JEE Advanced · Physics · 8. Rotational Motion
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.
- A Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
- B Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
- C Statement 1 is true, Statement 2 is false.
- D Statement 1 is false, Statement 2 is true
Answer & Solution
Correct Answer
(D) Statement 1 is false, Statement 2 is true
Step-by-step Solution
Detailed explanation
In case of pure rolling on inclined plane,
\(a=\frac{g \sin \theta}{1+I / m R^2} \)
\( I_{\text {solid }} < I_{\text {hollow }} \)
\( \therefore a_{\text {solid }}>a_{\text {hollow}}\)
\(\therefore\) 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.
\(\therefore(\mathrm{KE})_{\text {Hollow}}=(\mathrm{KE})_{\text {solid}}=\) \(\text {decrease in } \mathrm{PE}=m g h\)
\(\therefore \text {correct option is }(\mathrm{d})\)
\(a=\frac{g \sin \theta}{1+I / m R^2} \)
\( I_{\text {solid }} < I_{\text {hollow }} \)
\( \therefore a_{\text {solid }}>a_{\text {hollow}}\)
\(\therefore\) 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.
\(\therefore(\mathrm{KE})_{\text {Hollow}}=(\mathrm{KE})_{\text {solid}}=\) \(\text {decrease in } \mathrm{PE}=m g h\)
\(\therefore \text {correct option is }(\mathrm{d})\)
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