In a hydraulic machine, the two pistons have an area of cross-section in the ratio 1:5, what force should be applied on the narrow piston to get an output of 50 N on the wider piston?

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10N

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20N

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15N

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Solution

The correct option is B
10N

A $_{1}$ : A $_{2}$ =1:5 , F $_{2}$ = 50.

We know, Pressure on narrow piston = Pressure on the wider piston.

P $_{1}$ = P $_{2}$

F $_{1}$ /A $_{1}$ =F $_{2}$ /A $_{2}$

F $_{1}$ = $\frac{F_{2} \times A_{1}}{A_{2}}$ = $\frac{50 \times 1}{5}$ = 10 N

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In a hydraulic machine, the two pistons have an area of cross-section in the ratio 1:5, what force should be applied on the narrow piston to get an output of 50 N on the wider piston?

The correct option is B 10N A1: A2 =1:5 , F2= 50. We know, Pressure on narrow piston = Pressure on the wider piston. P1 = P2 F1/A1=F2/A2 F1= F2×A1A2 =50×15 = 10 N

The correct option is B
10N

A $_{1}$ : A $_{2}$ =1:5 , F $_{2}$ = 50.

We know, Pressure on narrow piston = Pressure on the wider piston.

P $_{1}$ = P $_{2}$

F $_{1}$ /A $_{1}$ =F $_{2}$ /A $_{2}$

F $_{1}$ = $\frac{F_{2} \times A_{1}}{A_{2}}$ = $\frac{50 \times 1}{5}$ = 10 N