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  • One end of a uniform rod of mass M and crosssectional area A is suspended from rigid support and an equal mass M is suspended from the other end. The stress at the mid-point of the r

One end of a uniform rod of mass M and crosssectional area A is suspended from rigid support and an equal mass M is suspended from the other end. The stress at the mid-point of the rod will be

Option: 1

\mathrm{\frac{2 M g}{A}}


Option: 2

\mathrm{\frac{3 M g}{2 A}}


Option: 3

\mathrm{\frac{M g}{A}}


Option: 4

\mathrm{Zero}


Answers (1)

best_answer

Since the rod is uniform, half its weight can be taker to act at its mid-point. Therefore, stress at mid-poin is

\mathrm{\frac{\text { weight of suspended mass }+ \text { weight of half the rod }}{\text { cross-sectional area }}}

\mathrm{ =\frac{M g+\frac{1}{2} M g}{A}=\frac{3 M g}{2 A}, \text { which is choice (b). } }

Posted by

Ritika Harsh

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