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 $=396 \times \frac{50}{100}$

$= Rs.\ 198$

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Updated on: 10-Oct-2022

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