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The diameter of roller $1.5\ m$ long is $84\ cm$. If it takes $100$ revolutions to level a play¬ground, find the cost of levelling this ground at the rate of $50$ paise per square metre.
Given:
The diameter of roller $1.5\ m$ long is $84\ cm$.
It takes $100$ revolutions to level a playground.
To do:
We have to find the cost of levelling the ground at the rate of $50$ paise per square metre.
Solution:
Diameter of the cylindrical roller $= 1.5\ m$
Radius of the roller $=\frac{1.5}{2}$
$=0.75\ m$
$= 75\ cm$
Length of the cylinder $(h) = 84\ cm$
Therefore,
Curved surface area of the roller $=2 \pi r h$
$=2 \times \frac{22}{7} \times 75 \times 84$
$=39600 \mathrm{~cm}^{2}$
Area covered in 100 revolutions $=39600 \times 100$
$=3960000 \mathrm{~cm}^{2}$
$=\frac{3960000}{100 \times 100}\ m^2$
$=396 \mathrm{~m}^{2}$
Rate of levelling the ground $=50$ paise per $\mathrm{m}^{2}$
Total cost of levelling $=396 \times \frac{50}{100}$
$= Rs.\ 198$