For the situation shown in the figure, pulleys are light and smooth. Pulleys C and D are fixed, but A and B are movable. Find the value of m (in kg) if m is in equilibrium. (take, m1 = 2 kg and m2 =1 kg)

Let tension  of m2 is T and tension of  m1 is 2TLet acceleration of m1 is a and acceleration of  m2 is 2aFor m2T−m2g=m2(2a)  ⇒T-1(10)=1(2a)......(i)Free body diagram (FBD) of m2 For m1: m1g−2T=m1(a)    ⇒2(10)-2T=2a ......(ii)Free body diagram (FBD) of m1From Eqs. (i)2 and (ii), we get2T-20=4a⇒20-2T=2a ⇒ a=0m/s2 Or substituting in (1) ⇒T-10=2(0)⇒T=10N Free body diagram of m For m,2T=mg⇒2(10)=m(10)⇒m=2kg