The dimensions of four wires of the same mate | vedclass

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(D) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sec

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Length $50 \, cm$,Diameter $0.5 \, mm$

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(D) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the change in length.Rearranging for $\Delta L$,we get $\Delta L = \frac{F \cdot L}{A \cdot Y}$.Since the material is the same,$Y$ is constant. For the same tension $F$,we have $\Delta L \propto \frac{L}{A}$.Since $A = \pi \cdot r^2 = \pi \cdot (d/2)^2 = \frac{\pi \cdot d^2}{4}$,we have $\Delta L \propto \frac{L}{d^2}$.Calculating the ratio $\frac{L}{d^2}$ for each option:$(a)$ $\frac{100}{1^2} = 100$$(b)$ $\frac{200}{2^2} = \frac{200}{4} = 50$$(c)$ $\frac{300}{3^2} = \frac{300}{9} \approx 33.33$$(d)$ $\frac{50}{0.5^2} = \frac{50}{0.25} = 200$Since the ratio is maximum for option $(d)$,the increase in length is maximum for wire $(d)$.

The dimensions of four wires of the same material are given below. In which wire will the increase in length be maximum when the same tension is applied?

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