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WBJEEStatement 1 Two cylinders, one hollow (metal) and the other solid (wood) with...
Statement 1 Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
and Statement 2 By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
and Statement 2 By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
Solution:
In case of pure rolling on inclined plane,
\(\)
\begin{aligned}
& a=\frac{g \sin \theta}{1+I / m R^2} \\
& I_{\text {solid }} < I_{\text {hollow }} \\
\therefore \quad & a_{\text {solid }}>a_{\text {hollow }}
\end{aligned}
\(\)
\(\therefore\) solid cylinder will reach the bottom first.
Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant.
\(\)
\begin{aligned}
& \therefore(\mathrm{KE})_{\text {Hollow }}=(\mathrm{KE})_{\text {solid }}=\text { decrease in } \mathrm{PE}=m g h \\
& \therefore \text { correct option is }(\mathrm{d})
\end{aligned}
\(\)
\(\)
\begin{aligned}
& a=\frac{g \sin \theta}{1+I / m R^2} \\
& I_{\text {solid }} < I_{\text {hollow }} \\
\therefore \quad & a_{\text {solid }}>a_{\text {hollow }}
\end{aligned}
\(\)
\(\therefore\) solid cylinder will reach the bottom first.
Further, in case of pure rolling on stationary ground, work done by friction is zero. Therefore, mechanical energy of both the cylinders will remain constant.
\(\)
\begin{aligned}
& \therefore(\mathrm{KE})_{\text {Hollow }}=(\mathrm{KE})_{\text {solid }}=\text { decrease in } \mathrm{PE}=m g h \\
& \therefore \text { correct option is }(\mathrm{d})
\end{aligned}
\(\)
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