Question

16. A uniform half metre rule can be balanced at the
29.0 cm mark when a mass 20 g is hung from its one
end.
(a) Draw a diagram of the arrangement.
(b) Find the mass of the half metre rule.
(c) In which direction would the balancing point
shift if 20 g mass is shifted inside from its one
end ?
​

Answer 1

Answer:

Let the weight of the half metre rule be x gf and it acts at the 25 cm mark (centre of gravity).

Now,

Acticlockwise moment = x×4

Clockwise moment = 20×21

In equilibrium,

Anticlockwise moment = Clockwise moment

4x=20×21

x=105 gf.

Thus, the weight of the metre rule is 105gf.

We know, one gf is the force acting on the body of mass 1 grams, thus mass of the body is 105grams.

Answer 2

Answer:

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Hello Dear.

Here is the answer---

•¶ Refers to the attachment for the Question.

Let the weight of the half metre rule be x gf and it acts at the 25 cm mark (centre of gravity).

Now,

•Acticlockwise moment = x * 4

•Clockwise moment = 20 * 21

In equilibrium,

Anticlockwise moment = Clockwise moment

•4x = 20*21

x = 105 gf.

Thus, the weight of the metre rule is 105 gf.

•We know, one gf is the force acting on the body of mass 1 grams, thus mass of the body is 105 grams.