Work Energy Power

JEE Main level, numerical answer, medium

88 of 183

Question 88 of 183

A rolling wheel of 12kg12kg is on an inclined plane at position PP and connected to a mass of 3kg3kg through a string of fixed length and pulley as shown in figure.
Consider PRPR as friction free surface.
The velocity of centre of mass of the wheel when it reaches at the bottom QQ of the inclined plane PQPQ will be 12xghms−1\dfrac{1}{2}\sqrt{xgh}m{s}^{-1}. The value of xx (rounded off to the nearest integer) is _____.

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Answer: 3

If we take the wheel as a ring then its moment of inertia will be, I=Mr2I=M{r}^{2}.
Using mechanical energy conservation,
Mgh−mgh=12mv2+12Mv2+12Iω2⇒12gh−3gh=123v2+1212v2+1212r2v2r2Mgh-mgh=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}M{v}^{2}+\dfrac{1}{2}I{\omega }^{2} \Rightarrow 12gh-3gh=\dfrac{1}{2}3{v}^{2}+\dfrac{1}{2}12{v}^{2}+\dfrac{1}{2}12{r}^{2}\dfrac{{v}^{2}}{{r}^{2}}
⇒9gh=15v2+12v22\Rightarrow 9gh=\dfrac{15{v}^{2}+12{v}^{2}}{2}
⇒v=23gh=1283gh\Rightarrow v=\sqrt{\dfrac{2}{3}gh}=\dfrac{1}{2}\sqrt{\dfrac{8}{3}gh}
Therefore, x=83≃3x=\dfrac{8}{3}≃3

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