A uniform meter scale weighs 75 g. It is pivoted at the 50 cm mark. A 40 g mass is placed at the 20 cm mark and a 50 g mass is placed at the 90 cm mark. Where should a 20 g mass be placed so that the scale is in equilibrium?

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RRB ALP Fitter 21 Jan 2019 Official Paper (Shift 2)

Answer 1

Calculation:

Let 20 g mass is placed at the x cm mark from pivot (X).

Taking moment about a pivot (X)

\(\sum{M_X}=0\)

(40 × 30) + (20 × x) = ( 50 × 40)

Then, \(x=\frac{2000-1200}{20}=\frac{800}{20}\) = 40 cm

So, the 20 g mass should be placed at 40 cm from the pivot or 10 cm mark from the 0 cm mark point to maintain the equilibrium of the scale.

Thus, the answer is a 10 cm mark from starting of the scale.

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