Question

The diameter of a roller 120 cm long is 84 cm. If it takes 500 completed revolutions to level a play ground, determine the cost of levelling it at the rate of Rs. 2.50 per square metre.

Answer 1

Roller is in the form of a cylinder as shown in the image.

Given, $h=120\text{cm}$,

Radius $r=\frac{\text{Diameter}}{2}=\frac{84}{2}=42\text{cm}$

Area covered by the roller in one revolution $=$ Curved Surface Area of the cylinder $=2\pi rh$
$=2\times \frac{22}{7}\times 42\times 120$

$=31,680{\text{cm}}^2$

$\therefore$ Area covered in $500$ revolutions $=(31,680\times 500){\text{cm}}^2=15,84,000{\text{cm}}^2$

Area of playground in ${\text{m}}^2=\frac{15840000}{100\times 100}{\text{m}}^2=1584{\text{m}}^2$

Cost of levelling the playground per ${\text{m}}^2=\text{Rs.}2.50$

$\therefore$ Cost of levelling the playground$=1584\times 2.50=\text{Rs.}3960$