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Question

One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire in attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Find the longitudinal strain in both the wires. Area of cross section of each wire is 0.005cm2 and Young modulus of the metal is 2.0 × 1011 N m2. Take g=10ms2.

A
105,5×105
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B
103,2×103
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C
104,3×104
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D
None of the above
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Solution

The correct option is C 104,3×104
The situation is described in figure (14-W2). As the 1 kg mass is in equilibrium, the tension in the lower wire equal the weight of the load.

Thus T1=10N
Stress =10N/0.005cm2
=2 × 107 N m2.

Longitudinal strain =stressY=2×107N m22×1011N m2=104.

Considering the equilibrium of the upper block, we can write,

T2=20N+T1,or,T2=30N.

Stress =30N/0.005cm2

=6×107Nm2.

Longitudinal strain =6×107Nm22×1011Nm2=3×104.

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