JAMB Physics 2023 Paper

Concept:For a body in equilibrium, the sum of clockwise moments about the pivot equals the sum of anticlockwise moments.Explanation:Weight of the block:$$W = mg = 0.5 \times 10 = 5\text{ N}$$The metre rule is light, so its weight is neglected.Take moments about the pivot at E, which is the 90 cm mark.Distance from E to the 0.5 kg mass at the 40 cm mark is:$$90 - 40 = 50\text{ cm}$$Distance from E to the string at D, the 60 cm mark, is:$$90 - 60 = 30\text{ cm}$$Only the vertical component of the tension, $T \sin 30^\circ$, produces a moment about E.Using the equilibrium condition:$$T \sin 30^\circ \times 30 = 5 \times 50$$Since $\sin 30^\circ = 0.5$,$$T \times 0.5 \times 30 = 250$$$$15T = 250$$$$T = \frac{250}{15} = 16.67\text{ N}$$Answer:The tension in the string is $16.67\text{ N}$.Therefore, the correct option is A.