Question -
Answer -
Given,
Diameter of the well =3 m
So, the radius of thewell = 3/2 m = 1.5 m
Depth of the well (h)= 14 m
Width of theembankment (thickness) = 4 m
So, the radius of theouter surface of the embankment = (4 + 1.5) m = 5.5 m
Let the height of theembankment be taken as h m
We know that theembankment is a hollow cylinder
Volume of theembankment = π (R2 – r2) × h
= π (5.52 –1.52) ×h ….. (i)
Volume of earth dugout = π × r2 × h
= π × 22 ×14 ….. (ii)
On equating both (i)and (ii) we get,
π (5.52 –1.52) × h = π × (3/2)2 × 14
(30.25 – 2.25) x h = 9x 14/ 4
h = 9 x 14/ (4 x 28)
h = 9/8 m
Therefore, the heightof the embankment is 9/8 m