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) \[\frac{3L}{4}\]

Correct Answer: C

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 \[x=\frac{2l}{3}\]