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Standard X

Physics

Translational and Rotational Equilibrium

A metre scale...

Question

A metre scale is supported at the centre. It is balanced by two weights. A and B, find the distance of B from the pivot clockwise and anticlockwise moment about 50 cm division are equal.

2.6 cm

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40 cm

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80 cm

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9.5 cm

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Solution

The correct option is C 40 cm
$20 \times d = 40 \times 20$
So, $d = \frac{800}{20} = 40 c m$
Hence, the 20 N force of B is acting from 90 cm walk.

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Similar questions

The diagram in the figure shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate:

(i) the total anticlockwise moment about O,

(ii) the total clockwise moment about O,

(iii) the difference of anticlockwise and clockwise moments,

and (iv) the distance from O where a 100 gf weight should be placed to balance the metre rule.

Q. The diagram shows a uniform metre rule weighing

100 g f

, pivoted at its centre

O

. Two weights

150 g f

and

250 g f

hang from the points

A

and

B

of the metre rule such that

O A = 40 c m

and

O B = 20 c m

. Calculate:
(i) the total anticlockwise moment about

O

,
(ii) the total clockwise moment about

O

;
(iii) the total clockwise and anticlockwise moments; and
(iv) the distance from

O

where a

100 g f

weight should be placed to balance the metre rule.

A meter rule is supported at its centre. It is balanced by two weights, A and B. If A and B are placed at a distance 30 cm, and 60 cm from the centre of scale, find the weight of B. The weight of A is 50N.

Q. A uniform metre scale has two weights of 10 gf and 8 gf suspended at the 10 cm and 80 cm marks respectively. If the metre scale itself weights 50 gf, find where must the weight be, so that the metre scale stays balanced?

Q. A metre scale is balanced on a knife-edge at its centre. When a mass of

10

kg is kept at the

12

cm mark, the scale is balanced at

45

cm. What is the mass of the metre scale?

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A metre scale is supported at the centre. It is balanced by two weights. A and B, find the distance of B from the pivot clockwise and anticlockwise moment about 50 cm division are equal.

The correct option is C 40 cm20×d=40×20So, d=80020=40cmHence, the 20 N force of B is acting from 90 cm walk.

The correct option is C 40 cm
$20 \times d = 40 \times 20$
So, $d = \frac{800}{20} = 40 c m$
Hence, the 20 N force of B is acting from 90 cm walk.