A uniform metre rule is balanced on a knife edge at the 50 cm mark. A 2 N weight hangs at the 20 cm mark. To restore balance, a 3 N weight should hang at the:
A60 cm mark
B70 cm mark
C80 cm mark
D75 cm mark
Answer & Solution
Correct answer: B. 70 cm mark
1. By the principle of moments, anticlockwise moment about pivot equals clockwise moment.
2. Pivot is at 50 cm. The 2 N weight is at 20 cm, so its arm is 50 − 20 = 30 cm.
3. Anticlockwise moment = 2 × 30 = 60 N·cm.
4. Let the 3 N weight hang at x cm to the right of pivot, with arm (x − 50) cm.
5. Clockwise moment = 3 × (x − 50). Setting it equal to 60 gives x − 50 = 20.
6. So x = 70 cm.
_Source: Selina Concise Physics ICSE Class 10, Ch 1 'Force' (icsehelp.com chapter extract)_
Related questions
Two children push on opposite ends of a uniform 3 m board pivoted at its centre. One pusheA see-saw of length 3 m is pivoted at its centre. A 40 kg boy sits 1.0 m from the pivot. WA uniform metre rule pivoted at 60 cm carries a 5 N weight at 20 cm. What single force appA heavy wheel is to be lifted using a crowbar. To minimise effort, the fulcrum should be pA uniform beam of length 4 m and weight 60 N rests on two supports, one at each end. A 100A force of 10 N acts at 60° to a 0.4 m long lever, measured from the pivot end. What is thWhich combination of conditions correctly describes equilibrium of an extended rigid body A uniform metre rule weighing 1 N is suspended at the 30 cm mark. A 4 N weight hangs at th