A) \[\frac{g}{13l}\]
B) \[\frac{g}{2l}\]
C) \[\frac{g}{3l}\]
D) \[\frac{7g}{3l}\]
Correct Answer: A
Solution :
\[\tau \,\,about\,\,P=5\,\,{{M}_{0}}g\,\ell \,\,2\,{{M}_{0}}g\,\times \,2\ell \] \[={{M}_{0}}g\,\ell \] \[\tau \,\,about\text{ }P={{I}_{p\alpha }}\] \[{{M}_{0}}g\,\ell =\left[ 5\,{{M}_{0}}{{\ell }^{2}}+2{{M}_{0}}\,\times \,4{{\ell }^{2}} \right]\alpha \] \[{{M}_{0}}g\,\ell \,\,=\,\,13\,{{M}_{0}}{{\ell }^{2}}\alpha \] \[\alpha =\frac{g}{13\,\ell }\]You need to login to perform this action.
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