The graph shows the stress–strain lines for two wires $P$ and $Q$ within their elastic limits. The wires have equal lengths and the area of cross-section of $P$ is twice that of $Q$. When the same load is hung from each wire, the ratio of the extension of $P$ to that of $Q$ is
Answer: (A) $1 : 8$
Concept. The slope of a stress–strain line is Young's modulus $Y$. Extension $\Delta L = \dfrac{FL}{AY}$.
Step 1. Young's moduli from the graph.
$$Y_P = \frac{4 \times 10^8}{2 \times 10^{-3}} = 2 \times 10^{11}\ \text{N/m}^2,\qquad Y_Q = \frac{2 \times 10^8}{4 \times 10^{-3}} = 5 \times 10^{10}\ \text{N/m}^2$$
So $Y_P = 4Y_Q$.
Step 2. Ratio of extensions. Same $F$ and $L$:
$$\frac{\Delta L_P}{\Delta L_Q} = \frac{A_QY_Q}{A_PY_P} = \frac{1}{2 \times 4} = \frac18$$
Answer. $1 : 8$ (Option A).
Solution by Sreeraj P, M.Sc Physics