“`html

Physics Numericals

Q1: A hammer exerts a force of 1.5 N on each of the two nails A and B. The area of cross-section of tip of nail A is 2 mm² while that of nail B is 6 mm². Calculate pressure on each nail in pascal.

Solution:
Given:
( F = 1.5 , text{N} )
Area of nail A, ( A_A = 2 , text{mm}^2 = 2 times 10^{-6} , text{m}^2 )
Area of nail B, ( A_B = 6 , text{mm}^2 = 6 times 10^{-6} , text{m}^2 )
Formula: ( P = frac{F}{A} )
Calculations:
Pressure on nail A: ( P_A = frac{1.5}{2 times 10^{-6}} = 7.5 times 10^5 , text{Pa} )
Pressure on nail B: ( P_B = frac{1.5}{6 times 10^{-6}} = 2.5 times 10^5 , text{Pa} )

Q2: A block of iron of mass 7.5 kg and dimensions 12 cm × 8 cm × 10 cm is kept on a table on its base of side 12 cm × 8 cm. Calculate (a) thrust and (b) pressure exerted on the table top. Take 1 kgf = 10 N.

Solution:
Given:
Mass ( m = 7.5 , text{kg} )
Base area ( A = 12 , text{cm} times 8 , text{cm} = 96 , text{cm}^2 = 96 times 10^{-4} , text{m}^2 )
Acceleration due to gravity ( g = 9.8 , text{m/s}^2 )
Formula:
Thrust ( W = m cdot g )
Pressure ( P = frac{W}{A} )
Calculations:
Thrust: ( W = 7.5 times 9.8 = 75 , text{N} )
Pressure: ( P = frac{75}{96 times 10^{-4}} = 7812.5 , text{Pa} )

Q3: A vessel contains water up to a height of 1.5 m. Taking the density of water ( 10^3 , text{kg/m}^3 ), acceleration due to gravity ( 9.8 , text{m/s}^2 ) and area of base of vessel 100 cm², calculate: (a) the pressure and (b) the thrust at the base of vessel.

Solution:
Given:
Height ( h = 1.5 , text{m} )
Density ( rho = 10^3 , text{kg/m}^3 )
Area of base ( A = 100 , text{cm}^2 = 100 times 10^{-4} , text{m}^2 )
Gravitational acceleration ( g = 9.8 , text{m/s}^2 )
Formula:
Pressure ( P = rho cdot g cdot h )
Thrust ( F = P cdot A )
Calculations:
Pressure: ( P = 10^3 cdot 9.8 cdot 1.5 = 1.47 times 10^4 , text{Pa} )
Thrust: ( F = 1.47 times 10^4 cdot 100 times 10^{-4} = 147 , text{N} )

Q4: The area of base of a cylindrical vessel is 300 cm². Water (density ( 10^3 , text{kg/m}^3 )) is poured into it up to a depth of 6 cm. Calculate: (a) the pressure and (b) the thrust of water on the base. (( g = 10 , text{m/s}^2 ))

Solution:
Given:
Area ( A = 300 , text{cm}^2 = 300 times 10^{-4} , text{m}^2 )
Depth ( h = 6 , text{cm} = 0.06 , text{m} )
Density ( rho = 10^3 , text{kg/m}^3 )
Gravitational acceleration ( g = 10 , text{m/s}^2 )
Formula:
Pressure ( P = rho cdot g cdot h )
Thrust ( F = P cdot A )
Calculations:
Pressure: ( P = 10^3 cdot 10 cdot 0.06 = 600 , text{Pa} )
Thrust: ( F = 600 cdot 300 times 10^{-4} = 18 , text{N} )

Q5: Calculate the height of a water column which will exert on its base the same pressure as the 70 cm column of mercury. Density of mercury is ( 13.6 , text{g/cm}^3 ).

Solution:
Given:
Height of mercury ( h_{text{mercury}} = 70 , text{cm} = 0.7 , 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{mercury}} cdot g cdot h_{text{mercury}} = rho_{text{water}} cdot g cdot h_{text{water}} )
( h_{text{water}} = frac{rho_{text{mercury}} cdot h_{text{mercury}}}{rho_{text{water}}} )
Calculation:
( h_{text{water}} = frac{13.6 cdot 0.7}{1} = 9.52 , text{m} )

“`