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?
Text Solution
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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`.
The area of cross section of the two arms of a hydraulilc press are 1 cm^2 and 10 cm^2 respectively figure. A force of 5N is applied on the water in the thinner arm. What force should be applied on the water in the thicker arm so that the water may remain in equilibrium? ,
The area of cross section of the two arms of a hydraulilc press are 1 cm^2 and 10 cm^2 respectively figure. A force of 5N is applied on the water in the thinner arm. What force should be applied on the water in the thicker arm so that the water may remain in equilibrium? ,
The area of cross-section of the two vertical arms of a hydraulic press are 1 cm ^(2) and 10 cm^(2) respectively. A force of 10N applied. As shown in the figure, to a tight fitting light piston in the thinner arm balances a force F applied to the corresponding piston in the thicker arm. Assuming that the levels of water in both the arms are the same, we can conclude:-
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A area of cross section of a large tank is 0.5m^2 . It has an opening near the bottom having area of cross section 1cm^2 . A load of 20 kg is applied on the water at the top. Find the velocity of the water coming out of the opening at the time when the height of water level is 50 cm above the bottom. Take g=10ms^-2
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