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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2008 · Paper 1 · Physics · Rotational Motion

Statement 1 Two cylinders, one hollow (metal) and the other solid...

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.

Options

  1. AStatement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
  2. BStatement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
  3. CStatement 1 is true, Statement 2 is false.
  4. DStatement 1 is false, Statement 2 is true

Correct answer

D. Statement 1 is false, Statement 2 is true

Step-by-step solution

In case of pure rolling on inclined plane, a=gsin⁡θ1+I/mR2Isolid <Ihollow ∴asolid >ahollow \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}\thereforesolidcylinderwillreachthebottomfirst.Further,incaseofpurerollingonstationaryground,workdonebyfrictioniszero.Therefore,mechanicalenergyofboththecylinderswillremainconstant.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.∴(KE)Hollow =(KE)solid = decrease in PE=mgh∴ correct option is (d)\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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