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: 
$$F = 51.29 + 125.71 = 177 \, \text{N}.$$