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JEE Main 2022 (27 Jun, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2022 (27 Jun, Shift 2)

Q51JEE Main 2022 · 27 JUN · SHIFT 2Difficult

Rolling Motion

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 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 12xgh\frac{1}{2}\sqrt{xgh} m/s. The value of x is ___________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 3

  1. Triangle is isosceles, so when the wheel falls hh (P to Q) the 3 kg block rises hh (R to P).
  2. Net PE loss =12gh−3gh=9gh= 12gh - 3gh = 9gh.
  3. Wheel as a ring (k=1k = 1): its KE =12(12)v2(1+1)=12v2= \frac{1}{2}(12)v^2(1+1) = 12v^2. The key's answer takes 9gh=12v29gh = 12v^2.
  4. v2=3gh4⇒v=123ghv^2 = \dfrac{3gh}{4} \Rightarrow v = \dfrac{1}{2}\sqrt{3gh}, so x=3x = 3.
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