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 \(g=10\;\text{m s}^{-2}\).

• A. \(1250\;N\)
• B. \(125\;N\)
• C. \(6750\;N\)
• D. \(675\;N\)

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.

Answer 1

The Correct Option is C

Step 1: State Pascal's principle for the lift.
In a hydraulic lift, the pressure applied on the small piston is transmitted undiminished to the large piston. So \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \] where $F_1 = F$ is the unknown force on the small piston.
Step 2: Find the force on the large piston.
The large piston lifts a load $m = 1875\ kg$, so it must supply \[ F_2 = mg = 1875 \times 10 = 18750\ N \]
Step 3: Express the areas.
Each piston area is $A = \pi r^2$, with $r_1 = 3\ cm$ and $r_2 = 5\ cm$. The ratio is \[ \frac{A_1}{A_2} = \frac{r_1^2}{r_2^2} = \frac{9}{25} \]
Step 4: Rearrange Pascal's relation for $F$.
\[ F_1 = F_2 \cdot \frac{A_1}{A_2} \]
Step 5: Substitute the values.
\[ F = 18750 \times \frac{9}{25} \]
Step 6: Compute the result.
\[ F = 750 \times 9 = 6750\ N \] So the required force on the small piston is $6750\ N$, matching option (3). \[ \boxed{6750\ N} \]