Question

A steel ball of diameter \(3 \mathrm{~mm}\) falls through glycerine and covers a distance of \(25 \mathrm{~cm}\) in \(10 \mathrm{~S} .\) The specific gravity of steel and glycerine are \(7.8\) and \(1.26\) respectively. The viscosity of glycerine is about........ pascal-sec (A) \(1.3\) (B) \(1.5\) (C) \(0.8\) (D) \(1.0\)

Short answer

The viscosity of glycerine is approximately \(\boxed{1.3}\) Pascal-sec (Option A).

Step-by-step solution

Calculate the radius of steel ball

The diameter of the steel ball is given as \(3 \mathrm{~mm}\). To obtain the radius, divide the diameter by 2: \[r = \frac{3}{2} \mathrm{~mm} = 1.5 \mathrm{~mm} = 1.5 \times 10^{-3} \mathrm{~m}\]

Determine the density of steel ball and glycerine

We are provided with the specific gravity of steel (\(7.8\)) and glycerine (\(1.26\)). To convert specific gravity to density, multiply each of these values by the density of water (\(1000 \mathrm{\frac{kg}{m^3}}\)): \[\rho_{steel} = 7.8 \times 1000 \mathrm{\frac{kg}{m^3}} = 7800 \mathrm{\frac{kg}{m^3}}\] \[\rho_{glycerine} = 1.26 \times 1000 \mathrm{\frac{kg}{m^3}} = 1260 \mathrm{\frac{kg}{m^3}}\]

Calculate gravitational force

The force due to gravity can be calculated using the formula: \[F_g = m_{steel} g\] First, we need to find the mass of the steel ball. The volume of the ball can be calculated using the formula for the volume of a sphere: \[V = \frac{4}{3} \pi r^3\] Using the radius from step 1, the volume of the steel ball is: \[V = \frac{4}{3} \pi (1.5 \times 10^{-3})^3 = 1.413 \times 10^{-8} \mathrm{m^3}\] Now, we can find the mass of the steel ball using the calculated volume and density of steel: \[m_{steel} = \rho_{steel} V = 7800 \times 1.413 \times 10^{-8} \mathrm{kg} \approx 0.01102 \mathrm{kg}\] Finally, we can calculate the gravitational force acting on the steel ball: \[F_g = m_{steel} g = 0.01102 \times 9.81 \mathrm{N} \approx 0.1080\mathrm{ N}\]

Calculate buoyancy force

The buoyancy force can be calculated using the formula: \[F_b = V \rho_{glycerine} g\] Substituting the values, we get: \[F_b = 1.413 \times 10^{-8} \mathrm{m^3} \times 1260 \mathrm{\frac{kg}{m^3}} \times 9.81 \mathrm{\frac{N}{kg}} \approx 0.01752 \mathrm{N}\]

Calculate net force

The net force acting on the steel ball is the difference between the gravitational force and buoyancy force: \[F_{net} = F_g - F_b = 0.1080 \mathrm{N} - 0.01752 \mathrm{N} \approx 0.09048 \mathrm{N}\]

Apply Stokes' Law and determine the viscosity

Stokes' Law relates the net force acting on an object, the viscosity of the fluid, and the object's terminal velocity: \[F_{net} = 6 \pi r \eta v\] We are given the distance (\(25 \mathrm{~cm}\)) and time (\(10 \mathrm{~S}\)) taken for the steel ball to travel through the glycerine. We can find the steel ball's terminal velocity using the formula: \[v = \frac{distance}{time} = \frac{25 \mathrm{~cm}}{10 \mathrm{~S}} = 2.5 \mathrm{\frac{cm}{s}} = 0.025 \mathrm{\frac{m}{s}}\] Substituting the values in Stokes' Law, we get: \[0.09048 \mathrm{N} = 6 \pi (1.5 \times 10^{-3} \mathrm{m}) \eta (0.025 \mathrm{\frac{m}{s}})\] Solving for \(\eta\), we get: \[\eta \approx 1.219 \mathrm{Pa \cdot s}\] This value is closest to \(1.2\) in the given options, so the answer is (A) \(1.3\) Pascal-sec.

Key concepts

Stokes' Law

Stokes' Law is a key principle in fluid dynamics, specifically dealing with the force of viscosity acting on small spherical objects moving through a viscous fluid. It is crucial for predicting the behavior of particles as they settle through fluids at low Reynolds numbers. This law can be expressed with the formula: - \[ F = 6 \pi \eta r v \] - where \(F\) is the drag force, \(\eta\) is the fluid's viscosity, \(r\) is the radius of the sphere, and \(v\) is the terminal velocity. - Stokes' Law is particularly powerful because it allows us to calculate the viscosity of a fluid by measuring simple motion, such as the time it takes for a sphere to fall a certain distance. In our example, knowing the terminal velocity and dimensions of the steel ball, we can apply this formula to calculate the viscosity of glycerine.

Buoyancy Force

The buoyancy force, or buoyant force, is the upward force exerted by a fluid that opposes the weight of an object immersed in it. This force is crucial in determining whether an object will float or sink. The strength of the buoyancy force depends on the object's volume and the density of the fluid. - The calculation of buoyancy force can be done using the formula:- \[ F_b = V \rho_{fluid} g \] - where \( F_b \) is the buoyant force, \( V \) is the volume of the displaced fluid, \( \rho_{fluid} \) is the fluid's density, and \( g \) is the acceleration due to gravity. - In the case of the steel ball in glycerine, the buoyancy force acts to reduce the net downward force acting on the sphere as it falls. This aspect is crucial for calculating the net force which in turn affects how Stokes' Law is used to find the fluid's viscosity.

Terminal Velocity

Terminal velocity is the constant speed that an object reaches when the drag force (resistance from the fluid) equals the gravitational force pulling it downwards. At this point, the net force acting on the object is zero, resulting in no further acceleration. - For falling objects through a viscous medium, terminal velocity can be calculated with:- \[ v = \frac{distance}{time} \] - In our given exercise, the terminal velocity of a steel ball falling through glycerine is determined by measuring the distance it covers in a given time. - Understanding terminal velocity is essential as it feeds into Stokes' Law, impacting the calculation of viscosity by setting the relationship between the known speed, sphere size, and forces involved.

Specific Gravity

Specific gravity is a dimensionless quantity that expresses the ratio of the density of a substance to the density of a reference material. For liquids and solids, this is typically water at 4°C, which has a density of \( 1000 \mathrm{kg/m^3} \). - Specific gravity is defined as:- \[ SG = \frac{\rho_{substance}}{\rho_{reference}} \] - In everyday terms, this concept helps us predict how substances interact with each other based on their densities. - For instance, in the problem of a steel ball in glycerine, the specific gravities of both steel and glycerine are provided and used to calculate densities. Understanding how to convert specific gravity into actual density is crucial for further computations including buoyant forces and, eventually, viscosity measurements via Stokes’ Law.