Answer

Verified

456.3k+ 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

What number is 20 of 400 class 8 maths CBSE

Which one of the following numbers is completely divisible class 8 maths CBSE

What number is 78 of 50 A 32 B 35 C 36 D 39 E 41 class 8 maths CBSE

How many integers are there between 10 and 2 and how class 8 maths CBSE

The 3 is what percent of 12 class 8 maths CBSE

Find the circumference of the circle having radius class 8 maths CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

One cusec is equal to how many liters class 8 maths CBSE