Question

A horizontal metallic rod of mass $m$ and length $l$ is supported by two vertical identical springs of spring constant $K$ each and natural lengths $l_0$. A current $i$ is flowing in the rod in the direction shown. If the rod is in equilibrium, then the length of each spring in this state is-

• A. $l_0+\frac{ilB-mg}{K}$
• B. $l_0+\frac{ilB-mg}{2K}$
• C. $l_0+\frac{mg-ilB}{2K}$
• D. $l_0+\frac{mg-ilB}{K}$

Answer 1

The correct option is B $l_0+\frac{ilB-mg}{2K}$

Given, $\text{Natural length of each spring}=l_0$



At the equilibrium condition, balancing forces acting on the rod, we get,

$ilB=mg+2Kx$

$x=\frac{ilB-mg}{2K}$

The final length of each spring would be,

$l_f=l_0+x=l_0+\frac{ilB-mg}{2K}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.