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Q.

The area of cross section of the two arms of a hydraulic press are $1 {cm}^{2} and 10 {cm}^{2}$ respectively (as shown). A force of 5 N is applied on the water in the thinner arm. The force should be applied on the water in thicker arm so that the water may remain in equilibrium is

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a

40 N

b

50 N

c

60 N

d

30 N

answer is C.

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Detailed Solution

A_{1} = 1 {cm}^{2} {,A}_{2} = 10 {cm}^{2}, F_{1} = 5N

Here pressure is same

$P_{1} = \frac{F_{1}}{A_{1}} = 5$

$P_{2} = \frac{F_{2}}{A_{2}}$

$F_{2} = P_{2} A_{2}$

$F_{2} = 5 \times 10$

$F_{2} = 50 N$

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The area of cross section of the two arms of a hydraulic press are 1 cm2 and 10 cm2 respectively (as shown). A force of 5 N is applied on the water in the thinner arm. The force should be applied on the water in thicker arm so that the water may remain in equilibrium is