Question:
A hydraulic lift is shown in the figure. The movable pistons \(A\), \(B\) and \(C\) are of radius \(10\,\text{cm}\), \(100\,\text{cm}\) and \(5\,\text{m}\) respectively. If a body of mass \(2\,\text{kg}\) is placed on piston \(A\), the maximum masses that can be lifted by piston \(B\) and \(C\) are respectively
A hydraulic lift is shown in the figure. The movable pistons \(A\), \(B\) and \(C\) are of radius \(10\,\text{cm}\), \(100\,\text{cm}\) and \(5\,\text{m}\) respectively. If a body of mass \(2\,\text{kg}\) is placed on piston \(A\), the maximum masses that can be lifted by piston \(B\) and \(C\) are respectively
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In a hydraulic lift, pressure remains same in all pistons:
\[
\frac{F}{A}=\text{constant}
\]
Since \(A=\pi r^2\), the lifted mass is proportional to \(r^2\).
Updated On: Jun 22, 2026
- \(200\,\text{kg}\) and \(500\,\text{kg}\)
- \(20\,\text{kg}\) and \(50\,\text{kg}\)
- \(200\,\text{kg}\) and \(5000\,\text{kg}\)
- \(2000\,\text{kg}\) and \(5000\,\text{kg}\)
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The Correct Option is C
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, \[ \frac{F_A}{A_A}=\frac{F_B}{A_B}=\frac{F_C}{A_C} \] Since \[ F=mg, \] we can write \[ \frac{m_Ag}{A_A}=\frac{m_Bg}{A_B}=\frac{m_Cg}{A_C} \] Cancelling \(g\), \[ \frac{m_A}{A_A}=\frac{m_B}{A_B}=\frac{m_C}{A_C} \]
Step 2: Use area of circular piston.
Area of a circular piston is \[ A=\pi r^2 \] Thus, mass lifted is proportional to the area of piston: \[ m\propto r^2 \]
Step 3: Calculate mass lifted by piston \(B\).
Given, \[ m_A=2\,\text{kg} \] Radius of piston \(A\): \[ r_A=10\,\text{cm} \] Radius of piston \(B\): \[ r_B=100\,\text{cm} \] Using \[ \frac{m_B}{m_A}=\frac{r_B^2}{r_A^2} \] \[ \frac{m_B}{2}=\frac{(100)^2}{(10)^2} \] \[ \frac{m_B}{2}=\frac{10000}{100} \] \[ \frac{m_B}{2}=100 \] \[ m_B=200\,\text{kg} \]
Step 4: Calculate mass lifted by piston \(C\).
Radius of piston \(C\): \[ r_C=5\,\text{m}=500\,\text{cm} \] Using \[ \frac{m_C}{m_A}=\frac{r_C^2}{r_A^2} \] \[ \frac{m_C}{2}=\frac{(500)^2}{(10)^2} \] \[ \frac{m_C}{2}=\frac{250000}{100} \] \[ \frac{m_C}{2}=2500 \] \[ m_C=5000\,\text{kg} \]
Step 5: Final conclusion.
Hence, the maximum masses lifted by pistons \(B\) and \(C\) are respectively \[ \boxed{200\,\text{kg} \text{ and } 5000\,\text{kg}} \]
In a hydraulic lift, pressure applied at one piston is transmitted equally throughout the liquid.
Therefore, \[ \frac{F_A}{A_A}=\frac{F_B}{A_B}=\frac{F_C}{A_C} \] Since \[ F=mg, \] we can write \[ \frac{m_Ag}{A_A}=\frac{m_Bg}{A_B}=\frac{m_Cg}{A_C} \] Cancelling \(g\), \[ \frac{m_A}{A_A}=\frac{m_B}{A_B}=\frac{m_C}{A_C} \]
Step 2: Use area of circular piston.
Area of a circular piston is \[ A=\pi r^2 \] Thus, mass lifted is proportional to the area of piston: \[ m\propto r^2 \]
Step 3: Calculate mass lifted by piston \(B\).
Given, \[ m_A=2\,\text{kg} \] Radius of piston \(A\): \[ r_A=10\,\text{cm} \] Radius of piston \(B\): \[ r_B=100\,\text{cm} \] Using \[ \frac{m_B}{m_A}=\frac{r_B^2}{r_A^2} \] \[ \frac{m_B}{2}=\frac{(100)^2}{(10)^2} \] \[ \frac{m_B}{2}=\frac{10000}{100} \] \[ \frac{m_B}{2}=100 \] \[ m_B=200\,\text{kg} \]
Step 4: Calculate mass lifted by piston \(C\).
Radius of piston \(C\): \[ r_C=5\,\text{m}=500\,\text{cm} \] Using \[ \frac{m_C}{m_A}=\frac{r_C^2}{r_A^2} \] \[ \frac{m_C}{2}=\frac{(500)^2}{(10)^2} \] \[ \frac{m_C}{2}=\frac{250000}{100} \] \[ \frac{m_C}{2}=2500 \] \[ m_C=5000\,\text{kg} \]
Step 5: Final conclusion.
Hence, the maximum masses lifted by pistons \(B\) and \(C\) are respectively \[ \boxed{200\,\text{kg} \text{ and } 5000\,\text{kg}} \]
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