A hydraulic jack is used to compress a spring as shown fig. Stiffness of spring is 10^5 N/m, by applying a pressure 'p' in the hydraulic, the spring gets compressed by 10mm. The cross sectional area of the piston is 25cm^2. The applied pressure 'p' is
- A. 4 \times 10^5 Pascals
- B. 40 Pascals
- C. 250 Pascals
- D. 25 Pascals
Answer: A. 4 \times 10^5 Pascals
Explanation: Force F = kx = 10^5 * 0.01 = 1000 N. Pressure P = F/A = 1000 / (25 * 10^-4) = 40 * 10^4 = 4 * 10^5 Pa.