Question

While determining the coefficient of viscosity of the given liquid, a spherical steel ball sinks by a distance \( x = 0.8 \, \text{m} \). The radius of the ball is \( 2.5 \times 10^{-3} \, \text{m} \). The time taken by the ball to sink in three trials are tabulated as shown: 

• A. 14 Pa.s
• B. 0.28 Pa.s
• C. 1.5 Pa.s
• D. 0.14 \( \times 10^3 \) Pa.s

Hint:

To find the viscosity of a liquid using a falling ball, ensure that all parameters like ball radius, time taken, and densities are correctly used in the formula.

Answer 1

The Correct Option is D

The viscosity of the liquid can be calculated using the formula for the time taken by a spherical ball to sink through a fluid:
\[ \eta = \frac{2r^2 (\rho_s - \rho_l) g}{9v} \] where: - \( \eta \) is the viscosity,
- \( r \) is the radius of the ball,
- \( \rho_s \) is the density of the ball,
- \( \rho_l \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( v \) is the velocity of the ball, which is calculated from the time taken to sink.
Using the data provided, we compute the velocity from the given times and then apply the formula. The final result for the coefficient of viscosity is \( 0.14 \times 10^3 \) Pa.s.