A mass of 0.1 ~kg is hung at the 20 ~cm mark from a 1 ~m rod weighing 0.25 ~kg pivoted at its centre. The rod will not topple if

Step-by-step solution

Given, m=0.1 ~kg Let x be the position of mass of 0.1 ~kg 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=30 ~cm =0.3 ~m The rod will not topple if net torque on it is zero. So, balancing the moments about its centre, we get m₁ g x=m₂ g x^ m₂ x=0.1 0.3=0.03 ~kg - m From options, it is possible only in case of ( c ) . m₂=0.15 ~kg x^ = 0.03 0.15 =0.2 ~m =20 ~cm The second mass of 0.15 ~kg should be hanged at (50+20=) 70 ~cm