The cross - section of two pistons in a hydraulic lift are 100 cm 2 and 800 cm 2 respectively. What is the minimum force required to lift 5000 kg -wt on large piston surface? (in kN)

Question

The cross - section of two pistons in a hydraulic lift are 100 cm 2 and 800 cm 2 respectively. What is the minimum force required to lift 5000 kg -wt on large piston surface? (in kN) (A) 392 (B) 400 (C) 412 (D) 500

Tardigrade Admin · Accepted Answer

As we know F 1 A 1= F 2 A2 F 1= F 2 × ( A 2/ A 1)=5000 × 9.8 × (800/100) F1=5000 × 9.8 × 8 F 1=392000 N

Q. The cross - section of two pistons in a hydraulic lift are $100 c m^{2}$ and $800 c m^{2}$ respectively. What is the minimum force required to lift $5000 k g$ -wt on large piston surface? (in kN)

392

400

412

500

As we know $F_{1} A_{1} = F_{2} A_{2}$
$F_{1} = F_{2} \times \frac{A _{2}}{A _{1}} = 5000 \times 9.8 \times \frac{800}{100}$
$F_{1} = 5000 \times 9.8 \times 8$
$F_{1} = 392000 N$

Q. The cross - section of two pistons in a hydraulic lift are 100cm2 and 800cm2 respectively. What is the minimum force required to lift 5000kg -wt on large piston surface? (in kN)

392

400

412

500

As we know F1​A1​=F2​A2​ F1​=F2​×A1​A2​​=5000×9.8×100800​ F1​=5000×9.8×8 F1​=392000N

Q. The cross - section of two pistons in a hydraulic lift are $100 c m^{2}$ and $800 c m^{2}$ respectively. What is the minimum force required to lift $5000 k g$ -wt on large piston surface? (in kN)

As we know $F_{1} A_{1} = F_{2} A_{2}$
$F_{1} = F_{2} \times \frac{A _{2}}{A _{1}} = 5000 \times 9.8 \times \frac{800}{100}$
$F_{1} = 5000 \times 9.8 \times 8$
$F_{1} = 392000 N$