KCET 2025 Physics Paper

To determine the coefficient of viscosity of a liquid, a spherical steel ball sinks a distance of $h=0.9 \text{ m}$. The ball's radius is $r=\sqrt{3} \times 10^{-3} \text{ m}$. The time taken by the ball to sink as measured in three trials is shown in the table below:

Trial No.	Time to sink $h$ (in seconds)
1	2.75
2	2.65
3	2.70

The density difference between the steel ball and the liquid is $7000 \text{ kg/m}^3$, and the gravitational acceleration $g=10 \text{ m/s}^2$. We aim to calculate the coefficient of viscosity of the liquid at room temperature.Calculations:Average Time: $t_{\text{avg}}=\frac{2.75+2.65+2.70}{3}=2.7 \text{ seconds}$Terminal Velocity:$\text{Terminal velocity}=\frac{h}{t_{\text{avg}}}=\frac{0.9}{2.7}=\frac{1}{3} \text{ m/s}$ Coefficient of Viscosity:$\eta=\frac{2}{9}\frac{\pi r^2 g(\rho_{\text{steel}}-\rho_{\text{ilquid}})}{v_t}$Substituting the values,$\eta=\frac{2}{9}\times\frac{\pi\times(1.732\times10^{-3})^2\times10\times7000}{\frac{1}{3}}$ Simplifying these terms, we find:$\eta=0.14 \,\text{Pa}\cdot\text{s}$The coefficient of viscosity of the liquid is 0.14 Pa.s.