A brass rod of cross sectional area \(1 \mathrm{~cm}^{2}\) and length $0.2 \mathrm{~m}\( is compressed length wise by a weight of \)5 \mathrm{~kg}$. If young's modulus of elasticity of brass is $1 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\( and \)\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)$ Then what will be increase in the energy of rod ? (A) \(10^{-5} \mathrm{~J}\) (B) \(2.5 \times 10^{-5} \mathrm{~J}\) (C) \(5 \times 10^{-5} \mathrm{~J}\) (D) \(2.5 \times 10^{-4} \mathrm{~J}\)

To calculate the stress acting on the rod, first, we need to find the force applied to the rod due to the weight. The force is given by \(F = m \times g\), where \(m\) is the mass of the weight and \(g\) is the acceleration due to gravity. Given, \(m = 5\ \mathrm{kg}\) and \(g = 10\ \mathrm{m/s^2}\), thus: \[F = 5\ \mathrm{kg} \times 10 \ \mathrm{m/s^2} = 50\ \mathrm{N} \] Next, calculate the stress acting on the rod with formula \(\sigma = \frac{F}{A}\), where \(\sigma\) represents stress, \(F\) is force, and \(A\) is the cross-sectional area. Here, we are given \(A = 1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}\). So: \[\sigma = \frac{50\ \mathrm{N}}{10^{-4}\ \mathrm{m^2}} = 5 \times 10^5\ \mathrm{N/m^2} \]

Next, we will calculate the strain on the rod using Young's modulus formula: \[Y = \frac{\sigma}{\epsilon} \] where \(Y\) is Young's modulus, \(\sigma\) is stress, and \(\epsilon\) is the strain. Rearranging for \(\epsilon\): \[\epsilon = \frac{\sigma}{Y} \] Given, \(Y = 1 \times 10^{11}\ \mathrm{N/m^2}\) and \(\sigma = 5 \times 10^5\ \mathrm{N/m^2}\). Thus: \[\epsilon = \frac{5 \times 10^5\ \mathrm{N/m^2}}{1 \times 10^{11}\ \mathrm{N/m^2}} = 5 \times 10^{-6} \]

Now, we need to calculate the increase in the energy stored in the rod due to compression: Using the formula for elastic potential energy, which states that the energy stored in a rod is given by: \[U = \frac{1}{2} V \times \sigma \times \epsilon\] Let's first calculate the volume, \(V\) of the brass rod using the formula \(V = A \times l\), where \(l\) is the length of the rod. Given, \(l = 0.2\ \mathrm{m}\) and \(A = 10^{-4}\ \mathrm{m^2}\): \[V = 10^{-4}\ \mathrm{m^2} \times 0.2\ \mathrm{m} = 2 \times 10^{-5}\ \mathrm{m^3} \] Now, plugging the values of \(\sigma\), \(\epsilon\), and \(V\) into the potential energy formula: \[U = \frac{1}{2} \times 2 \times 10^{-5}\ \mathrm{m^3} \times 5 \times 10^5\ \mathrm{N/m^2} \times 5 \times 10^{-6} \] \[U = 2.5 \times 10^{-5}\ \mathrm{J} \] The increase in energy of the rod is \(2.5 \times 10^{-5}\ \mathrm{J}\). Therefore, the correct option is (B).