Question

Diameter of a roller is 2.4 m and it is 1.68 m long. If it takes 1000 complete revolutions once over to level a field then the area of the field is:
A) 12672 sq m
B) 12671 sq m
C) 12762 sq m
D) 11768 sq m

Answer 1

Hint: This question is from the topic of surface areas and volumes. In this question, we have to find the area of the field using the given data. In solving this question, we will find the radius of the roller using the diameter of the roller. After that, we will find the circumference of the roller using the radius of the roller. After solving the further question, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we have to find the area of the field if the roller takes 1000 complete revolutions once over to level the same field. We have given that the diameter of the roller is 2.4 m and the length of the roller is 1.68 m.
Let us first find out the radius of the roller.
As we know that the radius is always half of the diameter, so we can write
\[r=\dfrac{D}{2}=\dfrac{2.4}{2}=1.2\]
Hence, radius of the roller is 1.2 m
We can take reference from the following figure:

Now, let us find out the circumference of the roller using the formula \[C=2\pi r\], where C is the circumference and r is the radius.
So, \[C=2\pi r=2\times \pi \times 1.2\text{ m}\]
Now, we will find the area covered in one revolution. Let us take the area as A.
\[A=2\times \pi \times 1.2\times 1.68=12.672\text{ sq m}\]
So, for 1000 revolutions the area of the field will be
\[1000\times 12.672=12672\text{ sq m}\]
Hence, for 1000 complete revolutions once over to level a field, the area of the field is 12672 sq m.

Note: We should have a better knowledge in the topic surface area and volumes to solve this type of question easily. Remember the following formulas for solving this type of question:
\[C=2\pi r\], where C is the circumference and r is the radius of the circle
We can solve this question by different methods.
For that, we will use the following formula:
The curved surface area of the roller is \[2\pi rh\] sq m, where r is radius of the roller and h is the height of the roller.
So, the curved surface area of the roller will be
\[A=2\pi rh=2\times \pi \times r\times h=2\times \pi \times 1.2\times 1.68=12.672\text{ sq m}\]
Hence, for 1000 revolutions, area of the roller will be
\[1000\times 12.672=12672\text{ sq m}\]
Hence, we got the same answer. So, we can use this method too.
Remember that the value of \[\pi \] is \[\dfrac{22}{7}\] that is 3.14285.