[amp_mcq option1=”50 cm” option2=”55 cm” option3=”56 cm” option4=”60 cm” correct=”option3″]
The correct answer is $\boxed{\text{B) 55 cm}}$.
The average depth of annual precipitation over the basin is the total precipitation divided by the area of the basin. The total precipitation is the sum of the precipitation in each isohyetal band. The area of each isohyetal band is given by the formula $A = b \times l$, where $b$ is the width of the band and $l$ is the length of the band. The width of each band is constant, so the area of each band is proportional to its length. The length of the band between the isohyets 45 cm and 55 cm is 100 km, and the length of the band between the isohyets 55 cm and 65 cm is 150 km. Therefore, the total precipitation is $P = 100 \times 45 + 150 \times 55 = 10250$ cm. The area of the basin is 250 km$^2$, so the average depth of annual precipitation is $P/A = 10250/250 = 55$ cm.
Option A is incorrect because it is the average depth of precipitation in the band between the isohyets 45 cm and 55 cm. Option C is incorrect because it is the average depth of precipitation in the band between the isohyets 55 cm and 65 cm. Option D is incorrect because it is the average depth of precipitation in the entire basin.