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

Answer 2

Step 1: Write down the given data
Radius of tube \( r = 2\,\text{cm} = 0.02\,\text{m} \)
Height of mercury column \( h = 30\,\text{cm} = 0.3\,\text{m} \)
Density of mercury \( \rho = 1.36 \times 10^4\,\text{kg/m}^3 \)
Acceleration due to gravity \( g = 10\,\text{m/s}^2 \)
Atmospheric pressure \( P_0 = 10^5\,\text{N/m}^2 \)
\(\pi = \dfrac{22}{7}\)

Step 2: Determine total pressure on the bottom surface
The total pressure at the bottom of the mercury column is the sum of atmospheric pressure and the hydrostatic pressure due to the mercury column: \[ P_{\text{total}} = P_0 + \rho g h \] Substitute the values: \[ P_{\text{total}} = 10^5 + (1.36 \times 10^4)(10)(0.3) \] \[ P_{\text{total}} = 10^5 + 4.08 \times 10^4 = 1.408 \times 10^5\,\text{N/m}^2 \]

Step 3: Calculate the area of the bottom of the tube
The cross-sectional area \( A = \pi r^2 \) \[ A = \frac{22}{7} \times (0.02)^2 = \frac{22}{7} \times 0.0004 = 0.001257\,\text{m}^2 \]

Step 4: Find the total force on the bottom
Force exerted by mercury on the bottom surface: \[ F = P_{\text{total}} \times A \] \[ F = (1.408 \times 10^5) \times 0.001257 \] \[ F = 177\,\text{N (approximately)} \]

Step 5: Physical explanation
The total force on the bottom comes from both the weight of the mercury column and the atmosphere pressing on the top surface. If only gauge pressure (excluding atmospheric pressure) were considered, the answer would be smaller, but since the question asks for the total force exerted, atmospheric pressure must be included.

Final answer

177