Question

A roller takes 500 complete revolution to level a playground find the cost of levelling the ground at the rate of 75 paise per sq meter if the diameter and length of the roller are 84 cm and 120 cm respectively ​

Answer 1

2376 Rupees

Step-by-step explanation:

radius=84cm=.84m

height=120cm= 1.2m

paise=75 paise / square metre

CSA of cylinder = 2×π×R×H

= 2×22/7×.84m×1.2m =6.3360 m

total distance= 6.3360m ×500= 3168 m

total paise = 3168 ×75 = 237600 paise= 2376 rupees

Answer 2

ANSWER:-

$roller \: is \: of \: form \: of \: cyllinder$

$radius = \frac{diameter}{2} = \frac{84}{2} = 42cm$

Area of roller = curved surface area of cylinder

=2πrh

$=( 2 \times \frac{22}{7} \times 42 \times 120) \: {cm}^{2}$

$= 31680 \: {cm}^{2}$

$area \: of \: 1 \: revolution = area \: of \: roller = 31680 {cm}^{2}$

$area \: of \: 500 \: revolution = 500 \times 31680 {cm}^{2}$

$area \: of \: playground = 15480000 {cm}^{2}$

$= 15840000 \times \frac{1}{100} \times \frac{1}{100} {m}^{2} = 1548 {m}^{2}$

$cost \: of \: 1 {m}^{2} = 75paise$

$cost \: of \: 1548 {m}^{2} = 1548 \times 75 = 116100paise = \frac{116100}{100} = 1161rs.$

hope it's clarify u mate

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