Question

2. The diameter of a road roller is 98 cm. It takes 100 complete revolutions to level the ground of area 308 sq.m. Find the length of the roller.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Diameter of a road roller = 98 cm

So, radius of road roller, r = 49 cm

Let assume that length of the roller be h cm.

We know, Area covered by road roller in 1 revolution is curved surface area of cylinder of radius r and height h.

[ Road roller is in the shape of cylinder ]

$\rm \: Area_{(Covered \: in \: 1 \: revolution)}= CSA_{(Cylinder)}$

$\rm \: Area_{(Covered \: in \: 1 \: revolution)}= 2\pi \: rh$

$\rm\implies \: Area_{(Covered \: in \: 100 \: revolution)}= 200\pi \: rh$

According to statement,

$\rm \: Area_{(Covered \: in \: 100 \: revolution)} = 308 \: {m}^{2} = 3080000 \: {cm}^{2}$

So,

$\rm \: 200\pi \: rh \: = \: 3080000$

$\rm \: 200 \times \dfrac{22}{7} \times 49 \times h \: = \: 3080000$

$\rm \: 200 \times 22 \times 7\times h \: = \: 3080000$

$\rm\implies \:h \: = \: 100 \: cm \: = \: 1 \: m$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Solution−

Given that,

Diameter of a road roller = 98 cm

So, radius of road roller, r = 49 cm

Let assume that length of the roller be h cm.

We know, Area covered by road roller in 1 revolution is curved surface area of cylinder of radius r and height h.

[ Road roller is in the shape of cylinder ]

\begin{gathered}\rm \: Area_{(Covered \: in \: 1 \: revolution)}= CSA_{(Cylinder)} \\ \end{gathered}

Area

(Coveredin1revolution)

=CSA

(Cylinder)

\begin{gathered}\rm \: Area_{(Covered \: in \: 1 \: revolution)}= 2\pi \: rh \\ \end{gathered}

Area

(Coveredin1revolution)

=2πrh

\begin{gathered}\rm\implies \: Area_{(Covered \: in \: 100 \: revolution)}= 200\pi \: rh \\ \end{gathered}

⟹Area

(Coveredin100revolution)

=200πrh

According to statement,

\rm \: Area_{(Covered \: in \: 100 \: revolution)} = 308 \: {m}^{2} = 3080000 \: {cm}^{2} Area

(Coveredin100revolution)

=308m

2

=3080000cm

2

So,

\begin{gathered}\rm \: 200\pi \: rh \: = \: 3080000 \\ \end{gathered}

200πrh=3080000

\begin{gathered}\rm \: 200 \times \dfrac{22}{7} \times 49 \times h \: = \: 3080000 \\ \end{gathered}

200×

7

22

×49×h=3080000

\begin{gathered}\rm \: 200 \times 22 \times 7\times h \: = \: 3080000 \\ \end{gathered}

200×22×7×h=3080000

\begin{gathered}\rm\implies \:h \: = \: 100 \: cm \: = \: 1 \: m \\ \end{gathered}

⟹h=100cm=1m

\rule{190pt}{2pt}

Additional information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★CSA

(cylinder)

=2πrh

★Volume

(cylinder)

=πr

2

h

★TSA

(cylinder)

=2πr(r+h)

★CSA

(cone)

=πrl

★TSA

(cone)

=πr(l+r)

★Volume

(sphere)

=

3

4

πr

3

★Volume

(cube)

=(side)

3

★CSA

(cube)

=4(side)

2

★TSA

(cube)

=6(side)

2

★Volume

(cuboid)

=lbh

★CSA

(cuboid)

=2(l+b)h

★TSA

(cuboid)

=2(lb+bh+hl)

• thanks