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JAMB Physics 2019 Paper

Question : 96

Total: 100

The diagram shows a uniform meter rule AB which balances horizontally at the 90cm mark when a mass of 0.2kg is suspended from B. Calculate the mass of the meter rule.

0.05kg
0.20kg
0.80kg
1.00kg

Validate

Solution:

Concept:

For a uniform meter rule in equilibrium, the sum of clockwise moments about the pivot equals the sum of anticlockwise moments.

Explanation:

The rule balances at the

90\,\mathrm{cm}

mark, so the pivot is at

90\,\mathrm{cm}

Since the rule is uniform, its weight acts at the centre, which is the

50\,\mathrm{cm}

mark.

The mass of the rule is represented by

M_r

Moment of the rule about the pivot is

M_r(90 - 50) = 40M_r

Moment of the suspended mass about the pivot is

0.2(100 - 90) = 0.2 \times 10 = 2

For balance, set the moments equal:

40M_r = 2

M_r = \frac{2}{40} = 0.05\,\mathrm{kg}

Answer:

0.05\,\mathrm{kg}

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