WAEC Physics 2024 Paper

Concept:At the lowest point of the circular path, the tension must provide the centripetal force and also balance the weight of the stone.Explanation:Convert the mass of the stone into kilograms: $200\ \mathrm{g} = 0.2\ \mathrm{kg}$.The radius of the circle is $r = 1.5\ \mathrm{m}$ and the speed is $v = 5\ \mathrm{m/s}$.Take the acceleration due to gravity as $g = 10\ \mathrm{m/s^2}$.The centripetal force is given by $F_c = \frac{mv^2}{r}$.Substitute the values into the formula: $F_c = \frac{0.2 \times (5)^2}{1.5}$.Simplify the expression: $F_c = \frac{0.2 \times 25}{1.5} = \frac{5}{1.5} = 3.33\ \mathrm{N}$.At the bottom of the circle, the stone's weight acts downward, so the tension acts upward and must also oppose the weight.Thus, the net force equation is $T = F_c + mg$.Calculate the weight of the stone: $mg = 0.2 \times 10 = 2.0\ \mathrm{N}$.Therefore, $T = 3.33\ \mathrm{N} + 2.0\ \mathrm{N} = 5.33\ \mathrm{N} \approx 5.3\ \mathrm{N}$.Answer:The tension in the string at the bottom of the circle is $5.3\ \mathrm{N}$, which is option A.