A cuboidal tank measuring 5m × 2.1 m × 4.5m is dug in one corner of a field measuring 13.5 m × 2.5m. The earth dug out is spread evenly over the remaining portion of the field. How much is the level of the field raised?
Solution
Correct Answer: Option C
We know that, volume of cuboid = length × breadth × height
∴ Volume of earth dug out from the tank = 5m × 2.1 m × 4.5m = 47.25 m3
Also, Volume of the cuboidal tank = volume of earth dug out from the tank.
Now,
the exposed area of the field upon which the earth dug out is spread evenly = area of the total field – area of the top face of the tank
Area of the total field = 13.5 m × 2.5m = 33.75 sq. m.
Area of the top face of the tank = 5m × 2.1 m = 10.5 sq. m.
The exposed area of the field = (33.75 – 10.5) sq. ft. = 23.25 sq. m.
Now, the earth dug out is spread evenly over 23.25 sq. m. area.
If level of field raises by L m. then from volume constancy we can say,
23.25 × L = 47.25
⇒ L ≈ 2.03 m
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