Question:
A hydraulic lift is shown in the figure. The radii of the movable pistons \(P_1\) and \(P_2\) are \(2\,\text{m}\) and \(5\,\text{m}\) respectively. If a block of mass \(x\) is placed on \(P_2\), then the minimum mass that should be kept on \(P_1\) to lift the block on \(P_2\) is
A hydraulic lift is shown in the figure. The radii of the movable pistons \(P_1\) and \(P_2\) are \(2\,\text{m}\) and \(5\,\text{m}\) respectively. If a block of mass \(x\) is placed on \(P_2\), then the minimum mass that should be kept on \(P_1\) to lift the block on \(P_2\) is
Show Hint
In hydraulic lift problems, use Pascal's law:
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]
and remember that piston area is proportional to the square of its radius.
Updated On: Jun 26, 2026
- \(0.4x\)
- \(0.16x\)
- \(0.8x\)
- \(0.25x\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use Pascal's law.
In a hydraulic lift, pressure transmitted through the liquid is same at both pistons.
So, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] Let the minimum mass placed on \(P_1\) be \[ m \] Then, \[ F_1=mg \] and \[ F_2=xg \] Thus, \[ \frac{mg}{A_1}=\frac{xg}{A_2} \] Cancel \(g\): \[ \frac{m}{A_1}=\frac{x}{A_2} \]
Step 2: Use the areas of circular pistons.
Area of a piston is \[ A=\pi r^2 \] For \(P_1\), \[ r_1=2\,\text{m} \] So, \[ A_1=\pi(2)^2=4\pi \] For \(P_2\), \[ r_2=5\,\text{m} \] So, \[ A_2=\pi(5)^2=25\pi \]
Step 3: Find the required mass.
From \[ \frac{m}{A_1}=\frac{x}{A_2}, \] we get \[ m=x\frac{A_1}{A_2} \] \[ m=x\frac{4\pi}{25\pi} \] \[ m=\frac{4}{25}x \] \[ m=0.16x \]
Step 4: Final conclusion.
Hence, the minimum mass that should be kept on \(P_1\) is \[ \boxed{0.16x} \]
In a hydraulic lift, pressure transmitted through the liquid is same at both pistons.
So, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] Let the minimum mass placed on \(P_1\) be \[ m \] Then, \[ F_1=mg \] and \[ F_2=xg \] Thus, \[ \frac{mg}{A_1}=\frac{xg}{A_2} \] Cancel \(g\): \[ \frac{m}{A_1}=\frac{x}{A_2} \]
Step 2: Use the areas of circular pistons.
Area of a piston is \[ A=\pi r^2 \] For \(P_1\), \[ r_1=2\,\text{m} \] So, \[ A_1=\pi(2)^2=4\pi \] For \(P_2\), \[ r_2=5\,\text{m} \] So, \[ A_2=\pi(5)^2=25\pi \]
Step 3: Find the required mass.
From \[ \frac{m}{A_1}=\frac{x}{A_2}, \] we get \[ m=x\frac{A_1}{A_2} \] \[ m=x\frac{4\pi}{25\pi} \] \[ m=\frac{4}{25}x \] \[ m=0.16x \]
Step 4: Final conclusion.
Hence, the minimum mass that should be kept on \(P_1\) is \[ \boxed{0.16x} \]
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