Question

A uniform metre scale of weight and 80gf is balanced at 70cm mark. From where must a weight of 60 gf is suspended to balance it
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Answer 1

You can see this from the center of the scale.

Then you can see that 50 gm is located at 10 cm and 100 gm is at 45 cm distance.

We can equate torques to find where the 80 gm is located to balance the scale.

Torque = Mass × Acceleration × Distance.

(50×10×10)+(100×45×10)=(80×10×Distance)

Since, Distance=12.5 cm from middle of the scale.

So, 80 gm is located at (50+12.5)=62.5 cm on the Meter SCALE

Answer 2

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