Question

The two light pistons A and B shown in the figure have area 'na' and 'a' respectively. The rod has cross sectional area S, Young's modulus Y and length L. A force F is applied on the piston B. (Assume liquid is incompressible)

• A. the force acting on the rod by the piston A is nF.
• B. the force acting on the rod by the piston A is F/n.
• C. the compression in the length of the rod due to the force exerted by the piston is $\frac{nFL}{SY}$
• D. the compression in the length of the rod due to the force exerted by the piston is $\frac{FL}{nSY}$

Answer 1

Answer: (A), (C)

The correct options are
A the force acting on the rod by the piston A is nF.
C the compression in the length of the rod due to the force exerted by the piston is $\frac{nFL}{SY}$

Pressure is same under both the pistons.
$\therefore$ Force acting on the piston is proportional to the area of
the pistons.
The force acting on the piston A by water is nF, which will be balanced by the force on the piston by the rod.
Hence, force acting on the piston by the rod is not $\frac{F}{n}$
Using properties of solid,
Young's modulus $Y=\frac{stress}{strain}=\frac{\frac{nF}{S}}{\frac{\Delta L}{L}}$
$\therefore \Delta L=\frac{nFL}{SY}$