Question

# A roller of diameter 1.4m and length 1.4m is used to press the ground having area 3080 sq.m. Find the number of revolutions that the roller will make to press the ground.(a) 700(b) 500(c) 1000(d) 800

Hint: Firstly calculate the area covered by roller during one revolution by using the formula, Surface area of roller $=\pi DL$ and then divide the complete area of ground by the value obtained from the formula to get the final answer.

To solve the problem we will write the given data first,
Diameter of roller = D = 1.4m ………………………………….. (1)
Length of roller = l = 1.4m ………………………………….. (2)
Area of Ground = 3080 sq.m. ………………………………….. (3)

To find the number of revolutions that roller will make to press the ground, first we have to calculate area pressed during one revolution and therefore we have to derive the formula of area of rolling surface only as shown below,

As we know that the roller has a cylindrical shape and therefore the formula of surface area of cylinder is given by,

Surface area of cylinder with diameter ‘D’ and height/length ‘L’ is given by,
Surface Area $=\dfrac{\pi {{D}^{2}}}{2}+\pi DL$

As we all know rolling does not require the circular surfaces of cylinder therefore the formula of surface area of roller required for rolling can be reduced to,
Surface area of roller $=\pi DL$

If we put the values of equation (1) and equation (2) in above equation we will get,
Surface area of roller $=\pi \times 1.4\times 1.4$
$\therefore$ Surface area of roller $=\pi \times 1.96$
$\therefore$ Surface area of roller $=1.96\pi$ sq.m.

As we know that the surface area of the roller required for rolling is ultimately the area pressed by it in one revolution. And therefore we can write,

Area pressed by roller in one revolution $=1.96\pi$ sq.m. ……………………………….. (4)
Now, to find the number of revolutions required to press the 3080 sq.m area we will divide it by area pressed by roller in one revolution therefore from equation (3) and equation (4) we can write,
Number of revolutions $=\dfrac{3080}{1.96\pi }$

$\therefore$ Number of revolutions $=\dfrac{3080}{1.96\times 3.142}$

$\therefore$ Number of revolutions $=\dfrac{3080}{6.15832}$

$\therefore$ Number of revolutions $=500.1$

$\therefore$ Number of revolutions $=500$

Therefore the 500 number of revolutions are required to press the ground of area 3080 sq.m.
Therefore the correct answer is Option (B)

Note: Do remember to reduce the area of two circular surfaces i. e. $\dfrac{\pi {{D}^{2}}}{2}$ OR $2\pi {{r}^{2}}$ from the surface area of roller required to press the ground and if you calculate the area with considering them then you will get the wrong answer. Also if you don’t know the formula of surface area in terms of diameter then you can derive it from the formula of it in terms of radius.