Question

The system shown in the figure consists of three springs and two rods of the same material as shown in the figure. The springs are initially relaxed and there is no friction. The temperature of the rods is increased by $\Delta T$. The coefficient of linear expansion of the material of the rods is $\alpha$. Then, energy stored in the spring $1$, when the system is in equilibrium, is:

• A. $\frac{75}{242}k{L}^2{\alpha }^2(\Delta T{)}^2$
• B. $\frac{27}{242}k{L}^2{\alpha }^2(\Delta T{)}^2$
• C. $\frac{81}{242}k{L}^2{\alpha }^2(\Delta T{)}^2$
• D. $\frac{49}{242}k{L}^2{\alpha }^2(\Delta T{)}^2$

Answer 1

The correct option is C $\frac{81}{242}k{L}^2{\alpha }^2(\Delta T{)}^2$

Given:
Temperature change $=\Delta T$
To find:
Energy stored in the spring $1$, $E=?$
Let us assume:
Compression in spring $1$, $2$ and $3$ is $x_1$, $x_2$ and $x_3$ respectively due to expansion of the rods on heating.


So, total expansion of the rods $=x_1+x_2+x_3$
$\Rightarrow L\alpha \Delta T+\left(\frac{L}{2}\right)\alpha \Delta T=x_1+x_2+x_3$
[ from formula of linear expansion ]
$\Rightarrow \frac{3}{2}L\alpha \Delta T=x_1+x_2+x_2$ ............(1)

The F.B.D of the rods is shown in the figure below.


Considering the equilibrium of rods,
we have, $x_1=2x_2=3x_3$ ............(2)
[ Spring force, $F=kx$ ]
From (1) and (2), we get
$x_1+\frac{x_1}{2}+\frac{x_1}{3}=\frac{3}{2}L\alpha \Delta T$
$\Rightarrow \frac{11}{6}{x}_1=\frac{3}{2}L\alpha \Delta T$
$\Rightarrow x_1=\frac{9}{11}L\alpha \Delta T$
So, energy stored in spring $1$,
$E=\frac{1}{2}k{x}_1^2=\frac{1}{2}k{\left[\frac{9}{11}L\alpha \Delta T\right]}^2$
$=\frac{81}{242}k{L}^2{\alpha }^2\Delta T^2$