Earth dug out on making a circular tank of radius 7 m is spread all around the tank uniformly to a width of 1 m to form an embankment of height 3.5 m.Calculate the depth of the tank.
A) \[\dfrac{16}{15}m\]
B) $14m$
C) $2m$
D) $\dfrac{15}{14}m$
Answer
634.5k+ views
Hint: The volume of a circular tank is $\pi {{r}^{2}}{{h}_{1}}$ where is radius of tank, is height of tank .Here the volume of circular tank is equal to volume of embankment.And volume of embankment is equal to $\pi {{r}_{1}}^{2}{{h}_{2}}-\pi {{r}_{2}}^{2}{{h}_{2}}$ whereis inner radius of embankment,is outer radius of embankment and is height of embankment.
Complete step-by-step answer:
From the problem it is given that,
A circular tank of radius $r=7m$
Depth or Height of tank =${{h}_{1}}$=?
Volume of the circular tank =${{V}_{1}}$=$\pi {{r}^{2}}{{h}_{1}}$=$\pi \times {{\left( 7 \right)}^{2}}\times {{h}_{1}}$⟶equation(1)
Also given ,
Inner radius of embankment =${{r}_{1}}=7m$(i.e., radius of circular tank)
Outer radius of embankment = ${{r}_{2}}=8m$(i.e., radius of circular of circular tank + its width=7+1=8)
Height of embankment=${{h}_{2}}$=$3.5m$
Now ,the volume of embankment = ${{V}_{2}}$=$\pi {{r}_{2}}^{2}{{h}_{2}}-\pi {{r}_{1}}^{2}{{h}_{2}}$
On substituting all the values in above volume formula we get
${{V}_{2}}=\pi {{h}_{2}}\left( {{r}_{2}}^{2}-{{r}_{1}}^{2} \right)$
${{V}_{2}}=\pi \times 3.5\left( {{8}^{2}}-{{7}^{2}} \right)$⟶equation(2)
The volume of circular tank = The volume of embankment (Since it is dug uniformly)
Therefore on equating R.H.S of the equation(1) & equation(2) we get as follows
$\pi \times {{\left( 7 \right)}^{2}}\times {{h}_{1}}$$=\pi \times 3.5\left( {{8}^{2}}-{{7}^{2}} \right)$
Here π gets cancelled out ,then
${{h}_{1}}=\dfrac{3.5\left( 64-49 \right)}{49}$m
${{h}_{1}}=\dfrac{3.5(15)}{49}$m
On simplifying we get
${{h}_{1}}=\dfrac{15}{14}$m
Hence, the height or depth of the tank is ${{h}_{1}}=\dfrac{15}{14}$m
The correct option is (D)
Note: Here depth is always equal to the height in earth dug which is only for uniformly spreading. The embankment width should include its outer radius usually.And make sure that the all units should have an absolute solution.
Complete step-by-step answer:
From the problem it is given that,
A circular tank of radius $r=7m$
Depth or Height of tank =${{h}_{1}}$=?
Volume of the circular tank =${{V}_{1}}$=$\pi {{r}^{2}}{{h}_{1}}$=$\pi \times {{\left( 7 \right)}^{2}}\times {{h}_{1}}$⟶equation(1)
Also given ,
Inner radius of embankment =${{r}_{1}}=7m$(i.e., radius of circular tank)
Outer radius of embankment = ${{r}_{2}}=8m$(i.e., radius of circular of circular tank + its width=7+1=8)
Height of embankment=${{h}_{2}}$=$3.5m$
Now ,the volume of embankment = ${{V}_{2}}$=$\pi {{r}_{2}}^{2}{{h}_{2}}-\pi {{r}_{1}}^{2}{{h}_{2}}$
On substituting all the values in above volume formula we get
${{V}_{2}}=\pi {{h}_{2}}\left( {{r}_{2}}^{2}-{{r}_{1}}^{2} \right)$
${{V}_{2}}=\pi \times 3.5\left( {{8}^{2}}-{{7}^{2}} \right)$⟶equation(2)
The volume of circular tank = The volume of embankment (Since it is dug uniformly)
Therefore on equating R.H.S of the equation(1) & equation(2) we get as follows
$\pi \times {{\left( 7 \right)}^{2}}\times {{h}_{1}}$$=\pi \times 3.5\left( {{8}^{2}}-{{7}^{2}} \right)$
Here π gets cancelled out ,then
${{h}_{1}}=\dfrac{3.5\left( 64-49 \right)}{49}$m
${{h}_{1}}=\dfrac{3.5(15)}{49}$m
On simplifying we get
${{h}_{1}}=\dfrac{15}{14}$m
Hence, the height or depth of the tank is ${{h}_{1}}=\dfrac{15}{14}$m
The correct option is (D)
Note: Here depth is always equal to the height in earth dug which is only for uniformly spreading. The embankment width should include its outer radius usually.And make sure that the all units should have an absolute solution.
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