Question

If a diameter of a road roller is 0.9m and it's length is 1.4m,how much area of a field will be pressed in its 500 rotation



Thanks!​

Answer 1

Step-by-step explanation:

lol I am in 6th and now the A lol

so - diameter of road roller - 0.9m

length - 1.5

radius(r) of roller = 0.9m

length(h) of roller =1.4m

To find:

area pressed in 500 rotations=?

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area pressed in 1 rotation= 2Πrh

area pressed in 500 rotations= 2Πrh×500

\begin{gathered} = > 2 \times \frac{22}{7} \times 0.2 \times 0.9 \times 500 \\ = > 2 \times \frac{22}{7} \times \frac{2}{10} \times \frac{9}{10} \times 500 \\ = > 2 \times \frac{22}{7} \times 2 \times 9 \times \=>\frac{3960{7}\=>565.714m.l\end{gathered} =>2× 722

×0.2×0.9×500=>2×722×102×109×500=>2× 722

×2×9×5=> 73960=>565.714m.

Hence, the area of field pressed in 500 rotations will be 565.714m.

Answer 2

Step-by-step explanation:

$given \: \: d = 0.9m \\ therefor \: radius r = \frac{0.9}{2} m \\ length \: l = 1.4m \\ area \: of \: field \: will \: be \: pressed \: in \: one \: rotation \: = area \: of \: cylindrical \: roller \\ = 2\pi \: r \: h \\ 2 \times \frac{22}{7} \times \frac{0.9}{2} \times 1.4 \: \: \: \: \: (here \: l = h) \\ = 3.96 {m}^{2} \\ area \: of \: field \: in \: 500 \: rotation \: is \: given \: by \\ = 500 \times 3.96 \\ = 1980 \: {m}^{2} \\ which \: is \: the \: required \: answer.$