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ICSE Class 10 Physics Question 8 of 20

Force — Question 10

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Question 10

A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100gf be suspended to keep the rule horizontal?

Answer

Let us assume that a 50 gf weight produces an anticlockwise moment about the middle point ( 50 cm ).

Now, if a weight of 100 gf produces a clockwise moment about the middle point and d cm be the distance from the middle.

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment.

50gf×50cm=100gf×dcm⇒d=50×50100⇒d=25cm50 gf \times 50cm = 100gf \times d cm \\[0.5em] \Rightarrow d = \dfrac{50\times50}{100}\\[0.5em] \Rightarrow d = 25 cm

Therefore, a weight of 100gf will be suspended at a distance of 25cm to keep the ruler balanced.