Two blocks of masses `m_(1)` and `m_(2)` are connected by a string of negligible mass which pass over a frictionless pulley fixed on the top of an inclined plane as shown in figure. The coefficient of friction between `m_(1)` and plane is `mu`.

Options

A. If `m_(1)=m_(2)`, the mass `m_(1)` first begin to move up inclined plane when the angle of inclination `theta`, then `mu=sec theta-tan theta`
B. If `m_(1)=m_(2)`, then mass `m_(1)` first begin to side down the plane if `mu=sec theta-tan theta`.
C. If `m_(1)=2m_(2)`, then mass `m_(1)` first begins to slide down the plane if `mu=2 tan theta`.
D. If `m_(1)=2m_(2)`, then mass `m_(1)` first begins to slide down the plane if `mu= tan theta - 1/2 sec theta`.

Answer

A, D

Explanation

The block `m_(1)` will just being to move up the plane if the downward force `m_(2)g` due to mass `m_(2)` trying to pull the mass `m_(1)` up the plane just equals the force
`m_(2)g = (m_(1)g sin theta + mum_(1)gcostheta)`
`mg = mg sin theta + mumg cos theta`
`mg(1-sin theta) = mumg cos theta mu = sec theta - tan theta`
So, choice (a) is correct.
(d) The block `m_(1)` will just begin to move down the plane if the downward force on `m_(1)` due to `m_(2)` if
`m_(2)g = m_(1)g(sin theta - mu cos theta)`
`m_(2)g = 2m_(2)g(sin theta - mu cos theta)`
`1/2 = sin theta - mu cos theta 1/(2 cos theta) = tan theta - mu`
`mu = tan theta - 1/2 sec theta` Choice (d) is correct.