• # question_answer A light rod of length L is suspended from a support horizontally by means of two vertical wires A & B of equal lengths as shown. CSA of A is half of that of B, and Young's modulas of A is double to that of B. A weight W is hung on the rod as shown. The value of x so that stress in A is same as that in B is A)  L/3                         B)  L/2       C) $\frac{2L}{3}$                                  D) $\frac{3L}{4}$

Solution :

Let tension in wires A & B are ${{T}_{A}}\And {{T}_{B}}$ respectively. ${{T}_{A}}+{{T}_{B}}=W$ ${{T}_{A}}\times x={{T}_{B}}(L-x)$ Solving above equation, ${{T}_{A}}=\frac{W(L-x)}{L},{{T}_{B}}=\frac{Wx}{L}$ Stress in $A=\frac{{{T}_{A}}}{{{A}_{A}}}$ where ${{A}_{A}}$ is CSA of wire A Stress in $B=\frac{{{T}_{B}}}{{{A}_{B}}}$ where ${{A}_{B}}$ is CSA of wire B It is given${{A}_{A}}=\frac{{{A}_{B}}}{2},\frac{{{T}_{A}}}{{{A}_{A}}}=\frac{{{T}_{B}}}{{{A}_{B}}}$which gives $x=\frac{2l}{3}$

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