Question

* Earth dug out on making a circular tank of radius 7 m is spread all round the tank uniformly to a wide
of 1 m to form an embankment of height 3.5 m. Then the depth of the tank is
​

Answer 1

et h1h1 be the height of circular tank Then

Volume of circular tank= Volume of embankment=πr2h1πr2h1.....(1)

Height of embankment, h2h2=3.5 m

Inner radius of embankment, r1r1=7m

Outer radius of embankment, r2r2=8m

∴∴ Volume of embankment=πr21h2−πr22h2πr12h2−πr22h2

=π(82×3.5−72×3.5)π(82×3.5−72×3.5)

=π(224−49×3.5)=52.5ππ(224−49×3.5)=52.5π

Substituting in (1) we get

π×72×h1=52.5ππ×72×h1=52.5π

or h1=52.549=7.57×22=1514mh1=52.549=7.57×22=15/14m

Answer 2

GIVEN:-

• $\rm{inner\:Radius\:of\:tank(R)=7cm}$

• $\rm{Height\:of\:embankment(h)=3.5m}$

• $\rm{Outer\:Radius\:of\:embankment(r)=8m}$

TO FIND:-

• Height of the tank(H)

FORMULAE USED:-

• ${\boxed{\rm{Volume\:of\:object=\pi r^2h}}}$

Where,

R= Radius

H = Height.

Now,

Volume of tank = Volume of embankment

So,

Volume of embankment= πr²h.......1

Now,

$\implies\rm{Volume\:of\:embankment=\pi R^{2}h-\pi r^2h}$

$\implies\rm{Volume\:of\: embankment= \pi ( 8^2\times{3.5}-7^2\times{3.5})}$

$\implies\rm{\pi(224-49\times{3.5})}$

$\implies\rm{(52.5\pi)}$

Substituting (1) in we get.

$\implies\rm{\pi r^2h=\dfrac{22}{7}\times{7^2}\times{h}}$

$\implies\rm{52.5\pi=\dfrac{22}{7}\times{49}\times{h}}$

$\implies\rm{52.5\pi=\dfrac{22}{\cancel{7}}\times{\cancel{49}}\times{h}}$

$\implies\rm{52.5\pi=22\times{7}\times{h}}$

$\implies\rm{52.5\times{\dfrac{22}{7}=154h}}$

$\implies\rm{1155=1078h}$

$\implies\rm{h=\dfrac{\cancel{1155}}{\cancel{1078}}}$

$\implies\rm{h=\dfrac{15}{14}}$

Hence, The height of tank is 15\14m