QNEET-2024-Ph-78: NEET 2024 Physics - Class 11 NEET-2024-Ph-78 Solution: A metallic bar of Young's modulus \(0.5 10^{11}\ {N m}^{-2}\)…

Given the properties and conditions of the metallic bar, we are required to calculate the compressive force developed due to heating. Key inputs include the Young's modulus (E), coefficient of linear thermal expansion (_), change in temperature ( _ T ), and the original dimensions of the bar. First, compute the linear expansion of the bar if it were free to expand. The change in length ( _ L ) due to thermal expansion can be computed through the formula: _ L = _ L 0 _ T Given: _ = 10 _ 5 _ C _ 1 L 0 = 1  m _ T = 100 _ C Thus: _ L = 10 _ 5 _ 1 _ 100 = 0.001  m This is the change in length that the bar would undergo if not constrained. However, in this scenario, the bar is constrained and does not actually expand. This constraint induces a compressive stress (constrained thermal stress) in the bar, which can be calculated using the formula relating stress, Young's modulus, and strain: _ = E _ Where the strain ( _ ) under constrained conditions due to thermal expansion is: _ = _ L L 0 = 0.001 1 = 0.001 Therefore: _ = 0.5 _ 10 11 _ 0.001 = 5 _ 10 7 N / m 2 This stress is the force per unit area. To find the compressive force, we need to multiply this stress by the cross-sectional area of the bar: F = _ A = 5 _ 10 7 N / m 2 _ 10 _ 3 m 2 = 5 _ 10 4 N Thus, the compressive force developed in the bar is 50 _ 10 3   N . Hence, the correct answer is Option B: 50 _ 10 3   N