In the system of pulleys shown what should be the value of m1 (in gram) such that 100 g remains at rest.  (Take g=10m/s2 )

Since 100 g is at rest,lf m1 is going up at acceleration a, we use2T−m1g=m1a   ……(1)For 100 g to be at rest, acceleration of 200 g mass is 2a if m1 is going up at a2−T=0.2(2a)2−1=0.4aa=104=52m/s2 From (1), we have 2(1)−10m1=m1522=10m1+5m124=25m1m1=425=0.16kg=160g