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IIT JEE Advanced 2012 Paper 1
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Section:
Physics
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© examsnet.com
Question : 1 of 60
Marks:
+1
,
-0
In the determination of Young’s modulus
(
Y
=
4
M
L
g
π
l
d
2
)
\left( Y = \frac{4MLg}{\pi l d^2} \right)
(
Y
=
π
l
d
2
4
M
Lg
)
by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their scale is 100. The contributions to the maximum probable error of the Y measurement
[JEE Adv 2012 P1]
due to the errors in the measurements of d and l are the same.
due to the errors in the measurements of d is twice that due to the error in the measurement of l.
due to the errors in the measurements of l is twice that due to the error in the measurement of d.
due to the errors in the measurements of d is four times that due to the error in the measurement of l.
Validate
Solution:
The least count is obtained as
0.5
100
\frac{0.5}{100}
100
0.5
= 0.005 mm
The contributions to the maximum probable error of the Y measurement are calculated as follows:
Δ
Y
Y
\frac{\Delta Y}{Y}
Y
Δ
Y
=
Δ
l
l
+
2
Δ
d
d
\frac{\Delta l}{l} + \frac{2\Delta d}{d}
l
Δ
l
+
d
2Δ
d
Δ
l
l
\frac{\Delta l}{l}
l
Δ
l
=
0.005
×
1
0
−
3
0.25
×
1
0
−
3
\frac{0.005 \times 10^{-3}}{0.25 \times 10^{-3}}
0.25
×
1
0
−
3
0.005
×
1
0
−
3
=
1
50
\frac{1}{50}
50
1
2
Δ
d
d
\frac{2\Delta d}{d}
d
2Δ
d
=
2
×
0.005
×
1
0
−
3
0.5
×
1
0
−
3
\frac{2 \times 0.005 \times 10^{-3}}{0.5 \times 10^{-3}}
0.5
×
1
0
−
3
2
×
0.005
×
1
0
−
3
=
1
50
\frac{1}{50}
50
1
© examsnet.com
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