MCQ
Two identical solid cylinders run a race starting from rest at the top of an inclined plane. If one cylinder slides and the other rolls
  • the sliding cylinder will reach the bottom first with greater speed
  • B
    the rolling cylinder will reach the bottom first with greater speed
  • C
    both will reach the bottom simultaneously with the same speed
  • D
    both will reach the bottom simultaneously but with different speeds

Answer

Correct option: A.
the sliding cylinder will reach the bottom first with greater speed
a
For sliding cylinder

$\mathrm{t}_{\mathrm{s}}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}} \text { and } \mathrm{v}_{\mathrm{s}}=\sqrt{2 \mathrm{gh}}$

For rolling cylinder,

$\mathrm{t}_{\mathrm{R}}=\frac{1}{\sin \theta} \sqrt{\beta\left(\frac{2 \mathrm{h}}{\mathrm{g}}\right)} \text { and } \mathrm{v}_{\mathrm{R}}=\sqrt{\frac{2 \mathrm{gh}}{\beta}}$

As $\beta\left(=1+\frac{\mathrm{I}}{\mathrm{MR}^{2}}\right)$ is greater than one, hence

$\mathrm{t}_{\mathrm{R}}>\mathrm{t}_{\mathrm{s}}$ but $\mathrm{v}_{\mathrm{R}}<\mathrm{v}_{\mathrm{s}}$

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