A hydraulic press consists of two cylinders of cross section 2 cm^2 and 50 cm^2. The piston of the smaller cylinder is pushed down with a force of 100 gf through a distance of 20 cm. Calculate : (1) The pressure transmitted by the fluid. (2) The force exerted by the piston in the large cylinder.
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Explanation:
By Pascal's law, the pressure increased using one piston is communicated everywhere in the fluid. Thus, the same pressure acts on the other piston.
Cross-sectional area of piston 1 is A
1
=100 cm
2
=0.01 m
2
Force on piston 1 is F
1
=10
7
dyne=100 N
Force on piston 2 is F
2
=2000 kg wt=2000×10=2×10
4
N
Let the cross-sectional area of Piston 2 be A
2
=A.
By Pascal's law, P
1
=P
2
⇒
A
1
F
1
=
A
2
F
2
⇒A
2
=
F
1
F
2
A
1
⇒A=
100
2×10
4
×0.01
=2 m
2
=2×10
4
cm
2
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