Question

Weights of 1 g, 2 g, ..., 100 g are suspended from the 1-cm, 2-cm, .., 100-cm marks respectively of a light metre scale. Where should it be supported for the system to be in equilibrium?

(a) 55-cm mark 

(b) 60-cm mark 

(c) 66-cm mark 

(d) 72-cm mark

Answer 1

Correct Answer is: (c) 66-cm mark

Let the scale be supported at the l-cm mark. 

N = g + 2 g + ... + 100 g = 5050 g.

Taking torque about the zero mark,

Nl = 1 x g + 2 x 2 g + ... 100 x 100 g.