A solid iron pole consists of cylinder of height 220 cm and his diameter 24.com
is surmounted by another cylinder of height 60 cm and radius 8 cm. Find mass of the
pole, given that 1 cm of iron =8gm mass
Answers
Answered by
3
Answer:
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
Answered by
5
Solution :-
Diameter of larger cylinder = 24cm
Radius of larger cylinder = 24/2 = 12cm
Height of the larger cylinder = 220 cm
Volume of the larger cylinder = πr²h
=> 3.14 × 12 × 12 × 220 cm³
=> 3.14 × 144 × 220 cm³
=> 99475.2 cm³
Radius of the smaller cylinder = 8 cm
Height of the smaller cylinder = 60 cm
volume of the smaller cylinder = πr²h
=> 3.14 × 8 × 8 × 60
=> 3.14 × 64 × 60
=> 12057.6 cm³
Volume of the iron pole :-
volume of smaller + volume of larger cylinder
=> 99475.2 cm³ + 12057.6 cm³
=> 111532.8 cm³
Mass in iron of 1 cm³ = 8g
Mass in iron of 111532.8 cm³ = 111532.8 × 8
=> 892,262.4 gm
=> 892262.4/1000
892,262.4
=> 892.26 kg
Hence, The mass of the iron pole is 892.26kg.
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