A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminum (wire B) of equal lengths (figure). The cross-sectional areas of wires A and B are 1.0 mm² and 2.0 mm² respectively.
$(Y_{A l} = 70 \times 10^{9} N / m^{2} and Y_{Steel} = 200 \times 10^{9} N / m^{2})$

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Mass m should be suspended close to wire A to have equal stress in both the wires.

Mass m should be suspended close B to have equal stresses in both the wires.

Mass m should be suspended close to wire A to have equal strain in both wires.

Mass m should be suspended at the middle of the wires to have equal stresses in both the wires

answer is B, D.

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Detailed Solution

Let T_A and T_B be the tensions in wire A and B, respectively.
For the rotational equilibrium of the system: $T_{A} x = T_{B} (l - x) \to (1)$
If stress is same in both wires
$\frac{T_{A}}{A_{A}} = \frac{T_{B}}{A_{B}} \Rightarrow \frac{T_{A}}{1} = \frac{T_{B}}{2} \Rightarrow T_{B} = 2 T_{A} \to (2)$
Substituting in (1),
$T_{A} x = 2 T_{A} (l - x) \Rightarrow x = 2 l - 2 x \Rightarrow x = \frac{2 l}{3}$
⇒ mass m should be hung closer to B for having same stress in both wires.
If strain is same in both wires
$\frac{T_{A}}{A_{A} Y_{Steel}} = \frac{T_{B}}{A_{B} Y_{A l}} \Rightarrow \frac{T_{A}}{1 \times 200 \times 10^{9}} = \frac{T_{B}}{2 \times 70 \times 10^{9}} \Rightarrow T_{B} = \frac{7}{10} T_{A} \to (3)$
Substituting in (1),
$T_{A} x = \frac{7}{10} T_{A} (l - x) \Rightarrow 10 x = 7 l - 7 x \Rightarrow 17 x = 7 l \Rightarrow x = \frac{7}{17} l$
⇒ mass m must be hung closer to wire A for having same strain in both wires.