ong>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