Question

a well of diameter 14 cm is dug 15cm deep the earth taken out of it has been spread evenly to form circular embankment Bank
all around the Wall of width 7 cm find the height of the embankment​

Answer 1

Answer:

Radius of the well, r = 7 m, and its depth, h = 15m

Volume of the earth dug out = volume of the well = (πr²h) cubic units

=(22/7×7×715)m3=2310m³

Width of the embankment = 7m ltbr.

External radius of the embankment =(7+7)m=14m

Internal radius of the embankment = 7m

Area of the embankment =π×[(14)²−7²]m²

=[22/7×(14+7)×(14−7)]m²

=(22/7×21×7)m²=462m²

Volume of the embankment = volume of the earth dug out

=2310m³

Height of the embankment =(volume of the embankment in m³/area of the embankment in m²)

=(2310/462)m

=5m

Hence, the height of the embankment formed = 5m

Answer 2

$\huge{\underline{\underline{\bf{\purple{Given}}}}}$

A well of diameter 14 cm is dug 15cm deep. The earth taken out of it has been spread evenly to form circular embankment Bank

all around the Wall of width 7 cm

$\huge{\underline{\underline{\bf{\purple{To\:Find}}}}}$

Find the height of the embankment

$\huge{\underline{\underline{\bf{\purple{Solution}}}}}$

Radius of well (r) = 14/2 = 7cm

Volume of well = πr²h

$\implies\sf π×7×7×15$

$\implies\sf 735πcm^3$

Radius of the well with embankment (R)

= Radius of well + width of embankment wall

= 7 + 7 = 14cm

Volume of circular embankment

$\implies\sf πR^2h-πr^2h$

$\implies\sf πh(R^2-r^2)$

$\implies\sf =πh(14^2-7^2)$

$\implies\sf =πh(196-49)$

$\implies\sf =147πhcm^3$

Volume of well = volume of circular embankment

$\implies\sf 735π=147πh$

$\implies\sf 735π=147πh$

$\implies\sf h=\frac{735π}{147π}$

$\implies\sf h=5cm$

$\large{\boxed{\bf{Required\:height=5cm}}}$