Aptitude - Volume and Surface Area - Discussion
Discussion Forum : Volume and Surface Area - General Questions (Q.No. 5)
5.
A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm3, then the weight of the pipe is:
Answer: Option
Explanation:
External radius = 4 cm,
Internal radius = 3 cm.
| Volume of iron |
|
||||||
|
|||||||
| = 462 cm3. |
cm3
Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.
Discussion:
60 comments Page 2 of 6.
Prasanth said:
6 years ago
Convert hollow cylinder into the metal sheet.
Hollow cylinder
h=21cm
Thickness=1cm
Metal sheet
b=21cm
h=1cm
loop length of the cylinder will become the length of the metal sheet.
The loop length=2*π*r, r=4cm.
The volume of metal sheet
= lbh = (2*π*4)*21*1=527.52 cu.cm.
weight=8*527.52=4220.16 gm=4.220 kg
is this wrong? if this so, explain me.
Hollow cylinder
h=21cm
Thickness=1cm
Metal sheet
b=21cm
h=1cm
loop length of the cylinder will become the length of the metal sheet.
The loop length=2*π*r, r=4cm.
The volume of metal sheet
= lbh = (2*π*4)*21*1=527.52 cu.cm.
weight=8*527.52=4220.16 gm=4.220 kg
is this wrong? if this so, explain me.
Rohan said:
6 years ago
How did you find out the DENSITY = 8? Please explain.
Mritunjay said:
6 years ago
1dm=how many metre
Deepak M said:
6 years ago
Thank you all.
Lavanya said:
6 years ago
Thank you, well explained @Dhinesh.
Dhinesh said:
6 years ago
Because, the VOLUME OF HOLLOW CYLINDER FORMULA is π(R^2-r^2)h.
Assume, (hollow iron pipe is in the shape of a hollow cylinder).
Here, we require RADIUS, but in question, it is in DIAMETER.
Outer dia=8cm[R=d/2=4cm];;then thickness=1cm(I.e, reduce diameter Of outer pipe by 2cm)..since thickness is both sides of pipe..(thickness = 1+1)
This gives diameter Of inner pipe = 6cm[r=d/2=3cm].
Then substitute R&r in formula....
Volume = 22/7 * (4^2 - 3^2) * 21 = 22 * (16 - 9) * 3 = 22*(7)*3 = 462 cm3.
Weight = density * volume.
= 8 * 462 = 3696 g.
= 3696/1000 = 3.696 kg.
Assume, (hollow iron pipe is in the shape of a hollow cylinder).
Here, we require RADIUS, but in question, it is in DIAMETER.
Outer dia=8cm[R=d/2=4cm];;then thickness=1cm(I.e, reduce diameter Of outer pipe by 2cm)..since thickness is both sides of pipe..(thickness = 1+1)
This gives diameter Of inner pipe = 6cm[r=d/2=3cm].
Then substitute R&r in formula....
Volume = 22/7 * (4^2 - 3^2) * 21 = 22 * (16 - 9) * 3 = 22*(7)*3 = 462 cm3.
Weight = density * volume.
= 8 * 462 = 3696 g.
= 3696/1000 = 3.696 kg.
(1)
Dhinesh said:
6 years ago
Because, the VOLUME OF HOLLOW CYLINDER FORMULA is π(R^2-r^2)h.
Assume, (hollow iron pipe is in the shape of a hollow cylinder).
Here, we require RADIUS, but in question, it is in DIAMETER.
Outer dia=8cm[R=d/2=4cm];;then thickness=1cm(I.e, reduce diameter Of outer pipe by 1cm).
This gives diameter Of inner pipe = 7cm[r=d/2=3.5cm].
Then substitute R&r in formula,
Volume = 22/7 * (4^2 - 3.5^2) * 21 = 22 * (16 - 12.25) * 3 = 22*(3.75)*3 = 247.5 cm3.
Weight = density * volume.
= 8 * 247.5 = 1980 g.
= 1980/1000 = 1.98 kg.
Assume, (hollow iron pipe is in the shape of a hollow cylinder).
Here, we require RADIUS, but in question, it is in DIAMETER.
Outer dia=8cm[R=d/2=4cm];;then thickness=1cm(I.e, reduce diameter Of outer pipe by 1cm).
This gives diameter Of inner pipe = 7cm[r=d/2=3.5cm].
Then substitute R&r in formula,
Volume = 22/7 * (4^2 - 3.5^2) * 21 = 22 * (16 - 12.25) * 3 = 22*(3.75)*3 = 247.5 cm3.
Weight = density * volume.
= 8 * 247.5 = 1980 g.
= 1980/1000 = 1.98 kg.
Dhana reddy said:
6 years ago
Why 4^2 -3^2? Please explain this step.
Indranath said:
6 years ago
Why not (4-3)^2?
Gautam said:
6 years ago
Isn't weight=mass*gravity?
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