Question

Weights of 1g, 2g...... 100g 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

Answer:

a

Explanation:

622:2728-262:7272--

Answer 2

It should be supported at 66 cm mark for the system to be in equilibrium.

Given-

• Weights are 1 g, 2 g , 3 g.........100 g
• Weights are suspended at a distance = 1 cm, 2 cm, 3 cm.......100 cm respectively to the masses.

For the system to be in equilibrium the point must lie in the center of mass.

So center of mass of the system can be calculated as

COM = (1² + 2² + 3².....100²)/ (1 + 2 + 3....+100)

COM = [(100)(101)(201)/6] / [(100) (101) /2]

COM = 201/3 = 67 cm

Hence at 67 cm from the origin or 66 cm from the first particle, at that point system will be supported to get equilibrium.