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:

Given :-

• A cylindrical roller of radius 3.5 m and 4 m long by making 10 rounds.

To Find :-

• What is the area of a ground that can be levelled by a cylindrical roller.

Formula Used :-

$\clubsuit$ LSA of Cylinder :

$\longmapsto \sf\boxed{\bold{\pink{L.S.A\: of\: Cylinder =\: 2{\pi}rh}}}$

where,

• L.S.A = Lateral Surface Area
• r = Radius
• h = Height

Solution :-

Given :

• Radius = 3.5 m
• Height = 4 m

Then, area covered in 1 rounds :

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

$\implies \sf \dfrac{44}{7} \times \dfrac{35}{10} \times 4$

$\implies \sf \dfrac{44}{7} \times \dfrac{14\cancel{0}}{1\cancel{0}}$

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

$\implies \sf 44 \times 2$

$\implies \sf\bold{\green{88\: {m}^{2}}}$

Then, area covered in 10 rounds :

$\implies \sf 10 \times 88$

$\implies\sf\bold{\red{880\: {m}^{2}}}$

$\therefore$ The area of a ground that can be travelled by a cylindrical roller is 880 m².

Answer 2

Answer:

$\huge \bf880 \: {m}^{2}$

Step-by-step explanation:

$\bf \: R = 3.5m$

$\bf \: H = 4m$

$\bf \: A = \bf ?$

$\bf Area \: covered \: in \: 1 \: round \: = LSA$

$\bf (LSA = Lateral \: surface \: area \: of \: cylinder)$

$\bf Area \: covered \: in \: 10 \: rounds$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =\bf10× 2× \frac{22}{7}\times 3.54$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = 20 \times \dfrac{\cancel{ {77}}^{11}}{\cancel{7}} \times 4$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= 80 \times 11$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= 880 \: {cm}^{2}$

∴ The area of a ground that can be travelled by a cylindrical roller is 880 m²