In the figure shown, the string and pulley are massless. The inclined plane is fixed and has coefficients of friction `mu_(s)=0.25` and `mu_(k)=0.2` . Consider two cases:
Case I : `m_(1)=2kg m_(2)=1.5 kg theta=37^(@) , "Case II":m_(1)=3kg m_(2)=2 kg theta =53^(@)`
Take `g=10 m//s^(2), sin 37^(@)=3/5, sin 53^(@)=4/5`

Options

A. In case I, both blocks are in equilibrium
B. In case I, friction on is 3 N, acting down the plane
C. In case II, both blocks are in equilibrium
D. In case II, friction on `m_(1)` is 3.6 N, acting up the plane

Answer

abc

Explanation

Assuming equilibrium `T=m_(2)g`
Taking friction up the plane , `T+f=m_(1) g sin theta ` and `N= m g cos theta`
put ` f=mu_(s)N` to get minimum allowed value of `m_(2) m_(2) ge m_(1) sin theta - mum_(1) cos theta `
reverser direction of friction to get maximum allowed value of `m_(2)m_(2)le m_(1)sin theta mum_(1) cos theta`
`rarr` again `m_(2)` is in the right range `rarr` static friction acts as `m_(2)ltm_(1) sin theta` friction acts up the plane
`rarr T+F =m_(1)g sin theta rarr T+F=m_(1) g sin theta -T = m_(1) g sin theta -m_(2) g =24 -20 =4N`