Question

A ruler of 30 cm length has been pivoted at it's middle. A weight of 5 gram is suspended at 5 cm mark of the ruler. At which mark of the ruler must another weight of 8 gram be suspended so that the ruler gets horizontally balanced.


P.S: I'll give brainliest to the most appropriate answer​

Answer 1

here is your answer

Hope this will help you

Answer 2

Answer:

The metre rule is pivoted at C, such that AC=50cm and CB=50cm. 50 gf weight is suspended near A.

Let D be the point of suspension of 100 gf weight at a distance of x from C, so that the rule is horizontal.

Left hand side moment,

LHM=50×50=2500gfcm(1)

Right hand side moment,

RHM=x×100=100×gfcm(2)

To keep the rule horizontal,

LHM=RHM

2500=100x⇒x=25cm.