Question

What is the area of a ground that can be levelled by a cylindrical roller of radius 3.5 m and 4 m long by making 10 rounds?​

Answer 1

Answer:

880m²

Solution

Area of a ground =?

Given R = 3.5m, n = 4m

Rounds = 10

Area covered by 1 revolution = curved Surface Area of the cylindral

=

π

Area =

Number of revolutions =

Total area = 879.2m

Total area

Answer 2

$\bf \: \underline{Given} :$ ↴

$\sf \: \implies Radius \: of \: cylindrical \: roller = 3.5 \: m$

$\sf \: \implies Length \: of \: cylindrical \: roller = 4 \: m$

$\sf \: \implies Number \: of \: rounds = 10$

$\bf \: \underline{To \: find} :$ ↴

$\sf \: \implies Area \: of \: ground \: that \: can \: be \: levelled = \: ?$

$\bf \: \underline{Formula \: to \: be \: used},$

$\implies \boxed{ \pink{\sf \: LSA \: of \: cylinder = 2 \pi r h }}$

$\sf \: Where,$

$\sf \: \implies LSA = Lateral \: surface \: area$

$\sf \:\implies \pi = \dfrac{22}{7} \: or \: 3.14$

$\sf \: \implies r = radius$

$\sf \: \implies h = height$

$\bf \:\underline{ \underline{ Solution}}$

$\sf \: Then, area \: to \: be \: covered \: in \: 1 \: round :$

$\sf \implies 2 \times \dfrac{22}{7} \times 3.5 \times 4$

$\sf \implies \dfrac{44}{7} \times 3.5 \times 4$

$\sf \implies \dfrac{44}{7} \times 14$

$\sf \implies \dfrac{44}{7} \times 14$

$\sf \implies \dfrac{616}{7}$

$\sf \implies 88$

$\underline{ \sf \: Area \: covered \: in \: 1 \: round = 88 \: m ^{2}}$

$\sf \: Now, area \: covered \: in \: 10 \: rounds :$

$\sf \implies 88 \times 10$

$\sf \implies 880$

$\underline{ \boxed{ \pink{\sf \: Answer = 880 \: m^ {2}}}}$