pressure in fluids and atmospheric pressure

“`html Physics Numericals

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:
\( h_{\text{water}} = \frac{13.6 \cdot 0.7}{1} = 9.52 \, \text{m} \)

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