Question

The diameter of a roller 1 m long is 70 cm. .If it takes 200 revolutions to level a playground, find the cost of levelling at the rate of 75 paisa per sq.m.

Answer 1

Roller is of form cylinder

Radius$=\frac{Diameter}{2}=\frac{1}{2}$m$

$=\frac{100}{2}=50cm$ ,since $1$m$=100$cm

Area of roller $=$Curved surface of the cylinder$=2\pi rh$

$=2\times \frac{22}{7}\times 50\times 70$

$=22,000‬$ sq.cm

Area of $1$ revolution$=$Area of roller$=22,000‬$sq.cm

Area of $200$ revolutions$=200\times 22,000‬=4,400,000‬sq.cm$

$=\frac{4400000}{100\times 100}$sq.m

$=440$sq.m

Cost of levelling the playground per sq.m$=75$ paisa

Cost of levelling the playground $440$ sq.m$=75$ paisa $\times 440=33,000$paisa

Since $100$ paisa$=1$ rupee,

$33000$ paisa$=\frac{33000}{100}=330$rupees

$\therefore$ Cost of levelling the playground $440$ sq.m is $330Rs$