question_answer

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 coefficient of friction for the lower half is given by

A) \[2\sin \phi \]               

B) \[2\cos \phi \]

C) \[2\tan \phi \]               

D) \[\tan \phi \]  

Correct Answer: C

Solution :

For the smooth portion BC, \[u=0,\,s=\ell ,a=g\sin \phi =v?\] From \[{{\upsilon }^{2}}-{{u}^{2}}=2as\] \[{{\upsilon }^{2}}-0=2g\sin \phi \times \ell \] For the rough portion CO \[u=v=\sqrt{2g\,\sin \phi .\ell }\] \[v=0,a=g(siin\phi -\mu \cos \phi )s=\ell \] From \[{{v}^{2}}-{{u}^{2}}=2as\] \[0-2g\ell \sin \phi =2g(sin\phi -\mu cos\phi )\ell \] \[-\sin \phi =\sin \phi -\mu \cos \phi \] \[\mu \cos \phi =2\sin \phi \] Hence, the correction option is [c].