Question:
In a hydraulic lift, compressed air exerts a force \(F\) on a small piston of radius \(3\;cm\). Due to this pressure, the second piston of radius \(5\;cm\) lifts a load of \(1875\;kg\). The value of \(F\) is
Take acceleration due to gravity \(g=10\;\text{m s}^{-2}\).
In a hydraulic lift, compressed air exerts a force \(F\) on a small piston of radius \(3\;cm\). Due to this pressure, the second piston of radius \(5\;cm\) lifts a load of \(1875\;kg\). The value of \(F\) is
Take acceleration due to gravity \(g=10\;\text{m s}^{-2}\).
Take acceleration due to gravity \(g=10\;\text{m s}^{-2}\).
Show Hint
In a hydraulic lift, pressure remains same in both pistons:
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]
Since \(A\propto r^2\), use the square of the radius ratio.
Updated On: Jun 22, 2026
- \(1250\;N\)
- \(125\;N\)
- \(6750\;N\)
- \(675\;N\)
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The Correct Option is C
Solution and Explanation
Step 1: Use Pascal's law.
In a hydraulic lift, pressure transmitted through the fluid is same everywhere.
Therefore, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] Here, \[ F_1=F \] Small piston radius: \[ r_1=3\;cm \] Large piston radius: \[ r_2=5\;cm \] Load lifted by large piston: \[ m=1875\;kg \] So, \[ F_2=mg \] \[ F_2=1875\times 10 \] \[ F_2=18750\;N \]
Step 2: Use area ratio.
Area of piston is \[ A=\pi r^2 \] Thus, \[ \frac{A_1}{A_2}=\frac{\pi r_1^2}{\pi r_2^2} \] \[ =\frac{r_1^2}{r_2^2} \] \[ =\frac{3^2}{5^2} \] \[ =\frac{9}{25} \] From Pascal's law, \[ F=F_2\frac{A_1}{A_2} \] \[ F=18750\times \frac{9}{25} \]
Step 3: Calculate \(F\).
\[ F=750\times 9 \] \[ F=6750\;N \]
Step 4: Final conclusion.
Hence, the force applied on the small piston is \[ \boxed{6750\;N} \]
In a hydraulic lift, pressure transmitted through the fluid is same everywhere.
Therefore, \[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \] Here, \[ F_1=F \] Small piston radius: \[ r_1=3\;cm \] Large piston radius: \[ r_2=5\;cm \] Load lifted by large piston: \[ m=1875\;kg \] So, \[ F_2=mg \] \[ F_2=1875\times 10 \] \[ F_2=18750\;N \]
Step 2: Use area ratio.
Area of piston is \[ A=\pi r^2 \] Thus, \[ \frac{A_1}{A_2}=\frac{\pi r_1^2}{\pi r_2^2} \] \[ =\frac{r_1^2}{r_2^2} \] \[ =\frac{3^2}{5^2} \] \[ =\frac{9}{25} \] From Pascal's law, \[ F=F_2\frac{A_1}{A_2} \] \[ F=18750\times \frac{9}{25} \]
Step 3: Calculate \(F\).
\[ F=750\times 9 \] \[ F=6750\;N \]
Step 4: Final conclusion.
Hence, the force applied on the small piston is \[ \boxed{6750\;N} \]
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