly in a dense fluid like glycerine, showcases how different forces interact. As the sphere descends, it accelerates until the drag force and buoyancy fully counter the gravitational pull. This results in a steady velocity called terminal velocity.
Factors affecting the sphere's motion in the fluid include its mass, the fluid's viscosity, and the sphere's diameter. With increased mass, terminal velocity increases, as heavier spheres can overcome greater drag forces.
For identical scenarios with different masses, heavier balls reach higher terminal velocities faster, leading to practical applications like in sedimentation processes or designing objects for safe descent in atmospheric layers.

Stokes' Law

Stokes' Law is crucial here to understand how drag force acts on a sphere moving through a fluid. This law helps us calculate the drag force based on the sphere's velocity, size, and the fluid's viscosity.
Formulated by George Stokes, the law states that the drag force, which is the resistive force opposing the motion, is directly proportional to the velocity of the sphere and can be expressed as:
\[ F_d = 6 \pi \eta rv \]
Here, \( F_d \) is the drag force, \( \eta \) is the fluid's viscosity, \( r \) is the radius of the sphere, and \( v \) is the velocity of the sphere.
Stokes' Law primarily applies to slow-moving spheres in highly viscous fluids like glycerine. Therefore, it's an excellent approximation for our scenario. This ensures that as a metallic sphere rolls down, each factor affecting the drag can be adequately measured and accounted for in our calculations.

Fluid Dynamics

Fluid dynamics is a significant area that investigates how liquids and gases move. For this exercise, it analyses how the metallic sphere interacts with the fluid, glycerine, through which it's falling.
Understanding fluid dynamics allows us to predict how the fluid will behave when a sphere is dropped into it. Factors like turbulence, viscosity, and density all play crucial roles.
In fluids like glycerine, which are dense and viscous, the flow tends to be more streamlined with less turbulence. This means the application of Stokes’ Law is feasible, providing a reliable prediction of terminal velocity.
In fluid dynamics, important equations like the Navier-Stokes equations describe the flow velocity of the fluid. However, for simple, small-scale events involving spheres and fluids, focus usually shifts to straightforward principles like those found in Stokes' Law.

Drag Force Calculation

The calculation of drag force is fundamental in determining how quickly a metallic sphere reaches terminal velocity when falling through a fluid. Drag force depends on factors like fluid viscosity, sphere velocity, and dimensions.
Applying Stokes' Law, drag force can be expressed as:
\[ F_d = 6 \pi \eta rv \]
In this formula, several factors influence the drag:

• Velocity \( v \) of the sphere: The faster it moves, the greater the drag force.
• Viscosity \( \eta \) of the fluid: Higher viscosity results in greater drag.
• Radius \( r \) of the sphere: Larger radii increase the drag.

Understanding drag force is essential for efficient design in industries like aerodynamics, where reducing this force on objects can greatly influence their performance.
In the specific exercise tackled, knowing how to calculate the drag force helped us determine how a more massive sphere achieves a higher terminal velocity by overcoming an increased drag force compared to a lighter sphere.