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Physics Numericals

Q11: (a) The diameter of neck and bottom of a bottle are 2 cm and 10 cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf, what force is exerted on the bottom of the bottle?

Solution: Given: Diameter of neck: \( d_{\text{neck}} = 2 \, \text{cm} \), Radius: \( r_{\text{neck}} = \frac{d_{\text{neck}}}{2} = 1 \, \text{cm} \) Diameter of bottom: \( d_{\text{bottom}} = 10 \, \text{cm} \), Radius: \( r_{\text{bottom}} = \frac{d_{\text{bottom}}}{2} = 5 \, \text{cm} \) Force applied: \( F_{\text{neck}} = 1.2 \, \text{kgf} = 1.2 \times 10 \, \text{N} = 12 \, \text{N} \) Formula: \[ \frac{F_{\text{neck}}}{A_{\text{neck}}} = \frac{F_{\text{bottom}}}{A_{\text{bottom}}} \] Area: \( A = \pi r^2 \) Calculation: Area of neck: \( A_{\text{neck}} = \pi (1)^2 = \pi \, \text{cm}^2 \) Area of bottom: \( A_{\text{bottom}} = \pi (5)^2 = 25\pi \, \text{cm}^2 \) \[ F_{\text{bottom}} = \frac{F_{\text{neck}} \cdot A_{\text{bottom}}}{A_{\text{neck}}} = \frac{12 \cdot 25\pi}{\pi} = 300 \, \text{N} \]

(b) Name the law/principle you have used to find the force in part (a).

Answer: Pascal’s Law.

Q12: A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.

Solution: Given: Force on smaller piston: \( F_1 = 50 \, \text{kgf} = 50 \times 10 = 500 \, \text{N} \) Diameter of smaller piston: \( d_1 = 5 \, \text{cm} \), Radius: \( r_1 = \frac{d_1}{2} = 2.5 \, \text{cm} \) Diameter of larger piston: \( d_2 = 25 \, \text{cm} \), Radius: \( r_2 = \frac{d_2}{2} = 12.5 \, \text{cm} \) Formula: \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \] Area: \( A = \pi r^2 \) Calculation: Area of smaller piston: \( A_1 = \pi (2.5)^2 = 6.25\pi \, \text{cm}^2 \) Area of larger piston: \( A_2 = \pi (12.5)^2 = 156.25\pi \, \text{cm}^2 \) \[ F_2 = \frac{F_1 \cdot A_2}{A_1} = \frac{500 \cdot 156.25\pi}{6.25\pi} = 12,500 \, \text{N} \]

Q13: Two cylindrical vessels fitted with pistons A and B of area of cross-section 8 cm² and 320 cm² respectively, are joined at their bottom by a tube and they are completely filled with water. When a mass of 4 kg is placed on piston A, find: (i) the pressure on piston A, (ii) the pressure on piston B, and (iii) the thrust on piston B.

Solution: Given: Mass on piston A: \( m = 4 \, \text{kg} \) Area of piston A: \( A_A = 8 \, \text{cm}^2 = 8 \times 10^{-4} \, \text{m}^2 \) Area of piston B: \( A_B = 320 \, \text{cm}^2 = 320 \times 10^{-4} \, \text{m}^2 \) Gravitational acceleration: \( g = 10 \, \text{m/s}^2 \) Formula: Pressure: \( P = \frac{F}{A} \), Thrust: \( F = P \cdot A \) Calculations: Force on piston A: \( F_A = m \cdot g = 4 \cdot 10 = 40 \, \text{N} \) (i) Pressure on piston A: \( P_A = \frac{F_A}{A_A} = \frac{40}{8 \times 10^{-4}} = 0.5 \, \text{kgf/cm}^2 \) (ii) Pressure on piston B (same as \( P_A \) since they are connected): \( P_B = 0.5 \, \text{kgf/cm}^2 \) (iii) Thrust on piston B: \( F_B = P_B \cdot A_B = 0.5 \cdot 320 = 160 \, \text{N} \)

Q14: What force is applied on a piston of area of cross-section 2 cm² to obtain a force of 150 N on the piston of area of cross-section 12 cm² in a hydraulic machine?

Solution: Given: Force on larger piston: \( F_2 = 150 \, \text{N} \) Area of larger piston: \( A_2 = 12 \, \text{cm}^2 \) Area of smaller piston: \( A_1 = 2 \, \text{cm}^2 \) Formula: \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \] Calculation: \[ F_1 = \frac{F_2 \cdot A_1}{A_2} = \frac{150 \cdot 2}{12} = 25 \, \text{N} \] “`