Question

A uniform meter rule of mass 80g is to be balanced at 60cm mark,where should be a mass of 100g be suspended to balance the rule​

Answer 1

Answer:

Weight of 100 g must be suspended at x = 68 cm mark

Explanation:

As we know that meter scale is uniform so its whole mass is to be concentrated at its mid point

So we know that torque due to weight of the scale about the support must be balanced by the torque due to extra weight suspended at x cm from support

So we will have

$80 \times (60 - 50) = 100 \times (x)$

so we have

x = 8 cm

so the suspended balance must be at x = 68 cm mark

#Learn

Topic : Torque balance

https://brainly.in/question/5800458

Answer 2

Answer:

Answer is fully wrong.