MHTCET 25 Apr 2025 Shift 1 Paper

Concept:The frequency of a stretched wire depends on its length and tension, and tension changes with effective gravity.Explanation:For a stretched wire fixed at both ends, the fundamental frequency is:$$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$Here, $L$ is the vibrating length, $T$ is the tension, and $\mu$ is the mass per unit length.The tension is produced by a suspended mass, so $T = Mg_{\text{eff}}$.At the poles, $g_{\text{eff}} = g$.At the equator, due to Earth's rotation, $g_{\text{eff}} = g - \omega^2 R$, which is smaller.Thus, for the same suspended mass, tension at the equator is less than at the poles.With the same length, lower tension gives a lower frequency, since $f \propto \sqrt{T}$.To match the fixed frequency of the tuning fork, the wire's frequency must be increased.Because $f \propto \frac{1}{L}$, reducing the vibrating length $L$ increases the frequency.Therefore, the length should be decreased.Answer:Option A: should be decreased.