Question

the diameter of a road roller is 90 cm and is 2m 10 CM in length. if it takes 400 revolution to level a playground, find the cost of levelling the ground at ₹2.25 per square meter.

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Answer 1

☈ Given :-

Diameter of the roller = 90cm

length= 2m 10 cm = 2.1 m

So , radius would be 45 cm or 0.45 m

☈ To find :-

Find the cost of levelling the ground at ₹2.25 per square meter.

☈ Solution :-

The part which will touch the ground will be the C.S.A. of roller

now , C.S.A. of roller = 2πrh

➞ 2 × (22/7) × 0.45 × 2.1

➞ 5.94 m²

It takes 400 revolutions , area of ground

➞ 5.94 × 400

➞ 2376 m²

Now cost to level = rate × area

➞ 2376 × 2.25

➞ 5346

➞ 5346 Rs.

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Answer 2

$\Large\sf\red{Given:-}$

The diameter of road roller = 90 cm

Length of road roller = 2m 10cm

$\Large\sf\red{To\:find:-}$

Cost of levelling the ground at Rs2.25 per square meter.

$\Large\sf\red{Solution:-}$

$\sf\blue{Radius\: of \:road \:roller =\dfrac{D}{2}=\dfrac{90}{2}=45cm}$

$\sf\blue{Length \:of \:road \:roller \:, h = 2m 10cm = 210cm}$

$\sf\blue{Revolution = CSA \:of \:road \:roller}$

$\Large\sf\pink{Formula\: used = 2\pi rh\:(CSA\:of\:Cylinder)}$

$\implies\sf\orange{2\times \dfrac{22}{7}\times 45\times 210}$

$\sf\implies\orange{=59400cm^2}$

$\sf\blue{No.\: of \:revolutions = 400}$

$\sf\green{Area\: covered \:by \:roller\: in\: 400 \:revolutions = 400\times 59400cm^2= 23760000cm^2}$

$\Large\tt\pink{10000cm^2 = 1sqm}$

$\implies\sf\orange{\dfrac{23760000}{10000} = 2376m^2}$

$\sf\blue{Cost\: of \:levelling \:1sq.m\: = Rs2.25 = \dfrac{225}{100}Rs}$

$\sf\blue{Cost \:of \:levelling \:the\: play ground =}$

$\sf\implies\orange {\dfrac{2375\times 225}{100} }$

$\sf\implies\orange{Rs.5346}$

$\Large\sf\green{Therefore\:cost\:of\:levelling = Rs.5346}$