Let x be the position of mass of 0.1kg from its centre and x′ be the position of second mass m2 that must be suspended to the other end to prevent the rod from toppling. So, x=50−20=30cm=0.3m The rod will not topple if net torque on it is zero. So, balancing the moments about its centre, we get m1gx=m2gx′ ⇒m2x=0.1×−.3=0.03kg−m From options, it is possible only in case of (c). m2=0.15kg x′=
0.03
0.15
=0.2m=20cm ∴ The second mass of 0.15kg should be hanged at (50+20)=70cm mark.