Question

In a hydraulic press, the small cylinder has a diameter of \(d_{1}\) cm, while the large cylinder has a diameter of \(d_{2}\) cm. If a force \(F_{1}\) is applied to a piston of small cylinder, then the force on the piston of large cylinder \(F_{2}\) is

• A. \(\frac{d_{1}^{2}}{d_{2}^{2}}\frac{1}{F_{1}}\)
• B. \(\frac{d_{1}^{2}}{d_{2}^{2}}F_{1}\)
• C. \(\frac{d_{2}^{2}}{d_{1}^{2}}\frac{1}{F_{1}}\)
• D. \(\frac{d_{2}^{2}}{d_{1}^{2}}F_{1}\)

Hint:

Pressure is transmitted equally, so force is proportional to piston area, which varies as the square of the diameter.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Pascal's law: pressure on both pistons is the same.
\(\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\)

Step 2: Areas
Area is proportional to the square of the diameter: \(\dfrac{A_2}{A_1}=\dfrac{d_2^2}{d_1^2}\).

Step 3: Result
\(F_2=F_1\dfrac{A_2}{A_1}=\dfrac{d_2^2}{d_1^2}F_1\)

Final Answer:
The force on the large piston is d2 squared over d1 squared times F1. \[ \boxed{\text{(D) }\dfrac{d_{2}^{2}}{d_{1}^{2}}F_{1}} \]