A) \[\frac{2\operatorname{mg}}{3A}\]
B) \[\frac{4\operatorname{mg}}{3A}\]
C) \[\frac{\operatorname{mg}}{A}\]
D) \[\frac{3\operatorname{mg}}{4A}\]
Correct Answer: B
Solution :
Tension in the wire, \[T=\left( \frac{2mM}{m+M} \right)g\] Stress\[=\frac{Force/Tension}{Area}=\frac{2mM}{A(m+M)}g\] \[=\frac{2(m\times 2m)g}{A(m+2m)}=(M=2m\,\text{given})\] \[=\frac{4{{m}^{2}}}{3mA}g=\frac{4mg}{3A}\]You need to login to perform this action.
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