Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1665: NEET 2012 Physics - Class 11 AIPM-2012-Ph-1665 Solution: A circular platform is mounted on a frictionless vertical axle. …

A circular platform is mounted on a frictionless vertical axle. Its radius R = 2 m and its moment of inertia about the axle is 200 kg m2200\ \text{kg m}^2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms−11\ \text{ms}^{-1} relative to the ground. Time taken by the man to complete one revolution is π s\pi\ \text{s} 3π2 s\frac{3\pi}{2}\ \text{s} 2π s2\pi\ \text{s} π2 s\frac{\pi}{2}\ \text{s}

Multiple Choice Options

  • A π\pi � s
  • B 3π2{{3\pi } \over 2} 2 3 � _ s
  • C 2 π\pi � s
  • D π2{\pi \over 2} 2 � _ s

Step-by-Step Verified Solution

Using conservation L i = 0 (Initial moment) L f = mvR � I _ (Final moment) According to the conservation of momentum L i = L f _ mvR � I. _ = 0 mvR = I. _ _ = ( 1 2 ) ( v + _ R ) t = 2 � R t ( 1 + 1 2 _ 2 ) = 2 � _ 2 t = 2 � sec.

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