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

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

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

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$\frac{M g}{A}$

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Zero

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Solution

The correct option is B
$\frac{3 M g}{2 A}$

At the mid-point, stress will develop because of half of the weight of the rod and that of the suspended weight.
Therefore, stress at mid-point is given by $\frac{M g + \frac{1}{2} M g}{A} = \frac{3 M g}{2 A}$ , which is choice (b).

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

The correct option is B 3Mg2A At the mid-point, stress will develop because of half of the weight of the rod and that of the suspended weight. Therefore, stress at mid-point is given by Mg+12MgA=3Mg2A, which is choice (b).

The correct option is B
$\frac{3 M g}{2 A}$

At the mid-point, stress will develop because of half of the weight of the rod and that of the suspended weight.
Therefore, stress at mid-point is given by $\frac{M g + \frac{1}{2} M g}{A} = \frac{3 M g}{2 A}$ , which is choice (b).