Question

the diameter of a road roller is 90 cm and 2m 10 cm length. If it takes 400 revolutions to level a playground, find the cost of levelling the ground at rs.2.25 per m^2

Answer 1

Solution

d=90cm=0.9m

r=45cm=0.45m

h(height/length)=2m10cm=2.10m

now ...the curved surface area of the roller is

=πdh

$= \frac{22}{7} \times 0.9 \times 2.10 \\ = 22 \times 0.9 \times 0.3 \\ = 5.94m {}^{2}$

therefore cost=(5.94×2.25)Rs=13.365 Rs

hope this helps you....xd

Answer 2

☈ 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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