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प्रश्न
Fig. 5 shows a uniform meter scale weighing 200 gf. Provided at its centre. Two weights 300 gf and 500 gf are suspended from the ruler as shown in the diagram. Calculate the resultant torque of the ruler and hence calculate the distance from mid-point where a 100 gf should be suspended to balance the meter scale.

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उत्तर
Resultant torque = sum of clockwise moments - sum of anticlockwise moments
Taking, moments about the mid point
Resulttant torque = (300 x 40) - (500 x 20)
or, Resultant torque = 12000 - 10000 = 2000 gf-cm
Let a mass of 100 gf be suspended at a distance 'd' from the mid point towards the right side,
so as to balance the metre scale.
Then, in balanced condition:
sum of clockwise moments = sum of anticlockwise moments
(300 x 40) = (500 x 20) + (100 x d)
or, 12000 = 10000 + 100d
or, 100d = 2000
or,d = 20cm to the right of the mid-point
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संबंधित प्रश्न
A half metre rod is pivoted at the centre with two weights of 20gf and 12gf suspended at a perpendicular distance of 6 cm and 10 cm from the pivot respectively as shown below.

1) Which of the two forces acting on the rigid rod causes clockwise moment?
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Assertion & Reasoning
- Assertion: The sum of the clockwise moments is equal to the sum of the anticlockwise moments.
- Reason: The principle of conservation of momentum is valid if the external force on the system is zero.
