BITSAT 2024 Shift 1 Paper

To calculate the elastic potential energy in the system, we consider two wires of identical material (both having Young's modulus $Y$ ) and length $L$, but with different radii: $R$ and $2R$. These wires are sequentially connected, and a weight $w$ is hung from them.First, determine the elongation of each wire due to the weight $w$ :For the first wire with radius $R$ :$\Delta l_1 = \frac{w L}{4 \pi R^2 Y}$For the second wire with radius $2R$ :$\Delta l_2 = \frac{w L}{\pi R^2 Y}$The elastic potential energy stored in a wire is given by:$U = \frac{1}{2} K (\Delta l)^2$where $K = \frac{Y A}{L}$ is the spring constant and $A$ is the cross-sectional area of the wire.Calculate the spring constants for both wires:For the first wire $(K_1)$ :$K_1 = \frac{Y \times \pi R^2}{L}$For the second wire $(K_2)$ :$K_2 = \frac{Y \times 4 \pi R^2}{L}$Now compute the potential energy for each wire and sum them up:$U = \frac{1}{2} \times \frac{Y \times 4 \pi R^2}{L} \times \left( \frac{w L}{4 \pi R^2 Y} \right)^2 + \frac{1}{2} \times \frac{Y \times \pi R^2}{L} \times \left( \frac{w L}{\pi R^2 Y} \right)^2$After performing the calculations, the total elastic potential energy in the system is:$U = \frac{5 w^2 L}{8 \pi R^2 Y}$