An uniform metre scale of weight 50 gf is balanced at 40 cm mark when a weight of hundred 100 gf is suspended at 5 cm mark. Where must a weight of 80 gf be suspended to balance meter scale?

Open in App

Solution

Step: 1 Given
Force = 50 N
Weight of meter scale = 50 gf
Balanced marked = 40 cm.
Weight of 100 gf at 5 cm mark.
let x is the distance of 80 gf load. from 40 cm
Step - 2
Diagram -
Step - 3
Formula used :
Moment of Force = Force x Perpendicular Distance
( 80 x X) + ( 50 x 10) = 100 x 35
80 X + 500 = 3500
80 X = 3000
X = 3000/80 = 37.5 cm
Distance in term of meter scale = 40 + 37 .5
= 77.5 cm

Suggest Corrections

Similar questions

A uniform meter rule balances horizontally on a knife-edge placed at the 58 cm mark when a weight of 20 gf is suspended from one end.

(i) Draw a diagram of the arrangement.

(ii) What is the weight of the rule?

Q. A uniform metre scale has two weights of 10 gf and 8 gf suspended at the 10 cm and 80 cm marks respectively. If the metre scale itself weights 50 gf, find where must the weight be, so that the metre scale stays balanced?

Q. Uniform metre scale is balance that 40 cm mark when weight of 20 gf and 5 gf are suspended at 5 cm and 75 cm respectively. calculate the weight of metre scale

Q. A uniform meter scale is put on a knife edge at 40 cm mark. It is found that this scale gets balanced when a weight of 20 gf is suspended from 20 cm mark. Find the weight of meter scale.

A meter scale is kept on a knife edge at $50 c m$ mark. A weight of $20 g f$ is hanging at $80 c m$ mark and a weight of $50 g f$ is hanging at $60 c m$ mark . Where should a weight of $55 g f$ be placed to make the meter scale balanced?

Join BYJU'S Learning Program