• # 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) $\tan \phi$

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].