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 \(r^2\).
Step 1: Recall Pascal's law for hydraulic systems. In a hydraulic lift, the pressure is transmitted equally throughout the fluid. So: \[ \frac{F_A}{A_A} = \frac{F_B}{A_B} = \frac{F_C}{A_C} \] where $F = mg$ is the weight and $A = \pi r^2$ is the piston area. Step 2: Write masses in terms of areas. Since $F = mg$, the relation becomes: \[ \frac{m_A}{A_A} = \frac{m_B}{A_B} = \frac{m_C}{A_C} \] So: \[ \frac{m_B}{m_A} = \frac{A_B}{A_A} = \frac{\pi r_B^2}{\pi r_A^2} = \left(\frac{r_B}{r_A}\right)^2 \] Step 3: Find the mass lifted by piston B. $r_A = 10$ cm, $r_B = 100$ cm, $m_A = 2$ kg: \[ m_B = m_A \left(\frac{r_B}{r_A}\right)^2 = 2 \times \left(\frac{100}{10}\right)^2 = 2 \times 100 = 200 \text{ kg} \] Step 4: Find the mass lifted by piston C. $r_C = 5$ m $= 500$ cm: \[ m_C = m_A \left(\frac{r_C}{r_A}\right)^2 = 2 \times \left(\frac{500}{10}\right)^2 = 2 \times 2500 = 5000 \text{ kg} \] Step 5: Verify the logic. The mechanical advantage increases as the square of the radius ratio. Piston B has 10 times the radius of A, so $10^2 = 100$ times the area, lifting 100 times the mass: $2 \times 100 = 200$ kg. Piston C has 50 times the radius of A, so $50^2 = 2500$ times the area: $2 \times 2500 = 5000$ kg. Step 6: State the final answer. The maximum masses lifted by pistons B and C are 200 kg and 5000 kg respectively. \[ \boxed{200 \text{ kg and } 5000 \text{ kg}} \]
