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Question

A steel rod of cross-sectional area 1 m2 is acted upon by forces as shown in the figure. The total elongation of the bar is (Take Y=2.0×1011 Nm2)


A
11×107 m
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B
13×107 m
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C
15×107 m
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D
17×107 m
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Solution

The correct option is B 13×107 m
The action of the forces on each part of the rod is shown below.


We know that,

Δl=F lA Y

So, we can calculate Δl for all parts of the rod.

ΔlAB=FAB lABA Y=60×103×1.51×2×1011

ΔlAB=4.5×107 m

ΔlBC=FBC lBCA Y=70×103×11×2×1011

ΔlBC=3.5×107 m

ΔlCD=FCD lCDA Y=50×103×21×2×1011

ΔlCD=5×107 m

Δl=ΔlAB+ΔlBC+ΔlCD

Δl=(4.5+3.5+5)×107 m

Δl=13×107 m

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