The dimensions of four wires of the same mate | vedclass

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(D) Young's Modulus is given by $Y = \frac{F/A}{\Delta L/L} = \frac{FL}{\Delta L A}$.Since $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$, we have $\

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

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(D) Young's Modulus is given by $Y = \frac{F/A}{\Delta L/L} = \frac{FL}{\Delta L A}$.Since $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$, we have $\Delta L = \frac{FL}{YA} = \frac{4FL}{Y \pi D^2}$.For wires of the same material ($Y$ is constant) subjected to the same force ($F$ is constant), the increase in length $\Delta L \propto \frac{L}{D^2}$.$(a)$ $\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)$ $\frac{50}{0.5^2} = \frac{50}{0.25} = 200$Comparing the values, the increase in length is maximum for option $(d)$.

The dimensions of four wires of the same material are given below. The increase in length is maximum in the wire of

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