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 \(8\,\text{m}\) respectively. If a body of mass \(2\,\text{kg}\) is placed on piston \(P_1\), then the force on piston \(P_2\) is \((\text{Ignore atmospheric pressure, acceleration due to gravity}=10\,\text{m s}^{-2})\)
A hydraulic lift is shown in the figure. The radii of the movable pistons \(P_1\) and \(P_2\) are \(2\,\text{m}\) and \(8\,\text{m}\) respectively. If a body of mass \(2\,\text{kg}\) is placed on piston \(P_1\), then the force on piston \(P_2\) is \((\text{Ignore atmospheric pressure, acceleration due to gravity}=10\,\text{m s}^{-2})\)
Show Hint
In a hydraulic lift,
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}.
\]
Since \(A=\pi r^2\), the force multiplication factor is
\[
\frac{F_2}{F_1}=\frac{r_2^2}{r_1^2}.
\]
Updated On: Jun 18, 2026
- \(320\,\text{N}\)
- \(80\,\text{N}\)
- \(1280\,\text{N}\)
- \(20\,\text{N}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Use Pascal's law.
In a hydraulic lift, pressure applied at one piston is transmitted equally throughout the liquid.
Therefore, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] where \(F_1\) is the force on piston \(P_1\), \(F_2\) is the force on piston \(P_2\), and \(A_1,A_2\) are their respective areas.
Step 2: Calculate the force on piston \(P_1\).
A body of mass \[ m=2\,\text{kg} \] is placed on piston \(P_1\).
So the force on \(P_1\) is its weight: \[ F_1=mg \] \[ F_1=2\times 10 \] \[ F_1=20\,\text{N} \]
Step 3: Calculate the ratio of areas.
The radius of piston \(P_1\) is \[ r_1=2\,\text{m} \] and the radius of piston \(P_2\) is \[ r_2=8\,\text{m}. \] Since area of a circular piston is \[ A=\pi r^2, \] we get \[ \frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} \] \[ = \frac{8^2}{2^2} \] \[ = \frac{64}{4} \] \[ =16 \]
Step 4: Find the force on piston \(P_2\).
From Pascal's law, \[ F_2=F_1\frac{A_2}{A_1} \] \[ F_2=20\times 16 \] \[ F_2=320\,\text{N} \]
Step 5: Final conclusion.
Therefore, the force on piston \(P_2\) is \[ \boxed{320\,\text{N}} \]
In a hydraulic lift, pressure applied at one piston is transmitted equally throughout the liquid.
Therefore, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] where \(F_1\) is the force on piston \(P_1\), \(F_2\) is the force on piston \(P_2\), and \(A_1,A_2\) are their respective areas.
Step 2: Calculate the force on piston \(P_1\).
A body of mass \[ m=2\,\text{kg} \] is placed on piston \(P_1\).
So the force on \(P_1\) is its weight: \[ F_1=mg \] \[ F_1=2\times 10 \] \[ F_1=20\,\text{N} \]
Step 3: Calculate the ratio of areas.
The radius of piston \(P_1\) is \[ r_1=2\,\text{m} \] and the radius of piston \(P_2\) is \[ r_2=8\,\text{m}. \] Since area of a circular piston is \[ A=\pi r^2, \] we get \[ \frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} \] \[ = \frac{8^2}{2^2} \] \[ = \frac{64}{4} \] \[ =16 \]
Step 4: Find the force on piston \(P_2\).
From Pascal's law, \[ F_2=F_1\frac{A_2}{A_1} \] \[ F_2=20\times 16 \] \[ F_2=320\,\text{N} \]
Step 5: Final conclusion.
Therefore, the force on piston \(P_2\) is \[ \boxed{320\,\text{N}} \]
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