A string of length 0.314 text{ m} and Young's | vedclass

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(B) The total extension $\Delta L$ is the sum of extensions of both strings: $\Delta L = \Delta L_A + \Delta L_B = \frac{F L_A}{Y_A A} + \frac{F L_B}{

Vedclass · Accepted Answer

$2$

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(B) The total extension $\Delta L$ is the sum of extensions of both strings: $\Delta L = \Delta L_A + \Delta L_B = \frac{F L_A}{Y_A A} + \frac{F L_B}{Y_B A}$.Given $F = mg = 0.8 	imes 10 = 8 	ext{ N}$.For string $A$: $L_A = 0.314 	ext{ m}$, $Y_A = 2 	imes 10^{10} 	ext{ N/m}^2$.For string $B$: $L_B = 2 L_A = 0.628 	ext{ m}$, $Y_B = 2 Y_A = 4 	imes 10^{10} 	ext{ N/m}^2$.The cross-sectional area $A = \pi r^2 = 3.14 	imes (0.2 	imes 10^{-3} 	ext{ m})^2 = 3.14 	imes 4 	imes 10^{-8} = 1.256 	imes 10^{-7} 	ext{ m}^2$.Calculating $\Delta L_A$: $\Delta L_A = \frac{8 	imes 0.314}{2 	imes 10^{10} 	imes 1.256 	imes 10^{-7}} = \frac{2.512}{2512} = 0.001 	ext{ m} = 1 	ext{ mm}$.Calculating $\Delta L_B$: $\Delta L_B = \frac{8 	imes 0.628}{4 	imes 10^{10} 	imes 1.256 	imes 10^{-7}} = \frac{5.024}{5024} = 0.001 	ext{ m} = 1 	ext{ mm}$.Total extension $\Delta L_{total} = \Delta L_A + \Delta L_B = 1 	ext{ mm} + 1 	ext{ mm} = 2 	ext{ mm}$.

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