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 1 of 6.
Ram said:
4 years ago
External diameter = 8 cm.
External radius = R =4 cm.
Given Thickness= 1 cm.
So, internal diameter= (8-1) cm = 7 cm.
Internal radius = r = 3.5 cm not 3 cms.
According to the volume of iron;
= 21*(22/7)*(16-12.25),
= 247.5 cm3.
Weight of iron = 247.5 * 8 = 1980 gram or 1.98 kg.
External radius = R =4 cm.
Given Thickness= 1 cm.
So, internal diameter= (8-1) cm = 7 cm.
Internal radius = r = 3.5 cm not 3 cms.
According to the volume of iron;
= 21*(22/7)*(16-12.25),
= 247.5 cm3.
Weight of iron = 247.5 * 8 = 1980 gram or 1.98 kg.
(7)
Roney said:
6 years ago
@Nazim.
From both side the 1cm will be deducted.
so dia = 6cm radius 3cm.
From both side the 1cm will be deducted.
so dia = 6cm radius 3cm.
(5)
Debosmita Misra said:
2 years ago
The external Diameter is given 8 cm and the thickness of the pipe is given 1 cm.
Hence, the External radius (R) would be 4 cm & Internal radius(r) would be (8 - 1 - 1)/2 = 3cm.
And the volume will be PI*(R*R - r*r)*h = (22/7)*(4*4 - 3*3)*21 = (22/7)*(16-9)*21 = 462 g.
Therefore, weight of the pipe = (462*8) g = 3696 g = 3.696 kg.
Hence, the External radius (R) would be 4 cm & Internal radius(r) would be (8 - 1 - 1)/2 = 3cm.
And the volume will be PI*(R*R - r*r)*h = (22/7)*(4*4 - 3*3)*21 = (22/7)*(16-9)*21 = 462 g.
Therefore, weight of the pipe = (462*8) g = 3696 g = 3.696 kg.
(4)
Dhinesh said:
8 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.
(3)
Bala said:
3 years ago
@Dinesh.
Thank You for your good explanation.
Thank You for your good explanation.
(2)
Dinesh Reddy said:
3 years ago
Can anyone explain this answer once again deeply?
(1)
Pema said:
4 years ago
Why cant it be =pi * 3^2 * 21?
(1)
Nazim Shaik K said:
6 years ago
External diameter = 8 cm.
External radius = R =4 cm.
Given Thickness= 1 cm.
So, internal diameter= (8-1) cm = 7 cm.
Internal radius = r = 3.5 cm not 3 cms.
According to the volume of iron;
= 21*(22/7)*(16-12.25),
= 247.5 cm3.
Weight of iron = 247.5* 8 = 1980 gram or 1.98 kg.
External radius = R =4 cm.
Given Thickness= 1 cm.
So, internal diameter= (8-1) cm = 7 cm.
Internal radius = r = 3.5 cm not 3 cms.
According to the volume of iron;
= 21*(22/7)*(16-12.25),
= 247.5 cm3.
Weight of iron = 247.5* 8 = 1980 gram or 1.98 kg.
(1)
Dhinesh said:
8 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.
(1)
Gautam said:
8 years ago
Isn't weight=mass*gravity?
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