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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 93
Total: 160
Two wires of equal length and equal cross sectional areas are suspended as shown in the figure. Then Young’s modulii are
Y
1
and
Y
2
respectively. The equivalent Young's modulus is
Y
1
+
Y
2
Y
1
+
Y
2
2
Y
1
Y
2
Y
1
+
Y
2
√
Y
1
Y
2
Validate
Solution:
Let the equivalent Young’s modulus of the combination is Y and the equivalent area of the combination will be 2A.
For parallel combination,
k
eq
=
k
1
+
k
2
Y
(
2
A
)
L
=
Y
1
A
L
+
Y
2
A
L
Y
=
Y
1
+
Y
2
2
© examsnet.com
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