A) \[m>dAL\]
B) \[m<dAL\]
C) \[m<dAL/2\]
D) \[m<dAL/4\]
Correct Answer: A
Solution :
[a] For equilibrium \[4dA\ell g=(dAL+m)g\] and\[\frac{\ell }{2}>{{Y}_{cm}}\] (for rotational equilibrium) \[{{Y}_{cm}}=\frac{m\times 0+dAL(L/2}{m+dAL}=\frac{dA{{L}^{2}}}{2(m+dAL)}\] \[\ell >\frac{A{{L}^{2}}d}{(ALd+m)}\Rightarrow \frac{m+dAL}{4dA}>\frac{A{{L}^{2}}d}{(ALd+m)}\] \[m>ALd\]
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