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15334794417361020634038.jpg A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of brass and its area of cross section 0.29*10-4 m^2 and the other is of steel of cross section 0.1*10^-4 m^2. In order to have equal stress in the two wires, a weight should be hung from the end A at a distance of.

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@Roshni Afrin

T_sa_s=T_ba_b\\\Rightarrow T_s=2T_b\\since\ the\ system\ is \ in\ equilibrium\ moments\ of\ force\ T_s\ and\ T_b\ about \ C\ will\ be\ equal\\\Rightarrow T_sx=T_b(2-x))\\\Rightarrow x=\frac{2}{3}m=66.6cm

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Safeer PP

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@Roshni Afrin 

sir could u plz explain me ... why u wrote Tx=T(2-x)

Let the point c is at a distance x from B(Si). Then the distance from A(Brass)=(total length of rod-x)=(2-x)

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Safeer PP

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