A) \[\frac{wl}{2aY}\]
B) \[\frac{2wl}{aY}\]
C) \[\frac{3wl}{2aY}\]
D) \[\frac{2wl}{3aY}\]
Correct Answer: A
Solution :
The situation is shown in figure below. Consider the small length of the rod at distance x from the ceiling. If the tension in the rod is T, then\[T=\frac{w}{l}(l-x)\] Now, under this tension elongation in the small part of rod is\[\Delta l=\frac{Tdx}{aY}=\frac{(l-x)wdx}{laY}\] \[\therefore \]Total elongation \[=\int_{{}}^{{}}{\Delta l=\int_{0}^{l}{\frac{(l-x)wdx}{laY}}}\] \[=\frac{w}{laY}\left[ lx-\frac{{{x}^{2}}}{2} \right]_{0}^{l}=\frac{wl}{2aY}\]You need to login to perform this action.
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