Question

A uniform metre rule of mass 100g balances at the 40cm mark by suspending an unknown mass m at the mark 20cm, find the value of m.​

Answer 1

Since, uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm.

so, distance of c.o.m. of rule from balanced point is 10cm.

Net moment at balancing point should be zero.

So, m×20 = 100×10 m = 50g

if m is moved to mark of 10cm then rule will tilt to the side where m is suspended.

Now, for balance this

50×30 = 100×10+ 50×(x-40) (x-40) = (1500- 1000)/50 = 10cm X = 50cm.

So, another 50g mass will be suspended at 50cm mark

Answer 2

Answer:

hey here is your answer

Explanation:

Since, uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm.

so, distance of c.o.m. of rule from balanced point is 10cm.

Net moment at balancing point should be zero.

So, m×20 = 100×10 m = 50g

if m is moved to mark of 10cm then rule will tilt to the side where m is suspended.

Now, for balance this

50×30 = 100×10+ 50×(x-40) (x-40) = (1500- 1000)/50 = 10cm X = 50cm.

So, another 50g mass will be suspended at 50cm mark