Question

Mercury is filled in a tube of radius \(2 \, \text{cm}\) up to a height of \(30 \, \text{cm}\). The force exerted by mercury on the bottom of the tube is ___N.
(Given, atmospheric pressure = \(10^5 \, \text{N/m}^2\), density of mercury = \(1.36 \times 10^4 \, \text{kg/m}^3\), \(g = 10 \, \text{m/s}^2\), \(\pi = \frac{22}{7}\))

Answer 1

Correct Answer: 177

The force exerted by mercury on the bottom of the tube can be calculated as:
\[ F = P_0A + \rho_m ghA \] 
where: 
- \( P_0 = 10^5 \, \text{Nm}^{-2} \) (atmospheric pressure), 
- \( A = \pi r^2 = \frac{22}{7} \times (2 \times 10^{-2})^2 \) (area of the base), 
- \( \rho_m = 1.36 \times 10^4 \, \text{kg m}^{-3} \) (density of mercury), 
- \( g = 10 \, \text{ms}^{-2} \), 
- \( h = 30 \times 10^{-2} \, \text{m} \) (height of mercury column). 
Calculating \( A \): 
\[ A = \frac{22}{7} \times (2 \times 10^{-2})^2 \] 
Substitute into the force equation: 
\[ F = 10^5 \times \frac{22}{7} \times (2 \times 10^{-2})^2 + 1.36 \times 10^4 \times 10 \times (30 \times 10^{-2}) \times \frac{22}{7} \times (2 \times 10^{-2})^2 \] 
Solving this: 
\[ F = 51.29 + 125.71 = 177 \, \text{N}. \]