Question

The diameter of a roller 120cm long is 84cm. It it takes 500

complete revolutions to level a playground, find the cost of

levelling it at the rate of 30 paise per square meter.​

Answer 1

Answer:

hope it helps mark as brainliest

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\bold{\therefore Cost\:of\:levelling=475.2\:rupees}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\bold{\underline{Given : }}} \\ \implies \text{Diameter \: of \: roller = 84\: cm} \\ \\ \implies\text{Length \: of \: roller = 120 \: cm} \\ \\ \implies \text{Total \: no. \: of \: revolution = 500} \\ \\ \red{\bold{\underline{To \: Find: }}} \\ \implies \text{Cost \: of \: levelling \: ground = ?}$

• According to given question :

$\text{For \: area \: of \: one \: revolution} \\ \implies C.S.A \: of \: cylinder = 2\pi rh \\ \\ \implies C.S.A= 2 \times \frac{22}{7} \times 84 \times 60 \\ \\ \implies \text{C.S.A =31680 }cm^{2} \\ \\ \text{Total \: area \: covered \: in \: 500 \: revolution } \\ \implies Area = 31680 \times 500 \\ \\ \implies \text{Area = 15840000 cm}^{2} = 1584 \: {m}^{2} \\ \\ \bold{For \: cost \: of \: levelling \: ground } \\ \implies Cost \: of \: levelling \: {1}m^{2} \:ground = 0.3 \: rupees \\ \\ \implies Cost \: of \: levelling \: 1584 {m}^{2} \: ground = 0.3 \times 1584 \\ \\ \green{\implies Cost = 475.2 \: rupees}$