The diagram shows a uniform metre rule balanced at its mid-point. What is the weight W placed at the $20$ cm mark?

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Solution

The correct option is B $4.0$ N
As the metre rule is in the equilibrium
Equating the torque by both weights about mid-point

$W \times 30 = 6.0 \times 20$
$W = \frac{120}{30} = 4.0 N$

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The diagram shows a uniform metre rule balanced at its mid-point. What is the weight W placed at the 20cm mark?

The correct option is B 4.0 NAs the metre rule is in the equilibrium Equating the torque by both weights about mid-point W×30=6.0×20W=12030=4.0 N

The diagram shows a uniform metre rule balanced at its mid-point. What is the weight W placed at the $20$ cm mark?

The correct option is B $4.0$ N
As the metre rule is in the equilibrium
Equating the torque by both weights about mid-point

$W \times 30 = 6.0 \times 20$
$W = \frac{120}{30} = 4.0 N$