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JEE Main 16 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 12
Total: 90
In order to determine the Young's modulus of a wire of radius
0.2
c
m
(measured using a scale of least count
=
0.001
c
m
) and length
1
m
(measured using a scale of least count
=
1
m
m
), a weight of mass
1
k
g
(measured using a scale of least count
=
1
g
) was hanged to get the elongation of
0.5
c
m
(measured using a scale of least count
0.001
c
m
). What will be the fractional error in the value of Young's modulus determined by this experiment?
[16 Mar 2021 Shift 2]
0.14
%
0.9
%
9
%
1.4
%
Validate
Solution:
Young's modulus,
Y
=
Stress
Strain
=
F
L
A
l
⇒
Y
=
m
g
L
π
R
2
I
.....(i)
[
∵
F
=
m
g
and
A
=
π
R
2
]
To determine the fractional errors, we can write Eq. (i) as follows
∆
Y
Y
=
∆
m
m
+
∆
L
I
+
2
⋅
∆
R
R
+
∆
l
I
⇒
∆
Y
Y
×
100
=
∆
m
m
×
100
+
∆
L
L
×
100
+
2
∆
R
R
×
100
+
∆
l
l
×
100
⇒
∆
Y
Y
×
100
=
100
[
∆
m
m
+
∆
L
L
+
2
∆
R
R
+
∆
l
I
]
=
100
[
1
1000
+
1
1000
+
2
(
0.001
0.2
)
+
0.001
0.5
]
=
1
10
+
1
10
+
1
+
1
5
=
14
10
=
1.4
%
© examsnet.com
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