Question

The diameter of a garden roller is 2.1 m and it is 2.2 m long. How much area will it cover in 400 revolutions?

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Answer 1

Answer:

The diameter of a garden roller is 2.1 m and it is 2.2 m long. How much area will it cover in 400 revolutions?

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

Given,

$\: \: \: \: \: \rm{Diameter \: \: of \: \: the \: roller, d = 2.1 \: m} \\ \\ \rm{The \: \: roller \: \: is \: \: 2.2 \: m \: \: long \: \: and \: \: hence,} \\ \\ \: \: \: \: \rm{ \therefore \: Height \: \: of \: \: the \: \: roller, h = 2.2 \: m} \\ \\ \: \: \: \: \: \rm{No. \: \: of \: \: revolution, n = 400}$

To find :

$\: \: \rm{Area \: \: covered \: \: by \: \: the \: \: rolle \: \: r in \: \: the \: \: given \: \: revolution = \: ? }$

As we know,

$\: \: \: \: \: \: \: \sf{ \underline{ \red{ \: \: Radius = \frac{1}{2} \ \: of \: \: diameter} \: \: }} \\ \\ \rm{ \therefore \: Radius \: \: of \: \: the \: \: roller, r = \frac{2.1}{2} \: m}$

Now,

The area covered by the roller in 1 revolution is equal to the lateral (curved) surface area of the roller.

$\boxed{ \sf{ \red{Lateral \: (curved) \: \: surface \: \: area \: \: of \: \: a \: cylinder = 2 \pi r h }}}$

$\bold{ \: \therefore \: Lateral \: \: surface \: \: area \: \: of \: \: the \: \: roller = 2 \pi r h} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ =( \cancel{2} \times \frac{22}{ \cancel{7}}\times \frac{ \cancel{ \: \: 2.1} \: \: \large{ {}^{0.3} } \: \: } { \cancel{2}} \times 2.2) \: \: m {}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ = (22 \times 0.3 \times 2.2) \: \: m {}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{= 14.52 \: \: m{}^{2} }$

$\tt{Hence, } \\ \: \: \: \: \: \: \: \: \: \large{\tt{The \: \: area \: \: covered \: \: by \: \: the \: \: rolller \: \: in }} \\ \: \: \large{\tt{ 1 \: \: revolution = \pink{14.52 \: m {}^{2} }}}$

$\bold{ \therefore \: Area \: \: covered \: \: by \: \: the \: \: roller \: \: in \: \: 400 \: \: revolutions} \\ \\ \: \: \: \: \: \bold{ = Lateral \: \: surface \: \: area \times No. \: \: of \: \: revolution } \\ \\ \: \: \: \: \: \bold{ = (14.52 \times 400) \: m{}^{2} } \\ \\ \: \: \: \: \: \bold{ = 5808 \: m {}^{2} }$

$\: \: \: \tt{So,} \\ \: \: \: \: \: \: \: \: \:\large{ \tt{Area \: \: covered \: \: by \: \: the \: \: roller \: \: in \: \: 400}} \\ \: \: \: \large{\tt{ \: revolutions = \underline{ \pink {\: 5808 \: m {}^{2} }}}}$