Question

Two wires of the same material (Young’s modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy in the system is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$K_1=\frac{y\pi (2R{)}^2}{L}$
$K_2=\frac{y\pi (R{)}^2}{L}$
Equivalent $\frac{1}{K_1}+\frac{1}{K_2}=\frac{1}{4y\pi R^2}+\frac{L}{y\pi R^2}$
Since,$K_1{x}_1=K_2{x}_2=w$
Elastic potential energy of the system
U=$\frac{1}{2}{k}_1{x}_1+\frac{1}{2}{k}_2{x}_2^2$
$=\frac{1}{2}{k}_1(\frac{W}{k_1}{)}^2+\frac{1}{2}{k}_2(\frac{W}{k_2}{)}^2$
$=\frac{1}{2}{W}^2+(\frac{1}{k_1}+\frac{1}{k_2})$
$=\frac{1}{2}{W}^2+\left(\frac{5L}{4y\pi R^2}\right)$

U=$\left(\frac{5w^{2}L}{8\pi y{R}^2}\right)$