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(C) The equation for the fundamental mode is $L = \frac{1}{2f\sqrt{\mu}} \sqrt{T}$.Comparing this with the equation of a line $y = mx + c$,where $y =

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$6250 \pm 546.9 \ Hz$

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(C) The equation for the fundamental mode is $L = \frac{1}{2f\sqrt{\mu}} \sqrt{T}$.Comparing this with the equation of a line $y = mx + c$,where $y = L$ and $x = \sqrt{T}$,the slope $m = \frac{1}{2f\sqrt{\mu}}$.Given $m = 8.0 	imes 10^{-3}$ and $\mu = 1.0 	imes 10^{-4} \ kg/m$,we have:$8.0 	imes 10^{-3} = \frac{1}{2f \sqrt{1.0 	imes 10^{-4}}} = \frac{1}{2f 	imes 10^{-2}}$.Solving for $f$: $f = \frac{1}{2 	imes 10^{-2} 	imes 8.0 	imes 10^{-3}} = \frac{1}{16 	imes 10^{-5}} = \frac{10^5}{16} = 6250 \ Hz$.To find the error $\Delta f$,we use the relative error formula from $f = \frac{1}{2m\sqrt{\mu}}$:$\frac{\Delta f}{f} = \frac{\Delta m}{m} + \frac{1}{2} \frac{\Delta \mu}{\mu}$.Given $\Delta m = 0.3 	imes 10^{-3}$ and $\Delta \mu = 0.1 	imes 10^{-4}$:$\frac{\Delta f}{f} = \frac{0.3 	imes 10^{-3}}{8.0 	imes 10^{-3}} + \frac{1}{2} \left( \frac{0.1 	imes 10^{-4}}{1.0 	imes 10^{-4}} ight) = \frac{0.3}{8.0} + \frac{0.1}{2} = 0.0375 + 0.05 = 0.0875$.$\Delta f = 6250 	imes 0.0875 = 546.875 \ Hz \approx 546.9 \ Hz$.Thus,$f = (6250 \pm 546.9) \ Hz$.

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