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ICSE Class 10 Physics 2023 Paper

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A metre scale of weight 50 gf50\,\mathrm{gf} can be balanced at 40 cm40\,\mathrm{cm} mark without any weight suspended on it.
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Question : 52 of 76
Marks: +1, -0
What minimum weight placed on this metre ruler can balance this ruler when it is pivoted at its centre?
Solution:  
To balance the ruler with minimum weight of ' WW ' gf, the weight must be suspended from 100 cm100\,\text{cm} mark.
In equilibrium,
Sum of anticlockwise moment == sum of clockwise moment.
  ⇒50×10  gf  cm=W×50  gf  cm\; \Rightarrow 50 \times 10\;\text{gf}\;\text{cm} = W \times 50\;\text{gf}\;\text{cm}
  ∴  W=50050=10  gf\; \therefore \; W = \frac{500}{50} = 10\;\text{gf}
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