Question

The movable cylindrical pistons \(P_1\) and \(P_2\) of a hydraulic lift are of radii \(2\ \text{m}\) and \(R\) respectively. A body of mass \(32\ \text{kg}\) on piston \(P_2\) is supported by a body of mass \(2\ \text{kg}\) placed on piston \(P_1\). The value of \(R\) is

• A. \(8\ \text{m}\)
• B. \(32\ \text{m}\)
• C. \(2\ \text{m}\)
• D. \(16\ \text{m}\)

Hint:

In a hydraulic lift, pressures on both pistons are equal: \[ \frac{F_1}{A_1}=\frac{F_2}{A_2}. \] Since piston area is proportional to the square of radius, use \[ A=\pi r^2. \]

Answer 1

The Correct Option is A

Step 1: Use Pascal's law for hydraulic lift.
In a hydraulic lift, pressure is transmitted equally throughout the liquid.
Therefore, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] Here, \[ F_1=2g \] and \[ F_2=32g \] Also, the radius of piston \(P_1\) is \[ r_1=2\ \text{m} \] and the radius of piston \(P_2\) is \[ r_2=R \]

Step 2: Write the areas of the pistons.
Area of piston \(P_1\) is \[ A_1=\pi r_1^2 \] \[ A_1=\pi(2)^2 \] \[ A_1=4\pi \] Area of piston \(P_2\) is \[ A_2=\pi R^2 \]

Step 3: Substitute in Pascal's law.
Using \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] we get \[ \frac{2g}{4\pi}=\frac{32g}{\pi R^2} \] Cancel \(g\) and \(\pi\),
\[ \frac{2}{4}=\frac{32}{R^2} \] \[ \frac{1}{2}=\frac{32}{R^2} \]

Step 4: Solve for \(R\).
Cross multiplying, \[ R^2=64 \] Therefore, \[ R=8\ \text{m} \]

Step 5: Final conclusion.
Hence, the value of \(R\) is \[ \boxed{8\ \text{m}} \]