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 $\small m^2$.

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 = $1584\ m^2$

Related Chapters

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-