A Rolling Wheel of 12 kg Is on an... |JEE Main 2022

JEE MAIN 2022

27-06-2022 S2

Question

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 $\frac{1}{2} \sqrt{x g h} \mathrm{~m} / \mathrm{s}$The value of x is ______

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Solution

Net loss in PE = Gain in KE $$ \begin{aligned} & 12 g h-3 g h=\frac{1}{2} 3 v^2+\frac{1}{2} 12 v^2+\frac{1}{2}\left[12 r^2\right]\left(\frac{v}{r}\right)^2 \\ & 9 g h=\frac{1}{2}[3+12+12] v^2 \\ & v^2=\frac{2 g h}{3} \Rightarrow v=\frac{1}{2} \sqrt{\frac{8}{3} g h} \\ & x=\frac{8}{3} \approx 3 \end{aligned} $$

Question Tags

JEE Main

Physics

Easy

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