Question

A uniform metre scale of weight 50 gf, is balanced at 60 cm mark, when a weight of 15 gf is suspended at the 10 cm mark. Where must a weight 100 gf be suspended to balance the metre scale.

Answer 1

Given:

Weight of the scale = 50 gf

The scale is balanced at 60 cm mark.

Weight suspended at 10 cm mark = 15 gf

To find:

Where must a weight 100 gf be suspended to balance the metre scale.

Solution:

Let us balance the moment of torque about the balanced mark, i.e. at 60 cm mark.

Let the 100 gf weight be suspended at a distance of x from 60 cm marks.

The moment of torque due to the weight of scale will be 0.

Distance of 15 gf weight from 60 cm mark = 60 - 10 cm = 50 cm

So,

15* 50 = 100*x

x = 7.5 cm

Therefore the weight must be suspended at 6+ 7.5 cm = 13.5 cm mark.

The 100 gf weight mus be suspended at 13.5 cm mark.

Answer 2

Answer:

13.5 cm mark

pls follow

6+7.5