a field is 16m long and 12m wide In one corner of the field, a pit measuring 4m into 2m into 1m is dug and the sand removed from the pit is evenly spread on The remaining area of the field. find how much is the level of field raise?
Answers
Answer:
⇒Given, Dimension of the field = 16 m × 12 m
⇒Dimension of the pit = 8 m × 2.5m × 2m
⇒ Volume of earth removed from the pit = 8m × 2.5 m × 2 m = 40 m3
⇒ Area of the remaining field = Area of the field – Area of the pit
= 16 m × 12 m – 8 m × 2.5 m
= 460 m²
Since, the earth removed is evenly spread over the remaining area of the field.
∴ Increase in level of remaining field × Area of remaining field = Volume of earth removed from the pit
Increase in level of remaining field = 40/460 =11.5m
Given: dimensions of the field = 16 m × 12 m
dimensions of the pit = 4 m × 2 m × 1 m
To find: The rise in level of the field
Solution: The volume of the sand removed from the pit will be the same as the volume of the pit.
Hence, volume of the pit = V = l × b × h
= 4 m × 2 m × 1 m
= 8 m³
∴ The volume of sand removed = 8 m³.
This volume of sand is spread across the remaining area of the field.
Now, let's find the area of the field = l × b
= 16 m × 12 m
= 192 m²
The area of the pit = 4 m × 2 m = 8 m².
Therefore, the remaining area of the field = (192 - 8) m²
= 184 m²
We know that volume is given by the product of height and area of the base.
As such, let the rise in level of the field be h cm.
Accordingly,
V = A × h
⇒ 8 = 184 × h
⇒ h = 8/184
⇒ h = 0.04
Therefore, the rise in level of the field = 0.04 m.
Answer: 0.04 m