Answer
453.3k+ views
Hint: Capacity of the overhead tank is the volume of water it can store. So, we have to determine the volume of the overhead tank, i.e. a cylinder is given by $V=\pi {{r}^{2}}h$ and also the water capacity of the sump, i.e. a cuboid given by ${{V}_{c}}=l\times b\times h$.
We know that Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank), which is in the shape of a cuboid.
Now, we know that the overhead tank is in the shape of a cylinder whose radius is given as r = 60 cm and its height is given as h = 95 cm.
We know that the water holding capacity of the overhead tank is equal to the volume of the cylinder.
We also know that the volume of the cylinder is equal to $\pi {{r}^{2}}h$. So, let the volume of the cylindrical tank be V. Then we can write,
$V=\pi {{r}^{2}}h$
Putting the value of r and h, we get,
$V=\pi {{\left( 60 \right)}^{2}}\left( 95 \right)$
Using the value of $\pi =\dfrac{22}{7}$ as mentioned in the question, we get
$V=\dfrac{22}{7}\left( 3600 \right)\left( 95 \right)$
It is given in cubic cm, so we have to convert it into the cubic meter as below,
$V=\dfrac{22}{7}\left( 3600 \right)\left( 95 \right)\dfrac{1}{1000000}$
$V=\dfrac{22}{7}\left( 0.36 \right)\left( 0.95 \right)\text{ meter cubic}....\left( i \right)$
Now we know that the cylindrical overhead tank is filled by the sump, which is in cuboidal shape. The dimensions of the sump i.e. in the shape of the cuboid are given as 1.57m x 1.44m x 0.95m. Therefore, we can write that
Length of the cuboidal tank, l = 1.57 m.
The breadth of the cuboidal tank, b = 1.44 m.
Height of the cuboidal tank, h = 0.95 m.
The capacity of the tank is equal to the volume of the cuboid.
Let the volume of the cuboid be ${{V}_{c}}$. Then, we can write it as,
${{V}_{c}}=l\times b\times h$
Substituting the values, we get
${{V}_{c}}=1.57\times 1.44\times 0.95....\left( ii \right)$
Let the height of the water left in the sump after filling the overhead tank be ${{h}_{o}}$. Breadth and length of cuboid would remain the same after defiling the tank.
So, we can write that the volume of water left in the sump
\[=l\times b\times {{h}_{o}}=1.57\times 1.44\times {{h}_{o}}....\left( iii \right)\]
The volume of the water consumed from the cuboidal tank (sump) = volume of the cylindrical tank = V
So, we get,
Water left in the sump = volume of the cuboid - the volume of water consumed
Water left in the sump $={{V}_{c}}-V$
From (i) , (ii) and (iii)
$1.57\times 1.44\times {{h}_{o}}=1.57\times 1.44\times 0.95-\left( \dfrac{22}{7}\left( 0.36 \right)\left( 0.95 \right) \right)$
Now solving the equation to get the value of ${{h}_{o}}$, we get
$1.57\times 1.44\times {{h}_{o}}=1.0729$
${{h}_{o}}=0.47456m$
Rounding off up to two terms of decimal, we can write the height as
${{h}_{o}}=0.47m$
The height of the water left in the sump after the overhead tank has been filled with water from a sump that had been full is 0.47m or 47 cm.
Now we need to compare the capacity of the tank with that of the sump.
We know that,
The volume of the water consumed from the cuboidal tank (sump) = volume of the cylindrical tank
The volume of water consumed from the sump $=l\times b\times \left( 0.95-{{h}_{o}} \right)$
We have found the value of ${{h}_{o}}$, so we can substitute it in the above equation as,
The volume of water consumed from the sump $=l\times b\times \left( 0.95-0.47 \right)$
On comparing it with the capacity of the sump (volume of the cuboidal tank) $=\dfrac{l\times b\times (0.95-0.47)}{l\times b\times 0.95}$, l, b will get canceled and we will get the ratio as,
$=\dfrac{0.48}{0.95}=0.5052$
So, we can say that the capacity of the overhead tank is 0.5 of the capacity of the sump.
Therefore, the capacity of the overhead tank is half of the capacity of the sump.
Note: We can compare the water of the overhead tank with the sump in an alternate way also i.e. by taking directly into consideration the volume of the overhead tank which is in the form of a cylinder, so now we will compare as \[\dfrac{\dfrac{22}{7}(0.36)(0.95)}{1.57\times 1.44\times 0.95}\] canceling 0.95 and evaluating, we will get it as 0.5. Therefore, the capacity of the overhead tank is half of the capacity of the sump.
We know that Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (underground tank), which is in the shape of a cuboid.
Now, we know that the overhead tank is in the shape of a cylinder whose radius is given as r = 60 cm and its height is given as h = 95 cm.
We know that the water holding capacity of the overhead tank is equal to the volume of the cylinder.
We also know that the volume of the cylinder is equal to $\pi {{r}^{2}}h$. So, let the volume of the cylindrical tank be V. Then we can write,
$V=\pi {{r}^{2}}h$
Putting the value of r and h, we get,
$V=\pi {{\left( 60 \right)}^{2}}\left( 95 \right)$
Using the value of $\pi =\dfrac{22}{7}$ as mentioned in the question, we get
$V=\dfrac{22}{7}\left( 3600 \right)\left( 95 \right)$
It is given in cubic cm, so we have to convert it into the cubic meter as below,
$V=\dfrac{22}{7}\left( 3600 \right)\left( 95 \right)\dfrac{1}{1000000}$
$V=\dfrac{22}{7}\left( 0.36 \right)\left( 0.95 \right)\text{ meter cubic}....\left( i \right)$
Now we know that the cylindrical overhead tank is filled by the sump, which is in cuboidal shape. The dimensions of the sump i.e. in the shape of the cuboid are given as 1.57m x 1.44m x 0.95m. Therefore, we can write that
Length of the cuboidal tank, l = 1.57 m.
The breadth of the cuboidal tank, b = 1.44 m.
Height of the cuboidal tank, h = 0.95 m.
The capacity of the tank is equal to the volume of the cuboid.
Let the volume of the cuboid be ${{V}_{c}}$. Then, we can write it as,
${{V}_{c}}=l\times b\times h$
Substituting the values, we get
${{V}_{c}}=1.57\times 1.44\times 0.95....\left( ii \right)$
Let the height of the water left in the sump after filling the overhead tank be ${{h}_{o}}$. Breadth and length of cuboid would remain the same after defiling the tank.
So, we can write that the volume of water left in the sump
\[=l\times b\times {{h}_{o}}=1.57\times 1.44\times {{h}_{o}}....\left( iii \right)\]
The volume of the water consumed from the cuboidal tank (sump) = volume of the cylindrical tank = V
So, we get,
Water left in the sump = volume of the cuboid - the volume of water consumed
Water left in the sump $={{V}_{c}}-V$
From (i) , (ii) and (iii)
$1.57\times 1.44\times {{h}_{o}}=1.57\times 1.44\times 0.95-\left( \dfrac{22}{7}\left( 0.36 \right)\left( 0.95 \right) \right)$
Now solving the equation to get the value of ${{h}_{o}}$, we get
$1.57\times 1.44\times {{h}_{o}}=1.0729$
${{h}_{o}}=0.47456m$
Rounding off up to two terms of decimal, we can write the height as
${{h}_{o}}=0.47m$
The height of the water left in the sump after the overhead tank has been filled with water from a sump that had been full is 0.47m or 47 cm.
Now we need to compare the capacity of the tank with that of the sump.
We know that,
The volume of the water consumed from the cuboidal tank (sump) = volume of the cylindrical tank
The volume of water consumed from the sump $=l\times b\times \left( 0.95-{{h}_{o}} \right)$
We have found the value of ${{h}_{o}}$, so we can substitute it in the above equation as,
The volume of water consumed from the sump $=l\times b\times \left( 0.95-0.47 \right)$
On comparing it with the capacity of the sump (volume of the cuboidal tank) $=\dfrac{l\times b\times (0.95-0.47)}{l\times b\times 0.95}$, l, b will get canceled and we will get the ratio as,
$=\dfrac{0.48}{0.95}=0.5052$
So, we can say that the capacity of the overhead tank is 0.5 of the capacity of the sump.
Therefore, the capacity of the overhead tank is half of the capacity of the sump.
Note: We can compare the water of the overhead tank with the sump in an alternate way also i.e. by taking directly into consideration the volume of the overhead tank which is in the form of a cylinder, so now we will compare as \[\dfrac{\dfrac{22}{7}(0.36)(0.95)}{1.57\times 1.44\times 0.95}\] canceling 0.95 and evaluating, we will get it as 0.5. Therefore, the capacity of the overhead tank is half of the capacity of the sump.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)