Question -
Answer -
Given,
Radius of the circularcylinder (r) = 2/2 m = 1 m
Height of the well (h)= 14 m
We know that,
Volume of the solidcircular cylinder = π r2h
= π × 12×14 …. (i)
And,
The height of theembankment (h) = 40 cm = 0.4 m
Let the width of theembankment be (x) m.
The embankment is ahollow cylinder with external radius = 1 + x and internal radius = 1
Volume of theembankment = π × r2 × h
= π × [(1 + x)2 –(1)2]× 0.4 ….. (ii)
As the well is spreadevenly to form embankment then the volumes will be same.
So, on equatingequations (i) and (ii), we get
π × 12 ×14 = π × [(1 + x)2 – (1)2] x 0.4
14/0.4 = 1 + x2 +2x – 1
35 = x2 +2x
x2 +2x – 35 = 0
Solving byfactorization method, we have
(x + 7) (x – 5) = 0
So, x = 5 m can onlybe the solution as it’s a positive value.
Therefore, the widthof the embankment is 5 m.