A rod of length l and negligible mass is suspended at its two ends by two wires steel(A) and aluminium (B) of equal length as shown. The cross sectional areas of A and B are 1 and 2 mm² respectively. If Y_Al and Y_Steel are 70 and 200 GPa respectively,

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mass m should be suspended at the mid point to have equal stresses in the wires

mass m should be suspended close to A to have equal stresses in the wires

mass m should be suspended close to B to have equal strains in the wires

mass m should be suspended close to B to have equal stresses in the wires

answer is B.

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Detailed Solution

Stress _Al = Stress _St
$\Rightarrow \frac{F_{Al}}{A_{Al}} = \frac{F_{St}}{A_{St}} \Rightarrow \frac{F_{Al}}{F_{St}} = \frac{2}{1} τ = 0 \Rightarrow F_{Al} x_{Al} = F_{st} x_{st} \Rightarrow \frac{x_{Al}}{x_{st}} = \frac{F_{st}}{F_{Al}} = \frac{1}{2}$
x_Al < x_st $\to$ closer to B
Strains_Al = Strain_St
$\Rightarrow \frac{F_{Al}}{A_{Al} Y_{Al}} = \frac{F_{St}}{A_{St} Y_{St}} \Rightarrow \frac{F_{Al}}{F_{St}} = \frac{A_{Al}}{A_{St}} x \frac{Y_{Al}}{Y_{St}} = \frac{2}{1} x \frac{70}{200} = \frac{7}{10} = \frac{x_{St}}{x_{Al}}$
x_Al > x_st $\to$ closer to A