A road has a diameter $0.7\ m$ and its width is $1.2\ m$. Find the least number of revolutions that the roller must take in order to level a playground of size $120\ m\times 44\ m$.

Given: A road has a diameter $0.7\ m$ and its width is $1.2\ m$. Size $120\ m\times 44\ m$. 

To do: To find the least number of revolutions that the roller must take in order to level a playground.

Solution:

As given, Diameter of a road roller $=0.7\ m=70\ cm$

So, radius $( r)=\frac{70}{2}$

$=35\ cm=\frac{35}{100}$ m

And width $( h)=1.2\ m$

Now, curved surface area$=2\pi rh=2\times \frac{22}{7}\times \frac{35}{100}\times 1.2\ m^2$

$=\frac{264}{100}\ m^2$
 
Area of playground$=120\ m\times 44\ m=5280\ m^2$
 

Hence, the number of revolutions made by the road roller $=\frac{5280}{264}\times 100=2000$ revolutions.

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Updated on: 10-Oct-2022

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