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Question

A steel wire of original length 1 m and cross-sectional area 4.00mm2 is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression ?

Y of the steel = 2.0×1011Nm2.Takeg=10ms2.

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Solution

From figure cos θ=XX2+I2

=XI{1+X2I2}12

=XI ....(1)

Increase in length,

ΔL=(AC+CB)AB

Here, AC =(I2+X2)12

So, ΔL=2(I2+X2)12

Given that,A=4mm2

= 4×102cm2

Substituting the values , we have

Y =FALΔL

2×1012

=T×1000(4×102)×[2(502+X2)12100] ....(2)

From equation (i), (ii) and the freebody diagram

2T cosθ=mg

2T(X50)=2.16×103×980

2×(2×1012)×(4×102)×[2(502+X2 12)100×X]100×50

= (2.16)×103×980

Solving for X, we get X = 1.5 cm.


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