Three vertical wires 1, 2 are 3 are supporting a non uniform rod of mass M in horizontal position as shown in the figure. The wires are of equal length and equal cross-sectional area. It is given that Y₂= Y₃.The wires 1 and 3 are attached at the extreme ends of the rod. Assume that Mg force acts just below the wire-2. Choose the correct option(s).

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$T_{2} = T_{3}$

$2 T_{1} = T_{3}$

$T_{1} = \frac{Mg}{5}$

$T_{1} = 2 T_{3}$

answer is B, C, D.

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Detailed Solution

$Y = \frac{TL}{AΔL}$ here A, $ΔL and L$ are same for all the three wires.
$YαT$ ___________(i)
For rotational equilibrium $T_{1} (2 x) = T_{3} (x)$
$\begin{aligned} 2 T_{1} = T_{3} \\ 2 Y_{1} = Y_{3} \end{aligned} - - - - - (ii)$
Given, $Y_{2} = Y_{3}$ ,
$\begin{array}{l} T_{2} = T_{3} - - - - - - (i) \\ T_{1} + T_{2} + T_{3} = Mg \end{array}$
(For translational equilibrium)
$T_{1} + 2 T_{1} + 2 T_{1} = Mg \Rightarrow T_{1} = Mg / 5$