Physics Numericals
Q6: The pressure of water on the ground floor is 40,000 Pa, and on the first floor is 10,000 Pa. Find the height of the first floor. (Take: Density of water = \(1000 \, \text{kg/m}^3\), \( g = 10 \, \text{m/s}^2 \)).
Solution:
Given:
Pressure difference: \( \Delta P = 40,000 – 10,000 = 30,000 \, \text{Pa} \)
Density of water: \( \rho = 1000 \, \text{kg/m}^3 \)
Gravitational acceleration: \( g = 10 \, \text{m/s}^2 \)
Formula: \( h = \frac{\Delta P}{\rho \cdot g} \)
Calculation:
\[
h = \frac{30,000}{1000 \cdot 10} = 3 \, \text{m}
\]
Q7: A simple U-tube contains mercury to the same level in both arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in the other arm? (Given: Density of mercury = \(13.6 \times 10^3 \, \text{kg/m}^3\), Density of water = \(10^3 \, \text{kg/m}^3\)).
Solution:
Given:
Height of water: \( h_{\text{water}} = 13.6 \, \text{cm} = 0.136 \, \text{m} \)
Density of mercury: \( \rho_{\text{mercury}} = 13.6 \times 10^3 \, \text{kg/m}^3 \)
Density of water: \( \rho_{\text{water}} = 10^3 \, \text{kg/m}^3 \)
Formula:
\[
\rho_{\text{water}} \cdot h_{\text{water}} = \rho_{\text{mercury}} \cdot h_{\text{mercury}}
\]
Calculation:
\[
h_{\text{mercury}} = \frac{\rho_{\text{water}} \cdot h_{\text{water}}}{\rho_{\text{mercury}}}
\]
\[
h_{\text{mercury}} = \frac{10^3 \cdot 0.136}{13.6 \times 10^3} = 0.01 \, \text{m} = 1 \, \text{cm}
\]
Q8: In a hydraulic machine, a force of 2 N is applied on the piston of area of cross-section 10 cm². What force is obtained on its piston of area of cross-section 100 cm²?
Solution:
Given:
Force applied on smaller piston: \( F_1 = 2 \, \text{N} \)
Area of smaller piston: \( A_1 = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 \)
Area of larger piston: \( A_2 = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 \)
Formula:
\[
\frac{F_1}{A_1} = \frac{F_2}{A_2}
\]
Calculation:
\[
F_2 = \frac{F_1 \cdot A_2}{A_1} = \frac{2 \cdot 100 \times 10^{-4}}{10 \times 10^{-4}} = 20 \, \text{N}
\]
Q9: What should be the ratio of area of cross-section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of its brake shoe by exerting a force of 0.5 N on the pedal?
Solution:
Given:
Force on pedal: \( F_1 = 0.5 \, \text{N} \)
Force on brake shoe: \( F_2 = 15 \, \text{N} \)
Formula:
\[
\frac{A_2}{A_1} = \frac{F_2}{F_1}
\]
Calculation:
\[
\frac{A_2}{A_1} = \frac{15}{0.5} = 30
\]
Ratio: \( 1 : 30 \)
Q10: The areas of pistons in a hydraulic machine are 5 cm² and 625 cm². What force on the smaller piston will support a load of 1250 N on the larger piston? Assume no friction and no leakage of liquid.
Solution:
Given:
Load on larger piston: \( F_2 = 1250 \, \text{N} \)
Area of larger piston: \( A_2 = 625 \, \text{cm}^2 = 625 \times 10^{-4} \, \text{m}^2 \)
Area of smaller piston: \( A_1 = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \)
Formula:
\[
\frac{F_1}{A_1} = \frac{F_2}{A_2}
\]
Calculation:
\[
F_1 = \frac{F_2 \cdot A_1}{A_2} = \frac{1250 \cdot 5 \times 10^{-4}}{625 \times 10^{-4}} = 10 \, \text{N}
\]