test

yup answer is c only...

posting solution...

drawing fbd....

let the masses m₁and m₂ have relative accelerations a_o wrt the movable pulley...for m₁ upward and for m₂ downward...

let the mass M move with acceleration a..

thus the pulley moves with accn a downwards...

so net accn of m₁=a-a_o
and accn of m₂=a+a_o
now for mass m₁ to move with constant velocity..
a-a_o=0...or, a=a_o...

thus acceleration of m₂=2a

let tension in downward string be T....and upward string be T'
now.. 2T-T'=0.a....so T'=2T....(i)

thus for mass M, 2T=Ma....(ii)

for mass m₁, 1.g-T=0...or, T=g....(iii)
for mass m₂, 2.g-T=2.2a...(iv)
substituting value of g in (iv),
2T-T=4a..
or, T=4a.....(v)

dividing (i) by (v)
^2T/_T=^Ma/_4a
or, M=8g.....hence solved...[1]