Question

The diameter of a road roller of length 120cm is 84cm .if it takes 500 complete revolutions to level a play ground,then find the cost of levelling it at the cost of 75 paise per sq meter

Answer 1

Answer:1188 rupees

Step-by-step explanation:

curved surface area of cylinder =2πrh

Area =31680cm²

After 500 revolution =31680*500= 15840000cm²

=1584m²

1m²=75paise

1584m²=1584×75

=118800 paise

= 1188 rupees

Answer 2

The cost of levelling it at the cost of 75 paise per sq meter is Rs.1188

Step-by-step explanation:

Length of roller = 120 cm

Diameter of roller = 84 cm

Radius of roller =$\frac{84}{2}=42 cm$

Curved surface area of roller = $2 \pi r h = 2 \times \frac{22}{7} \times 42 \times 120=31680 cm^2$

Area covered in 1 revolution = $31680 cm^2$

Area covered in 500 revolutions =$31680 \times 500 =15840000 cm^2$

$1 m^2 = 10000 cm^2$

Area covered in 500 revolutions =$\frac{15840000}{10000}=1584 m^2$

Cost of leveling 1 sq.m. = 75 paisa = Rs.0.75

Cost of leveling 1584 sq.m. =$1584 \times 0.75=Rs.1188$

Hence the cost of levelling it at the cost of 75 paise per sq meter is Rs.1188

#Learn more:

The diameter of a roller 1m long is 70cm if it takes 200 revolution ro level a play ground find the cost of levelling at the rate of 75 paise per sq • m

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