Two wires of the same length and material but different radii $r_{1}$ and $r_{2}$ are suspended from a rigid support both carry the same load at the lower end. What is the ratio of the stress developed in the second wire to that developed in the first wire.

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Solution

Stress (P) = $\frac{L o a d}{A r e a}$
$∴$ for the first wire,
stress, $P_{1} = \frac{F}{π r_{1}^{2}}$
And for the second wire,
$P_{2} = \frac{F}{π r_{2}^{2}}$
$∴$ $\frac{P_{2}}{P_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}$

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Two wires of the same length and material but different radii r1 and r2 are suspended from a rigid support both carry the same load at the lower end. What is the ratio of the stress developed in the second wire to that developed in the first wire.

Stress (P) = LoadArea∴ for the first wire,stress, P1=Fπr21And for the second wire,P2=Fπr22∴ P2P1=r21r22

Two wires of the same length and material but different radii $r_{1}$ and $r_{2}$ are suspended from a rigid support both carry the same load at the lower end. What is the ratio of the stress developed in the second wire to that developed in the first wire.

Stress (P) = $\frac{L o a d}{A r e a}$
$∴$ for the first wire,
stress, $P_{1} = \frac{F}{π r_{1}^{2}}$
And for the second wire,
$P_{2} = \frac{F}{π r_{2}^{2}}$
$∴$ $\frac{P_{2}}{P_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}$