A) \[6.2\times {{10}^{-8}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]
B) \[2.1\times {{10}^{-4}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]
C) \[4.5\times {{10}^{-5}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]
D) \[9.3\times {{10}^{-6}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]
Correct Answer: D
Solution :
\[{{\tau }_{0}}\int\limits_{x=L}^{x=2L}{({{E}_{m}})(2L-x)}\] \[=\int\limits_{L}^{2L}{\left( \frac{{{\mu }_{0}}i}{2\pi x} \right)(i)(2L-x)}dx\] \[=\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi x}[2L\ln x-x]_{L}^{2L}\] \[{{\tau }_{0}}=\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi }[2L\ln (2)-L]\] \[=\frac{(0.386){{\mu }_{0}}{{i}^{2}}L}{2\pi }\] ? (i) \[=(0.386)(2\times {{10}^{-7}})(2)\] \[=3.1\times {{10}^{-7}}N-m\] \[{{I}_{0}}=\frac{m{{L}^{2}}}{3}\] \[=\frac{(0.1){{(1)}^{2}}}{3}=\frac{1}{30}\text{kg-}{{\text{m}}^{\text{2}}}\] \[\therefore \] \[\alpha =\frac{{{\lambda }_{0}}}{{{I}_{0}}}=9.3\times {{10}^{-6}}\text{rad/}{{\text{s}}^{\text{2}}}\]You need to login to perform this action.
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