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(C) The extension $\Delta L$ of a wire is given by the formula $\Delta L = \frac{F \cdot L}{A \cdot Y}$, where $F$ is the tension, $L$ is the length,

Vedclass · Accepted Answer

$L = 50 \ cm, d = 0.05 \ mm$

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(C) The extension $\Delta L$ of a wire is given by the formula $\Delta L = \frac{F \cdot L}{A \cdot Y}$, where $F$ is the tension, $L$ is the length, $A$ is the cross-sectional area, and $Y$ is Young's modulus.Since $A = \pi \cdot (d/2)^2 = \frac{\pi d^2}{4}$, we can write $\Delta L = \frac{4 F L}{\pi d^2 Y}$.Given that $F$ and $Y$ are constant for the same material and same tension, we have $\Delta L \propto \frac{L}{d^2}$.Calculating the ratio $\frac{L}{d^2}$ for each case:$A: \frac{200}{(0.5)^2} = \frac{200}{0.25} = 800$$B: \frac{300}{(1.0)^2} = \frac{300}{1} = 300$$C: \frac{50}{(0.05)^2} = \frac{50}{0.0025} = 20000$$D: \frac{100}{(0.2)^2} = \frac{100}{0.04} = 2500$Comparing these values, the ratio is maximum for option $C$.

Let $L$ be the length and $d$ be the diameter of the cross-section of a wire. Wires of the same material with different $L$ and $d$ are subjected to the same tension along the length of the wire. In which of the following cases will the extension of the wire be the maximum?

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