Question

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass.​

Answer 1

Step-by-step explanation:

Height (h

1

) of larger cylinder =220 cm

Radius (r

1

) of larger cylinder =

2

24

=12 cm

Height (h

2

) of smaller cylinder = 60 cm

Radius (r

2

) of smaller cylinder =8 cm

Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder

=πr

1

2

h

1

+πr

2

2

h

2

=(π(12)

2

×220)+(π(8)

2

×60)

=π[144×220+64×60]

=3.14[31,680+3,840]

=3.14×35520=111,532.8 cm

3

Mass of 1 cm

3

iron = 8 g

Mass of 111532.8 cm

3

iron = 11532.8×8=892262.4 g

solution

Answer 2

GIVEN:-

$\small\sf\pink{height\:(h1)\:of\:larger\:cylinder=220cm}$

$\small\sf\pink{radius\:(r1)\:of\:larger\:cylinder=12cm}$

$\small\sf\pink{height\:(h2)\:of\:smaller\:cylinder=60cm}$

$\small\sf\pink{radius\:(r2)\:of\:smaller\:cylinder=8cm}$

⠀

$\small\sf\blue{Total\:volume\:of\:pole=Volume\:of\:larger\:cylinder+}$

$\small\sf\blue{Volume\:of\:smaller\:cylinder}$

⠀⠀

$\longrightarrow$$\small\sf\red{π \: {r1}^{2} h + π \: {r2}^{2}h2}$

$\longrightarrow$$\small\sf\red{π{(12)}^{2} \times 220 + π {(8)}^{2} \times 60 }$

$\longrightarrow$$\small\sf\red{π(144 \times 220 + 64 \times 60)}$

$\longrightarrow$$\small\sf\red{35520 \times 3.14}$

$\longrightarrow$$\small\sf\red{{1,11,5328cm}^{3}}$

⠀⠀

$\small\sf\green{Mass\:of\:{1cm}^{3}\:of\:iron=8g}$

⠀⠀

$\therefore$$\large\sf\gray{Mass\:of\:{1,11,532.8cm}^{3}=}$

$\longrightarrow$$\small\sf\orange{1,11,532.8×8g}$

$\longrightarrow$$\small\sf\orange{892262.4g}$

$\longrightarrow$$\small\sf\orange{892.262kg}$