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

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1. 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

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