Two syringes of different cross section (without needle) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1 cm and 3 cm respectively. If a force of 10 N is applied to the smaller piston then the force exerted on the larger piston is

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30 N

60 N

90 N

100 N

answer is C.

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Detailed Solution

Since pressure is transmitted undiminished through out the water

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$

where $F_{1} and F_{2}$ are the forces on the surface and on the larger pistons respectively and $A_{1} and A_{2}$ are the respective areas.

$F_{2} = \frac{A_{2}}{A_{1}} F_{1} = \frac{π {(\frac{D_{2}}{2})}^{2}}{π {(\frac{D_{1}}{2})}^{2}} F_{1} {= (\frac{D_{2}}{D_{1}})}^{2} F_{1}$

= $\frac{{(3 \times 10^{- 2} m)}^{2}}{{(1 \times 10^{- 2} m)}^{2}} \times 10 N = 90 N$

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