In this chapter we will discuss questions related volume of cylinder.

The formula for volume of cylinder is given as;

Volume = \mathtt{\pi r^{2} h}

Where, r is radius and h is height of cylinder.

## Volume of cylinder – solved problems

**Question 01**

A vessel in the form of cylinder is placed in science laboratory. The cylinder is of radius 21 cm and height 60 cm. Find the volume of water that can filled inside the vessel.

**Solution**

Calculating the volume of cylinder.

Volume = \mathtt{\pi r^{2} h}

Putting values in the formula, we get;

\mathtt{Volume\ =\ \frac{22}{7} \times 21\times 21\times 60\ =83160\ cm^{3} \ }

Hence, 83160 cu. cm of water can be filled inside the vessel.

**Question 2**

Let “V” be the volume of cylinder. What will happen to volume if radius is halved and height is doubled.

**Solution**

Let “r” and “h” be the height of original cylinder.

New radius (R) = r/2

New height (H) = 2h

Calculating volume of new cylinder;

\mathtt{Volume\ =\ \pi R^{2} H}\\\ \\ \mathtt{Volume\ =\ \pi \times \left(\frac{r}{2}\right)^{2} \times 2h}\\\ \\ \mathtt{Volume\ =\ \pi \times \frac{r^{2}}{4} \times 2h=\frac{1}{2} \pi r^{2} h}

Hence, the volume of new cylinder is half of original cylinder.

**Question 03**

Two cylinders are present in a room. The ratio of their radius is 4 : 1 and ratio of height is 1 : 4. Find the ratio of their volumes.

**Solution**

Let r1 & h1 be the dimension of first cylinder.

Let r2 & h2 be the dimension of second cylinder.

Since ratio of radius is in ratio 4 : 1

r1 = 4x and r2 = x

Since ratio of height is in ratio 1 : 4

h1 = y and h2 = 4y

Now taking the ratio of their volumes

\mathtt{\frac{cylinder\ 1}{cylinder\ 2} =\frac{\pi \times ( 4x)^{2} \times y}{\pi \times x^{2} \times ( 4y)}}\\\ \\ \mathtt{\frac{cylinder\ 1}{cylinder\ 2} =\ \frac{16}{4} =\frac{4}{1}}

Hence, 4 : 1 is the ration of their volumes.

**Question 04**

Steve decided to dig a well of diameter 2 meter and height 14 meters in his backyard. If the cost of digging is $10 per sq meter, then find the total cost incurred by Steve.

**Solution**

Diameter (D) = 2 meter

Radius (r) = D/2 = 1 meter

Height = 14 meter

Since we are digging cylindrical well, the total volume of soil taken out is equal to the volume of cylinder.

\mathtt{Volume\ =\ \pi r^{2} h}\\\ \\ \mathtt{Volume\ =\ \frac{22}{7} \times ( 1)^{2} \times 14=44\ m^{3}}

So, total of 44 cu. meter of soil is taken out for well development.

Now let’s **calculate the cost of well development**.

1 sq. meter = 10$

44 sq meter = 44 x 10 = 440 $

Hence, Steve has to incur total cost of 440$.

**Question 05**

A cylindrical tank of radius 0.7 meter and height 5 meter is filled with water by cylindrical pipe of radius 0.1 meter through which water flows at the rate of 2 meter per second. Find the time taken to fill the cylindrical tank.

**Solution**

Volume of cylindrical tank = \mathtt{\pi r^{2} h}

\mathtt{Volume\ =\ \frac{22}{7} \times ( 0.7)^{2} \times 5=7.7\ m^{3}}

So, the volume of cylindrical tank is 7.7 sq meter.

**Now let’s calculate the volume of water filled by pipe in one second.**

Water filled by pipe = \mathtt{\pi r^{2}} \ \times ( water\ flow)

\mathtt{Water\ fill\ \ =\ \frac{22}{7} \times ( 0.1)^{2} \times 2=0.063\ m^{2}}

So the pipe throws 0.063 cubic meter of water.

**Now, let’s calculate the time taken to fill the whole tank.**

Time taken = total volume / volume filled by pipe in seconds

Time taken = 7.7 /0.063 = 122.22 seconds

Hence, it will take around 122 second to fill the whole tank.