Q.3 A well is dug out whose radius is 4.2m and depth is 38m. The earth taken out from well is spread all over a rectangular field whose dimensions are 130m * 115m . Find the rise in the level of field.
Answers
GIVEN :-
- A well is dug out whose radius is 4.2m and depth is 38m.
- It is then spread all over the rectangular field whose dimensions are 130m×115m.
TO FIND :-
- Rise (height) in the level of field .
TO KNOW :-
★ Volume of Cuboid = πr²h
★ Volume of Cuboid = l × b × h
SOLUTION :-
For well ,
- Radius(r) = 4.2m
- Height(H) = 38m
Well is in the form of Cylinder.
Volume of well dugged out = πr²H
ㅤㅤㅤㅤㅤㅤㅤㅤ= (22/7) × (4.2)² × 38
ㅤㅤㅤㅤㅤㅤㅤㅤ= (22/7) × 17.64 × 38
ㅤㅤㅤㅤㅤㅤㅤㅤ= 14747/7
ㅤㅤㅤㅤㅤㅤㅤㅤ= 2106.7 m³ -----(1)
__________________________
For field ,
- Length (l) = 130m
- Bredth (b) = 115m
Field is in the form of Cuboid.
Volume of field = l × b × h
ㅤㅤㅤㅤㅤ ㅤㅤ= 130 × 115 × h
ㅤㅤㅤㅤㅤㅤ ㅤ= 14950h m³ -------(2)
♦ When same amount of earth is digged out and spread all over the field , total volume in both the cases will be same.
So, equating (1) and (2) ,
→ 2106.7 = 14950h
→ h = 2106.7/14950
→ h = 0.14m = 14cm
Hence , there is a rise of 14cm or 0.14m in the field.
MORE TO KNOW :-
★ Volume of Cone = (1/3)πr²h
★ Volume of Cube = edge³
★ Volume of Sphere = (4/3)πr³
★ Volume of Hemisphere = (2/3)πr³