An electric string of length l is elastically stretched through a length e by a force F. The area of cross- section of the string is A and its Young’s modul...

Answer Details

The expression that is correct is F = EAe/l. Here's why: Young's modulus (E) is a measure of the stiffness of a material. The larger the value of E, the harder it is to stretch the material. The cross-sectional area (A) of the string is a measure of how much material there is in the string. A larger area means more material, which means a greater force is required to stretch the string. The length of the string (l) is the distance between the two points that the string is attached to. The length extension (e) is the amount by which the length of the string increases when a force F is applied. The formula for Young's modulus is E = stress/strain, where stress is the force applied per unit area and strain is the extension per unit length. In this problem, we can use Hooke's law to relate the force applied to the extension of the string: F = kx, where k is the spring constant and x is the displacement of the string. Since the string is elastically stretched, we can assume that Hooke's law applies. The strain in the string is e/l, and the stress is F/A. Therefore, we can write: E = (F/A)/(e/l) Rearranging this equation gives: F = EAe/l So the correct expression is F = EAe/l.