A uniform metre scale is in equilibrium position.At 5cm a weight of 40 gf is suspended. The scale is balanced at 30 cm.What is the weight of the scale.
Answers
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The weight of the scale is 50 gf.
Let the weight of the scale be 'm' gf
The metre scale is pivoted at 30cm.
The torque of 40gf about the pivot (clockwise moment)
T₁ = Mass × Acc. due to gravity × Distance of the weight from pivot.
As 40gf = mass × acc. due to gravity and,
Distance of this weight from pivot = (30 - 5)cm = 25cm
So, T₁ = 40 × 25 gf-cm = 1000 gf-cm
The weight of the scale acts on its centre of mass which is situated at the mid point (50 cm) of the rod.
Thus the distance of centre of mass of the scale from the pivot is = (50 - 30)cm = 20 cm [As the scale is balanced at 30cm]
So, the torque of the 'm' gf weight of the scale about the pivot (anti clockwise moment) is
T₂ = Weight of rod × Distance of the weight from pivot.
= m × 20
As the sum of clockwise moments = Sum of anti-clockwise moments. So
T₁ = T₂
Or, 1000 = 20m
⇒m = 1000/20 = 50gf
Weight is 50gf.