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The upper half of an inclined plane with inclination  \phi is perfectly smooth while the lower half is rough .A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

  • Option 1)

    2\tan \phi

  • Option 2)

    \tan \phi

  • Option 3)

    2\sin \phi

  • Option 4)

    2\cos \phi


Answers (1)


As we learnt in

Kinetic or Dynamic Friction -

f_{K}\;\alpha\ R

f_{K}=\mu_{K} R

f_{K}= kinetic friction 

\mu_{K}= coefficient of kinetic friction

R = reaction

- wherein


\therefore\ \mu_{K}<\mu_{s}

\mu_{K}=depends on the nature of surface in contact.




For upper half smooth incline, component of g down the incline = gsin\phi    

\therefore \; \; v^{2}=2(gsin\phi )\frac{l}{2}


For lower half rough incline, frictional retardation = \mu _{k}gcos\phi       

\therefore      Resultant acceleration = gsin\phi -\mu _{k}gcos\phi

\therefore \; \; \; \; \; 0=v^{2}+2(gsin\phi -\mu _{k}gcos\phi )\frac{l}{2}

or      0=2(gsin\phi )\frac{l}{2}+2g(sin\phi -\mu _{k}cos\phi )\frac{l}{2}

or\; \; \; 0=sin\phi +sin\phi-\mu _{k}cos\phi

or\; \; \; \mu _{k}cos\phi =2sin\phi

or\; \; \; \mu _{k}=2tan\phi

Correct option is 1.



Option 1)

2\tan \phi

This is the correct option.

Option 2)

\tan \phi

This is an incorrect option.

Option 3)

2\sin \phi

This is an incorrect option.

Option 4)

2\cos \phi

This is an incorrect option.

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