Question

The diameter of the roller $1.5m$ long is $84cm$ if it takes $100$ revolutions to level a playground, find the cost of levelling this ground at the rate of $50$ paise per square metre. $\left(m^2\right)$

Answer 1

Diameter of a roller $=84cm$

$\therefore$ Radius $=\frac{84}{2}=42cm$

and height $\left(h\right)=1.5m=1.5\times 100cm=150cm$ $[\because 1m=100cm]$

$\therefore$ Curved surface area of cylinder$=2\pi rh$

$=2\times \frac{22}{7}\times 42\times 150c{m}^2=39600c{m}^2$

Area covered in 1 revolution $=$ Curved surface area of the given cylinder

$\therefore$ Area covered in $100$ revolutions.

$=39600\times 100=3960000c{m}^2$

$=\frac{3960000}{100\times 100}=396m^2$

Rate of levelling the ground for $1m^2=50$ paise $=Rs.0.5$ [$\because Re.1=100$ paise ]

$\therefore$ Total cost of levelling $396m^2=396\times 0.50=Rs.198$