Question

A uniform meter rule is pivoted at the 30cm Mark.It balances horizontally when a mass of 45g is hung from the 5cm mark

Answer 1

Answer:

Explanation:

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ThereddevilsSamaritan

The weight of the scale is 50 gf.

Let the weight of the scale be 'm' gf

The metre scale is pivoted at 30cm.

The torque of 40gf about the pivot (clockwise moment)

T₁ = Mass × Acc. due to gravity × Distance of the weight from pivot.

As 40gf = mass × acc. due to gravity and,

Distance of this weight from pivot = (30 - 5)cm = 25cm

So, T₁ = 40 × 25 gf-cm = 1000 gf-cm

The weight of the scale acts on its centre of mass which is situated at the mid point (50 cm) of the rod.

Thus the distance of centre of mass of the scale from the pivot is = (50 - 30)cm = 20 cm                               [As the scale is balanced at 30cm]

So, the torque of the 'm' gf weight of the scale about the pivot (anti clockwise moment) is

T₂ = Weight of rod × Distance of the weight from pivot.

= m × 20

As the sum of clockwise moments = Sum of anti-clockwise moments. So

T₁ = T₂

Or, 1000 = 20m

⇒m = 1000/20 = 50gf

Weight is 50gf.

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Answer 2

The answer is 56.25 g.

Given:

A meter rule pivoted at 30 cm.

A mass of 45g hung at 5cm

To Find:

Mass of the meter rule

Solution:

As the ruler is pivoted at 30 cm, the torque will be balanced at this point.

There are two torques acting on the ruler, one  $45*10=450N$ at 5 cm, and one  $m*10=10m$ at 50 cm, where m is the mass of the ruler.

The force acts at the centre of mass of the ruler which is at 50 cm.

Hence equating the torques, we get

$450*[30-5]=10m[50-30]\\\\m=\frac{45*25}{20}\\\\m=56.25g$

The mass of the ruler is 56.25 g.

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