(a) \(d = 15 cm \implies r = \frac{15}{2} cm\)
\(V = \frac{4\pi r^{3}}{3}\)
\(V = \frac{4}{3} \times \frac{22}{7} \times (\frac{15}{2})^{3}\)
\(V = \frac{1125 \times 22}{7 \times 2}\)
\(V = 1,767.857 cm^{3}\)
(b)(i) Longitude difference : 105° + 35° = 140°
Distance along the parallel of latitude = \(\frac{140}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 28\)
= \(\frac{140 \times 44 \times 6400 \times 0.8829}{360 \times 7}\)
= \(13812.48 km\)
\(\approxeq 13800 km\)
(ii) Distance from the equator : \(\frac{28}{360} \times 2 \times \frac{22}{7} \times 6400\)
= \(3128.89 km\)
\(\approxeq 3130 km\)