Question

A horizontal cylindrical pipe consists of two coaxial sections, the section to the left has cross-sectional area A and that to the right has a cross-sectional area 3A. Some amount of water is enclosed in the pipe two airtight pistons. The smaller piston is connected to a fixed support with the help of a spring of force constant k and another spring of force constant 2k is connected with the larger piston as shown in the Figure. If the free end P of the spring connected to the larger piston is slowly shifted by a distance $\Delta x$, how much will the other spring be extended?

• A. $\frac{2}{3}\Delta x$
• B. $\frac{3}{2}\Delta x$
• C. $\Delta x$
• D. $\frac{1}{3}\Delta x$

Answer 1

Answer: (A)

$\frac{2}{3}\Delta x$

Answer 2

Let $x_1$ and $x_2$ be the extensions of the springs connected to the smaller piston (area $A$) and the larger piston (area $3A$), respectively. The pressure exerted by the smaller piston is $P = \frac{kx_1}{A}$. The pressure exerted by the larger piston is $P = \frac{2kx_2}{3A}$. Equating these pressures gives $\frac{kx_1}{A} = \frac{2kx_2}{3A}$, which simplifies to $x_1 = \frac{2}{3}x_2$. Therefore, $\Delta x_1 = \frac{2}{3}\Delta x_2$. Given that the larger piston is shifted by $\Delta x$, we have $\Delta x_2 = \Delta x$. Thus, $\Delta x_1 = \frac{2}{3}\Delta x$.