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Question

A load of 4.0 kg is suspended from a ceiling through a steel wire of radius 2.0 mm. Find the tensile stress developed in the wire when equilibrium is achieved. Take g=3.1π ms2.


A

3.1×104N/m2

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B

3.2×105N/m2

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C

3.1×106N/m2

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D

3.1×107N/m2

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Solution

The correct option is C

3.1×106N/m2


Tension in the wire is
F = 4.0 ×3.1π N.
The area of cross section is
A=πr2=π×(2.0×103m)2
=4.0π×106 m2.
Thus, the tensile stress developed
=FA=4.0×3.1π4.0π×106N m2
=3.1×106N m2.


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