The lengths of four wires A, B, C and D made | vedclass

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(A) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{\pi r^2 \cdot \Delta l}$.Rearranging for elongation $\Delta l$,

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(A) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{\pi r^2 \cdot \Delta l}$.Rearranging for elongation $\Delta l$, we get $\Delta l = \frac{F \cdot l}{Y \cdot \pi r^2}$.Since the material is the same ($Y$ is constant) and the tension $F$ is the same, $\Delta l \propto \frac{l}{r^2}$.For wire $A$: $\Delta l_A \propto \frac{1}{(0.2)^2} = \frac{1}{0.04} = 25$.For wire $B$: $\Delta l_B \propto \frac{2}{(0.4)^2} = \frac{2}{0.16} = 12.5$.For wire $C$: $\Delta l_C \propto \frac{3}{(0.6)^2} = \frac{3}{0.36} = 8.33$.For wire $D$: $\Delta l_D \propto \frac{4}{(0.8)^2} = \frac{4}{0.64} = 6.25$.Comparing the values, the elongation is greatest in wire $A$.

The lengths of four wires $A, B, C$ and $D$ made of the same material are $1 \,m, 2 \,m, 3 \,m$ and $4 \,m$ respectively. The radii of the wires $A, B, C$ and $D$ are $0.2 \,mm, 0.4 \,mm, 0.6 \,mm$ and $0.8 \,mm$ respectively. For the same applied tension, the elongation is more in which wire?

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