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(A) Let $r_A, r_B$ be the radii and $Y_A, Y_B$ be the Young's moduli of wires $A$ and $B$ respectively.Given: $\frac{r_A}{r_B} = \frac{2}{3}$ and $\fr

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$90 \ cm$ from $P$

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(A) Let $r_A, r_B$ be the radii and $Y_A, Y_B$ be the Young's moduli of wires $A$ and $B$ respectively.Given: $\frac{r_A}{r_B} = \frac{2}{3}$ and $\frac{Y_A}{Y_B} = \frac{3}{2}$.Let $L$ be the length of the wires (assumed equal) and $\Delta L$ be the equal elongation in both wires.From the formula for elongation,$\Delta L = \frac{F L}{A Y}$,where $F$ is the tension in the wire and $A = \pi r^2$ is the cross-sectional area.Since $\Delta L$ is the same for both wires,$\frac{F_A L}{\pi r_A^2 Y_A} = \frac{F_B L}{\pi r_B^2 Y_B}$.Thus,$\frac{F_A}{F_B} = \frac{r_A^2 Y_A}{r_B^2 Y_B} = \left(\frac{r_A}{r_B}ight)^2 \left(\frac{Y_A}{Y_B}ight) = \left(\frac{2}{3}ight)^2 	imes \left(\frac{3}{2}ight) = \frac{4}{9} 	imes \frac{3}{2} = \frac{2}{3}$.Let the weight be suspended at a distance $x$ from $P$. Taking torque about the point of suspension,$F_A x = F_B (150 - x)$.$\frac{F_A}{F_B} = \frac{150 - x}{x} = \frac{2}{3}$.$3(150 - x) = 2x \implies 450 - 3x = 2x \implies 5x = 450 \implies x = 90 \ cm$ from $P$.

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