A cylindrical road roller made of iron is $1 m$ long. Its internal diameter is $54 c m$ and the thickness of the iron sheet used in making the roller is $9 c m$ . Find the mass of the roller, if $1 {c m}^{3}$ of iron has $7.8 g$ mass (Use $π = 3.14$ )

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Solution

Given $r_{1} = 27 c m, t = 9 c m$ , $r_{2} = r_{1} + t = 36 c m$
$V = π (36^{2} - 27^{2}) \times 100$
$V = 178128.3 c m^{3}$
Mass of roller $= 178128.3 \times 7.8 = 1389.4 k g$

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A cylindrical road roller made of iron is 1m long.Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm.Find the mass of the roller, if 1 $c m^{3}$ of iron has 7.8 gm mass. $(U s e π = 3.14)$

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A cylindrical road roller made of iron is 1m long. Its internal diameter is 54cm and the thickness of the iron sheet used in making the roller is 9cm. Find the mass of the roller, if 1cm3 of iron has 7.8g mass (Use π=3.14)

Given r1=27cm,t=9cm, r2=r1+t=36cmV=π(362−272)×100V=178128.3cm3Mass of roller=178128.3×7.8=1389.4kg

A cylindrical road roller made of iron is $1 m$ long. Its internal diameter is $54 c m$ and the thickness of the iron sheet used in making the roller is $9 c m$ . Find the mass of the roller, if $1 {c m}^{3}$ of iron has $7.8 g$ mass (Use $π = 3.14$ )

Given $r_{1} = 27 c m, t = 9 c m$ , $r_{2} = r_{1} + t = 36 c m$
$V = π (36^{2} - 27^{2}) \times 100$
$V = 178128.3 c m^{3}$
Mass of roller $= 178128.3 \times 7.8 = 1389.4 k g$