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A hydraulic press has a load F placed on the larger piston which is balanced by a smaller piston where a load of $20\;kg$ is kept. (neglecting mass of position) The larger piston has a diameter of $0.35\;m$ and the smaller piston has diameter of $0.1\;m$ The press is filled with oil of density $750\;kg/m^3$ and the larger position is at a height $1.5\;m$ relative to smaller piston. Find F.

a) $4.0 \times 10^3 \;N$

b) $2.5 \times 10^3 \;N$

c) $1.3 \times 10^3 \;N$

d) $3.0 \times 10^3 \;N$

Can you answer this question?
 
 

1 Answer

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Difference in the pressure at the two positions is equal to the pressure due to liquid column of AB.
Pressure at larger position $P_1=\large\frac{F}{A_1}$
$\qquad= \large\frac{F}{\pi \bigg(\Large\frac{0.35}{2}\bigg)^2}$
Pressure at smaller position $P_2=\large\frac{20}{A_2}$
$\qquad= \large\frac{20 \times g}{\pi \bigg(\Large\frac{0.1}{2}\bigg)^2}$
$P_2-P_1=h \rho g$
$\large\frac{20 \times 9.8 }{\pi \times (5 \times 10^{-2})^2}- \large\frac{F}{\pi \times (17.5 \times 10^{-2})^2}$$= 1.5 \times 750 \times 9.8$
Solving we get,
$F= 1.3 \times 10^3\;N$
Hence c is the correct answer

 

answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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