Question

A uniform meter scale of weight 80 gf is balanced at 70 cm mark. From where must a weight of 60 gf be suspended to balanced it.
Pls answer it fast. It's urgent

Answer 1

see figure,

A weight of 80 gf is balanced at 70cm mark from A as shown in figure. due to meter scale is uniform, centre of mass of scale is situated at midpoint of metre scale. so, 80gf weight is balanced at 20cm from centre of mass.

Let 60gf weight is balanced at x cm( left sides) from centre of mass.

at centre of mass,

$\tau_{net}=0$

so, 80gf × 20cm - 60gf × x cm = 0

or, 80gf × 20cm = 60gf × x cm

or, x = 80/3 cm

hence, 60gf weight is balanced at (50 - 80/3) = 70/3 cm from point A.

Answer 2

Answer:

anticlockwise moments = clockworkwise moments

80×(70-50)=60×x

80×20=60×x

x=80/3 cm

x is the distance from 70 cm

• the mark will be x+70= 70+80/3=290/3=96.67cm