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Q: 4    The diameter of a roller is \small 84\hspace{1mm}cm and its length is \small 120\hspace{1mm}cm. It takes \small 500 complete revolutions to move once over to level a playground. Find the area of the playground in \small m^2.
 

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Given,

The diameter of the cylindrical roller = \small d = 84\hspace{1mm}cm

Length of the cylindrical roller = \small h = 120\hspace{1mm}cm

The curved surface area of the roller = 2\pi r h = \pi dh

\\ = \frac{22}{7}\times84\times120 \\ \\ = 31680\ cm^2

\therefore Area of the playground = Area\ covered\ in\ 1\ rotation \times500

\\ = 31680 \times500 \\ = 15840000 cm^2 \\ = 1584\ m^2

Therefore, the required area of the playground = 1584\ m^2

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HARSH KANKARIA

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