Two wires A and B of the same material have lengths in the ratio $1 : 2$ and diameters in the ratio $2 : 1$. They are subjected to the same tensile force. The ratio of their extensions $\Delta L_A : \Delta L_B$ is
A$1 : 8$
B$8 : 1$
C$1 : 2$
D$2 : 1$
Answer & Solution
Correct answer: A. $1 : 8$
$\Delta L = \dfrac{FL}{AY}$. Same material ⇒ same $Y$. $\dfrac{\Delta L_A}{\Delta L_B} = \dfrac{L_A}{L_B}\cdot\dfrac{A_B}{A_A} = \dfrac{1}{2}\cdot\dfrac{d_B^2}{d_A^2} = \dfrac{1}{2}\cdot\dfrac{1}{4} = \dfrac{1}{8}$.
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