If we take the wheel as a ring then its moment of inertia will be, I=Mr2.
Using mechanical energy conservation,
Mgh−mgh=21mv2+21Mv2+21Iω2⇒12gh−3gh=213v2+2112v2+2112r2r2v2
⇒9gh=215v2+12v2
⇒v=32gh=2138gh
Therefore, x=38≃3
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown…
JEE Main 2022 — Physics Mechanics
2022integermedium
A rolling wheel of 12kg is on an inclined plane at position P and connected to a mass of 3kg through a string of fixed length and pulley as shown in figure.
Consider PR as friction free surface.
The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be 21xghms−1. The value of x (rounded off to the nearest integer) is _____.
Official previous-year question
Held on 27 Jun 2022 · Verified 6 Jul 2026.
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