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 the last bit with explanation
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Answer 1

GiveN :-

• 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.

To Do :-

1. (a) Draw a diagram of the arrangement.
2. (b) Find the mass of the half metre rule.
3. (c) In which direction would the balancing point shift if 20 g mass is shifted inside from its one end ?

SolutioN :-

Let the mass of 20g be hanged at a distance of x from the pivot . From the diagram , we see that (Refer to attachment ) , the measure of BO = AO - AB = 29cm - 25cm = 4cm . Now according to the law of moments we know that the sum of clockwise moments is equal to the sum of anticlockwise moments. In other words the algebraic sum of all the moments acting along a point is 0 . Now we know that since the scale is uniform the centre of mass will act at 25 cm .

$\red{\bigstar}\underline{\underline{\boldsymbol{ According\ to \ Law \ of \ Moment :-}}}$

$\sf:\implies\pink{\sum (Clockwise\ moments)=\sum( Anticlockwise \ moments )} \\\\\sf:\implies (m_1)(d_1)=(m_2)(d_2) \\\\\sf:\implies (29cm-25cm)(m_{(scale)}= (20g)(50cm-29cm)\\\\\sf:\implies 4m_{(scale)} = 20g(21 cm) \\\\\sf:\implies m_{(scale)}=\dfrac{20\times 21}{4} \\\\\sf:\implies \underset{\blue{\sf Required\ Mass}}{\underbrace{\boxed{\pink{\frak{ Mass_{(scale)}= 105\ grams }}}}}$