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Question: Two syringes of different cross section (without needle) filled with water are connected with a tigh...

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

A

30 N

B

60 N

C

90 N

D

100 N

Answer

90 N

Explanation

Solution

Since pressure is transmitted undimished throughout the water

∴F1 A1=F2 A2\therefore \frac { \mathrm { F } _ { 1 } } { \mathrm {~A} _ { 1 } } = \frac { \mathrm { F } _ { 2 } } { \mathrm {~A} _ { 2 } }

Where F1\mathrm { F } _ { 1 } and are the force on the smaller and on the larger pistons respectively and and are the respective areas.

∴F2=A2 A1 F1=π(D2/2)2π(D1/2)2 F1=(D2D1)2 F1\therefore \mathrm { F } _ { 2 } = \frac { \mathrm { A } _ { 2 } } { \mathrm {~A} _ { 1 } } \mathrm {~F} _ { 1 } = \frac { \pi \left( \mathrm { D } _ { 2 } / 2 \right) ^ { 2 } } { \pi \left( \mathrm { D } _ { 1 } / 2 \right) ^ { 2 } } \mathrm {~F} _ { 1 } = \left( \frac { \mathrm { D } _ { 2 } } { \mathrm { D } _ { 1 } } \right) ^ { 2 } \mathrm {~F} _ { 1 }

=(3×10−2 m)2(1×10−2 m)2×10 N=90 N= \frac { \left( 3 \times 10 ^ { - 2 } \mathrm {~m} \right) ^ { 2 } } { \left( 1 \times 10 ^ { - 2 } \mathrm {~m} \right) ^ { 2 } } \times 10 \mathrm {~N} = 90 \mathrm {~N}