Question

A non uniform rod of mass 100 g is balanced as shown. Where should the 80g mass be placed if the rod is now balanced 50 cm from end A

Answer 1

Answer:

As 100g is balanced at fulcrum it will have a perpendicular distance of 40cm from itself and mass m will have 40−20=20cm distance.

Here moment of forces are equal that is balance each other.

Hence m×(40−20)=100×(50−40)

⇒m=100/2=50g.

Hence value of m is 50g.