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Q.

An automobile sack is lifted by a hydraulic jack that consists of two pistons . The larger piston is 1 m in diameter and small piston is 10 cm in diameter . If W be weight of car , the minimum force is needed on small piston to lift car is

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a

100 %

b

1 %

c

200 %

d

2 %

answer is B.

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Detailed Solution

Given that

Effort piston diameter = 10 cm

Radius = 5 cm

Load piston diameter = 1 m

Radius = 50 cm

We know $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$

Weight of car = W

Effort required = F

$\frac{F}{5^{2}} = \frac{W}{{(50)}^{2}}$

$F = \frac{W \times 25}{2500} = \frac{W}{100}$

$% F = \frac{\frac{W}{100}}{W} \times 100 = 1 %$

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An automobile sack is lifted by a hydraulic jack that consists of two pistons . The larger piston is 1 m in diameter and small piston is 10 cm in diameter . If W be weight of car , the minimum force is needed on small piston to lift car is