MHTCET 11 May 2024 Shift 1 Paper

Concept:The frequency of a stretched wire depends on its vibrating length and the tension, which changes with the acceleration due to gravity.Explanation:The frequency of a stretched wire is given by:$$n = \frac{1}{2l}\sqrt{\frac{T}{m}}$$Here, $l$ is the vibrating length, $T$ is the tension in the wire, and $m$ is the mass per unit length.For the same setting, the tension is produced by a fixed load $M$, so:$$T = Mg$$Thus, the frequency becomes:$$n = \frac{1}{2l}\sqrt{\frac{Mg}{m}}$$For resonance with the same tuning fork, the frequency $n$ must remain unchanged at both the poles and the equator.Therefore, $l \propto \sqrt{g}$.The value of $g$ is greater at the poles than at the equator:$$g_{\text{poles}} > g_{\text{equator}}$$Since $l \propto \sqrt{g}$, the vibrating length at the equator must be smaller than that at the poles to keep the frequency constant.Hence, the vibrating length of the wire should be decreased.Answer:The vibrating length of the wire should be decreased.Correct option: A. should be decreased.