The upper half of an inclined plane of inclination `theta` is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to rest at the bottom if the coefficient of friction between the block and the lower half of plane in given by

Options

A. `mu sin theta`
B. `mu = cot theta`
C. `mu = 2 cos theta`
D. `mu = 2 tan theta`

Answer

D

Explanation

The acceleration of the body on upper half, `a_(1) = (mg sin theta)/(m) = g sin theta`
Let v be the velocity of the block at the end of smooth half.
Using
`v^(2) = u^(2) + 2 a (1)/(2) = 2 (g sin theta) (1)/(2) = g l sin theta`
For lower half, the body comes back to rest at the bottom,
it means the body retards on this section.
The retardation of the body on lower half,
`a_(2) = ((mu m g cos theta - m g sin theta))/(m) = g (mu cos theta - sin theta)`
Again using `v^(2 ) = u^(2) + 2` as, we get :
` = u^(2) - 2g (mu cos theta - sin theta) (1)/(2) implies u^(2) = 2 g (mu cos theta - sin)(1)/(2)`
But velocity of the body at the starting of this section is equal to final velocity at end of smooth section
`:. g l sin theta = 2 g (mu cos theta - sin theta) (1)/(2) implies mu cos theta = 2 sin theta implies mu = 2 tan theta`