MHTCET 23 Apr 2025 Shift 1 Paper

Concept:For mass $M$ to remain stationary, the tension in the string must balance its weight, and the same tension provides the centripetal force needed by mass $m$ moving in a horizontal circle of radius $L$.Explanation:Let the tension in the string be $T$.Since mass $M$ stays at rest, the tension must exactly balance its weight:$$T = \mathrm{Mg}$$The same tension acts as the centripetal force for mass $m$, which revolves with angular speed $\omega$:$$T = mL\omega^{2}$$Equating the two expressions for $T$:$$mL\omega^{2} = \mathrm{Mg}$$$$\omega^{2} = \frac{\mathrm{Mg}}{mL}$$The frequency $f$ is related to angular speed by $\omega = 2\pi f$.Substituting this relation:$$(2\pi f)^{2} = \frac{\mathrm{Mg}}{mL}$$$$4\pi^{2}f^{2} = \frac{\mathrm{Mg}}{mL}$$$$f = \frac{1}{2\pi}\sqrt{\frac{\mathrm{Mg}}{mL}}$$Answer:The required frequency is $\frac{1}{2\pi}\sqrt{\frac{\mathrm{Mg}}{mL}}$.Hence, the correct option is B.