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The area of cross sectioin of the two arms of a hydraulic press is `1cm^(2)` and `20 cm^(2)` respectively. A force of `10 N` is applied on the water in the thinner arm. What force should be applid on the water in the thicker arms so that water may remain in equilibrium? |
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Answer» In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is `P_(0)` and a force `F` is applied to maintain the equilibrium the pressur just below the pistons are `P_(M)=P_(0)+(10N)/(1cm^(2))` and `P_(N)=P_(0)+F/(10cm^(2))` These pressures should be the same i.e. `P_(M)=P_(n)`which gives `F=200N`. |
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