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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}
]

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