The cross-section of a canal is in the shape of an isosceles trapezium which is 3 m wide at the bottom and 5 m wide at the top. If the depth of the canal is 2 m and it is 110 m long, what is the maximum capacity of this canal? (Take π = 22/7)

The canal is 3 m wide at the bottom and 5 m at the top, thus the canal is in the shape of a trapezium whose volume will be equal to the product of area of trapezium and the width.

Length of canal = 110 m

Volume of canal = $$(\frac{1}{2} \times $$(sum of parallel sides) $$\times$$ height $$) \times$$ length

=> Volume = $$(\frac{1}{2} \times (5+3) \times 2)\times 110$$

= $$8 \times 110 = 880$$ $$m^3$$

=> Ans - (D)