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Standard XII

Physics

Poisson's Effect

A light rod o...

Question

A light rod of length ‘L’ is suspended from a support horizontally by means of two vertical wires A and B of equal length as shown. Cross-section area of ‘A’ is half that of B. A weight ‘w’ is hung on the rod as shown. The value of ‘x’ so that the stress in ‘A’ is same as that in ‘B’, is

\frac{L}{3}

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\frac{L}{2}

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\frac{2 L}{3}

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\frac{3 L}{4}

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Solution

The correct option is C $\frac{2 L}{3}$

$T_{A} + T_{B} = W$
$(T_{A}) (x) = T_{B} (L - x)$
Solving the above equations
$T_{A} = \frac{w (L - x)}{L}; T_{B} = \frac{W x}{L}$
Stress in ‘A’ = $\frac{T_{A}}{A_{A}}$ Where $A_{A}$ is cross section area of wire ‘A’.
Stress in ‘B’ = $\frac{T_{B}}{A_{B}}$ where $A_{B}$ is cross section area of wire ‘B’
It is given $A_{A} = \frac{A_{B}}{2}, \frac{T_{A}}{A_{A}} = \frac{T_{B}}{A_{B}} \Rightarrow x = \frac{2 L}{3}$

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The correct option is C 2L3 TA+TB=W (TA)(x)=TB(L–x) Solving the above equations TA=w(L−x)L;TB=WxL Stress in ‘A’ = TAAA Where AA is cross section area of wire ‘A’. Stress in ‘B’ = TBAB where AB is cross section area of wire ‘B’ It is given AA=AB2,TAAA=TBAB⇒x=2L3