Problem 58 - Master Resource Book in JEE Main Physics

Question

A wide jar is filled with glycerine having specific gravity \(1.26\), in this jar, a steel ball of radius \(0.25 \mathrm{~cm}\) has been dropped. After some time it has been observed that ball is taking equal interval of time \((1.8 \mathrm{~s})\) to cover equal successive distances, of \(20 \mathrm{~cm} .\) [Take, \(\left.\rho_{\text {steel }}=7.8 \times 10^{3} \mathrm{~kg}-\mathrm{m}^{3}, g=9.81 \mathrm{~ms}^{-2}\right]\). The viscosity of glycerine is [in \(\mathrm{N}-\mathrm{sm}^{-2}\) ] (a) \(0.802\) (b) \(1.67\) (c) \(0.76\) (d) \(0.963\)

Step-by-Step Solution

Answer

The viscosity of glycerine is 0.963 N-s/m² (option d).

Step 1: Define the problem

We need to find the viscosity of glycerine when a steel ball moves through it at constant velocity, indicated by equal time intervals to cover equal distances.

Step 2: Identify known variables

- Radius of steel ball, \( r = 0.25 \) cm = \( 0.0025 \) m- Density of steel ball, \( \rho_{\text{steel}} = 7.8 \times 10^3 \) kg/m³- Specific gravity of glycerine = 1.26, thus \( \rho_{\text{glycerine}} = 1.26 \times 1000 \) kg/m³- Gravitational acceleration, \( g = 9.81 \) m/s²- Terminal velocity, \( v_t \) is found using distance and time (20 cm in 1.8 seconds). Thus, \( v_t = \frac{0.20}{1.8} \) m/s.

Step 3: Apply the Terminal Velocity Equation

When the ball reaches terminal velocity, the net force is zero. This yields:\[ 6 \pi \eta r v_t = \frac{4}{3} \pi r^3 (\rho_{\text{steel}} - \rho_{\text{glycerine}}) g \]where \( \eta \) is the viscosity of glycerine. Solving for \( \eta \), we get:\[ \eta = \frac{ (2/9) (r^2) (\rho_{\text{steel}} - \rho_{\text{glycerine}}) g } {v_t} \]

Step 4: Calculate Terminal Velocity

Using the distance and time, \( v_t = \frac{0.20}{1.8} \approx 0.1111 \) m/s.

Step 5: Plug in values and solve for viscosity

Substitute the known values into the formula for viscosity:\[ \eta = \frac{ (2/9) (0.0025)^2 (7800 - 1260) \times 9.81 }{ 0.1111 } \]Solve for \( \eta \):\[ \eta = \frac{ (2/9) (0.00000625) (6540) (9.81) }{ 0.1111 } \approx 0.963 \text{ N-s/m}^2 \]

Step 6: Match solution to options

The calculated viscosity \( \eta = 0.963 \) N-s/m², which matches option (d). Thus, the correct answer is (d) 0.963.

Key Concepts

terminal velocity

Terminal velocity is the constant speed that an object reaches when the force of gravity pulling it downward is balanced by the frictional force of the medium through which it moves. In this case, a steel ball falling through glycerine reaches terminal velocity when the gravitational force is equal to the viscous drag force of the glycerine. At this point, there is no net force acting on the ball, thus it moves with a constant velocity.

In our exercise, the ball reaches terminal velocity as it travels equal distances in equal time intervals. We calculate the terminal velocity using the given distance and time: \[ v_t = \frac{0.20}{1.8} \text{ m/s} \approx 0.1111 \text{ m/s}. \]The understanding of terminal velocity is crucial in determining the viscosity of glycerine using Stokes' Law, as it provides the constant rate of fall needed for calculations.

specific gravity

Specific gravity is a measure of the density of a substance compared to the density of water. It is a dimensionless quantity and helps in determining how heavy a liquid is compared to water. In our exercise, glycerine has a specific gravity of 1.26, meaning it is 1.26 times as dense as water.

Specific gravity is important when calculating the density of glycerine, which will then be used in determining the viscosity through Stokes' Law:\[ \rho_{\text{glycerine}} = 1.26 \times 1000 \text{ kg/m}^3. \]This concept is important in hydrodynamics and fluid mechanics, where the behavior of different fluids in relation to one another and to solids is studied.

hydrodynamics

Hydrodynamics deals with the movement and behavior of fluids in motion. It is a crucial area of study for understanding how different forces act upon objects moving through fluids.

In our scenario, hydrodynamics helps us understand how the steel ball interacts with the glycerine as it falls under gravity. The interplay of forces -- such as gravitational force and viscous drag -- needs to be balanced out for the ball to reach terminal velocity. This balance of forces is what hydrodynamics seeks to comprehend so we can predict the motion of objects through various fluids.

• The ball's density compared to the fluid is essential for determining its falling speed.
• The viscosity of glycerine affects how quickly the ball reaches terminal velocity.

Understanding hydrodynamics allows for precise calculations needed in many engineering and scientific applications, such as designing objects that move through fluids effectively.

Stokes' Law

Stokes' Law offers an essential mathematical relationship that allows us to calculate the viscosity of a fluid by considering the forces on a sphere moving through it.

The law states that the viscous drag force experienced by a small sphere moving through a fluid at steady state (constant velocity) is:\[ F_d = 6 \pi \eta r v_t, \]where:

• \( F_d \) is the drag force,
• \( \eta \) is the viscosity,
• \( r \) is the radius of the sphere,
• and \( v_t \) is the terminal velocity.

In our exercise, to find the viscosity \( \eta \), we rearrange this equation using known values and solve it based on the condition that the viscous force matches the gravitational settling force.

This technique allows scientists and engineers to measure and compare the flow characteristics of different fluids, making Stokes' Law a fundamental concept in fluid dynamics.