Question

Q no.2
3
A 21m deep well with diameter 6m is dig and the
earth from the digging is evenly spread to from a
platform 27m x 11m. Find the height of the
platform.​

Answer 1

Given,depth of well=21m

Radius of well=6/2=3

Volume of earth dugout from the well=πr^2h

22/7×3×3×21

=594m^3

Let the height of the platform=h m.

Volume of platform=volume of earth dugout

27×11×h=594

h=594/27×11=594/297

so height of the platform,h=2m.

Answer 2

Answer:

$\bigstar{\bold{Height\:of\:platform=2\:m}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Height of the well (h) = 21 m
• Diameter of the well (d) = 6 m
• Length of the platform (l) = 27 m
• Breadth of the platform (b) = 11 m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Height of the platform (H)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The volume of the cylindrical well = Volume of cuboidal platform

→ Volume of a cylinder is given by the equation,

  Volume of a cylinder = π r² h

→ Radius of the well = d/2 = 6/2 = 3 m

→ Substitute the given datas in the above equation,

  Volume of well = 22/7 × 3 × 3 × 21

  Volume of well = 22 × 3 × 3 × 3

  Volume of well = 594 m³

→ Volume of cuboid is given by the equation,

 Volume of a cuboid = l × b × H

→ Substitute the given datas,

   Volume of platform = 27 × 11 × H

   Volume of platform = 297 H

→ But we know that volume of platform = volume of well

→ Hence,

  297 H = 594

         H = 594/297

         H = 2 m

→ Hence height of the platform is 2 m

$\boxed{\bold{Height\:of\:platform=2\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The volume of a cylinder is given by the equation,

  Volume of cylinder = π r² h

→ Volume of a cuboid is give by the formula,

   Volume of cuboid = l  × b × h