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A uniform meter scale weighs 50 g. It is pivoted at the 70 cm mark. Where should a 40 g mass be placed so that the scale is in equilibrium?

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RRB ALP Fitter 23 Jan 2019 Official Paper (Shift 2).
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  1. At the 25 cm mark
  2. At the 5 cm mark
  3. At the 95 cm mark
  4. At the 45 cm mark

Answer (Detailed Solution Below)

Option 3 : At the 95 cm mark
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Detailed Solution

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Concept:

Principle of Moments:

  • If the algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium.
  • According to the principle of moments, in equilibrium: The sum of anticlockwise Moments = Sum of clockwise moments.

Calculation:

Given:

Meter scale, i.e. length of scale = 100 cm, weight of scale = 50 g

Let x be the required distance from the pivoted end. Then as the meter scale is uniform, its weight will act at its midpoint, i.e., the 50 cm mark. The mass of the meter scale produces an anticlockwise moment about O and the suspended weight produces a clockwise moment.

F6 Vinanti Engineering 16.02.23 D2

Anticlockwise moment of the weight of the scale about O = 50 × (70 - 50) = 50 × 20

Clockwise moment of mass 40 g about O = 40x

According to the principle of moments,

Anticlockwise moment = Clockwise moment

⇒ 50 × 20 = 40x

⇒ x = 25 cm

∴ 40 g mass should be placed at (70 + 25) = 95 cm so that the scale is in equilibrium

 

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