Question

A roller is 120 cm long and has diameter 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of 30 paise per sq. M.

Answer 1

given
l=120cm =120/100=1.2m
d=84cm=84/100=0.84m
revolution =500 of level the ground
cost of level=?
1m^2= 30paise

answer step;

circumference of cricle =pi×d=22/7×0.84=2.95m
b=2.95m
A=l×b=1.2×2.95=3.545m^2
1rev= 3.545m^2 of area
500rev = 500×3.545= 1772.5m^2 0f area

area of 1m^2 =30paise
area of 1772.5m^2= 1772.5×30= 53175 paise

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

${\boxed{\sf\:{Length\;of\;Roller\;(l)}}}$

= 120 cm

$\textbf{\underline{Convert\;it\;into\;metre}}$

$\tt{\rightarrow\dfrac{120}{100}}$

= 1.2m

Also,

${\boxed{\sf\:{Diameter\;of\;roller\;(d)}}}$

= 84 cm

$\textbf{\underline{Convert\;it\;into\;m}}$

$\tt{\rightarrow\dfrac{84}{100}}$

= 0.84 m

We know that :-

$\tt{\rightarrow Radius=\dfrac{Diameter}{2}}$

$\tt{\rightarrow Radius=\dfrac{0.84}{2}}$

= 0.42 m

$\textbf{\underline{Curved\;Surface\;area\;of\;roller\; (CSA)}}$

= 2πrh

$\tt{\rightarrow 2\times\dfrac{22}{7}\times 0.42\times 1.2}$

= 2 × 22 × 0.06 × 12

= 3.168 m²

$\textbf{\underline{Area\;levelled\;by\;roller\;in\;1\; revolution}}$

= 3.168 m²

Now,

$\textbf{\underline{Area\;levelled\;by\;roller\;in\;500\; revolution}}$

= 500 × 3.168 m²

= 1584 m²

Cost of levelling at 0.30 per m²

= 0.30 × 1584

= 475.20