A) The maximum value of Y that can be determined is \[{{10}^{14}}N/{{m}^{2}}\]
B) \[\frac{\Delta Y}{Y}\]?gets minimum contribution from the uncertainty in the length.
C) The figure of merit is the larges for the length of the rod.
D) \[\frac{\Delta Y}{Y}\]gets its maximum contribution from the uncertainty in strain.
Correct Answer: A
Solution :
\[\ell =1m\] \[r=5\times {{10}^{-3}}m\] \[F=50\pi \,\times {{10}^{3}}N\] \[\gamma =\frac{F/A}{\frac{\Delta \ell }{\ell }}\] \[\gamma =\frac{\Delta \ell }{\ell }=\frac{F}{A}\] \[\gamma =\frac{50\pi \times {{10}^{3}}}{\pi \times {{(5\times {{10}^{-3}})}^{2}}}\times \frac{\ell }{\Delta \ell }\] \[\gamma =\frac{50\times {{10}^{3}}}{25\times {{10}^{-6}}}\times \frac{1}{\Delta \ell }\Rightarrow \gamma =\frac{2\times {{10}^{9}}}{\Delta \ell }\] \[\gamma =\frac{2\times {{10}^{9}}}{\varepsilon }\] \[\gamma \max .=2\times {{10}^{9}}\]
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