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Standard XII

Physics

Newton's Second Law: Rotatory Version

A uniform rul...

Question

A uniform rule is pivoted at its mid point. A weight of $50 g f$ is suspended at one end of it. Where should a weight of $100 g f$ be suspended, to keep the rule horizontal?

10 c m

from mid point

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25 c m

from mid point

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20 c m

from mid point

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15 c m

from mid point

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Solution

The correct option is B $25 c m$ from mid point
To keep the rod horizontal, total torque should be zero
$\Rightarrow τ = 50 g f (0.5) - 100 g f (x) = 0$
$\Rightarrow 50 g f (0.5) = 100 g f (x)$
$\Rightarrow x = 0.25$
So weight should suspended from $25 c m$ from mid point.
Therefore, B is correct option.

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

The correct option is B 25 cm from mid pointTo keep the rod horizontal, total torque should be zero⇒τ=50gf(0.5)−100gf(x)=0⇒50gf(0.5)=100gf(x)⇒x=0.25So weight should suspended from 25cm from mid point.Therefore, B is correct option.

A uniform rule is pivoted at its mid point. A weight of $50 g f$ is suspended at one end of it. Where should a weight of $100 g f$ be suspended, to keep the rule horizontal?