If the radius of the circular ends of a truncated conical bucket of height 45 cm are 14 cm and 7 cm, then find the capacity in cubic cm of the bucket.
Answer & explanation
Correct answer: option 2
Volume of the bucket = \(\frac{1}{3}\) \(\pi \)H(r12 + r22 + r1r2)
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 45 (142 + 72 + 14 × 7)
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 45 × 343
= 16,170