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(B) The formula for Young's modulus is $Y = \frac{FL}{Ae} = \frac{4FL}{\pi D^2 e}$.Given: $L = 2 \,m$, $F = 1.0 	imes 9.8 \,N$, $e = 0.8 	imes 10^{-

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$(2.0 \pm 0.2) 	imes 10^{11} \,N/m^2$

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(B) The formula for Young's modulus is $Y = \frac{FL}{Ae} = \frac{4FL}{\pi D^2 e}$.Given: $L = 2 \,m$, $F = 1.0 	imes 9.8 \,N$, $e = 0.8 	imes 10^{-3} \,m$, $\Delta e = 0.05 	imes 10^{-3} \,m$, $D = 0.4 	imes 10^{-3} \,m$, $\Delta D = 0.01 	imes 10^{-3} \,m$.First, calculate the value of $Y$:$Y = \frac{4 	imes 9.8 	imes 2}{\pi 	imes (0.4 	imes 10^{-3})^2 	imes (0.8 	imes 10^{-3})} = \frac{78.4}{\pi 	imes 0.16 	imes 10^{-6} 	imes 0.8 	imes 10^{-3}} \approx 1.95 	imes 10^{11} \,N/m^2 \approx 2.0 	imes 10^{11} \,N/m^2$.Now, calculate the relative uncertainty $\frac{\Delta Y}{Y}$:$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta D}{D} + \frac{\Delta e}{e}$.Since $F$ and $L$ are exact, $\frac{\Delta F}{F} = 0$ and $\frac{\Delta L}{L} = 0$.$\frac{\Delta Y}{Y} = 2 \left( \frac{0.01}{0.4} ight) + \left( \frac{0.05}{0.8} ight) = 2(0.025) + 0.0625 = 0.05 + 0.0625 = 0.1125$.Absolute uncertainty $\Delta Y = 0.1125 	imes Y = 0.1125 	imes 1.95 	imes 10^{11} \approx 0.22 	imes 10^{11} \,N/m^2$.Rounding to one decimal place, $\Delta Y \approx 0.2 	imes 10^{11} \,N/m^2$.Thus, $Y = (2.0 \pm 0.2) 	imes 10^{11} \,N/m^2$.

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