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IIT JEE Advanced 2011 Paper 2
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© examsnet.com
Question : 3
Total: 60
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is
0.9%
2.4%
3.1%
4.2%
Validate
Solution:
For the screw gauge, we have
Least count =
Pitch
Number
of
divisions
on
head
scale
⇒ L.C. =
0.5
50
mm = 0.01 mm
Therefore, the diameter of the ball is
D = 2.5 + 0.01× 20 = 2.70 mm
The density of the ball is
ρ =
M
V
=
M
(
4
3
)
π
r
3
(
D
2
)
3
Therefore,
Δ
ρ
ρ
=
M
M
+
3
Δ
D
D
Now,
Δ
M
M
= 0.02 and D = Least count. Therefore,
Δ
ρ
ρ
= 0.02 +
3
(
0.01
2.7
)
= 0.031
or the percentage error in density is 3.1%.
© examsnet.com
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