A) \[3:\pi \]
B) \[\pi :3\]
C) \[1:\pi \]
D) \[\pi :1\]
Correct Answer: A
Solution :
[a] \[\delta =\frac{W{{\ell }^{3}}}{3YI}\], where\[W=load\], \[\ell =\]length of beam and \[I\]is geometrical moment of inertia for rectangular beam, \[I=\frac{b{{d}^{3}}}{12}\]where \[b=breadth\]and \[d=depth\] For square beam \[b=d\,\,\,\therefore \,\,\,{{I}_{1}}=\frac{{{b}^{4}}}{12}\] For a beam of circular cross-section \[{{I}_{2}}=\left( \frac{\pi {{r}^{4}}}{4} \right)\] \[\therefore \,\,\,\,{{\delta }_{1}}=\frac{W{{\ell }^{3}}\times 12}{3Y{{b}^{4}}}=\frac{4W{{\ell }^{3}}}{Y{{b}^{4}}}\] (for sq. cross section) \[and\,\,{{\delta }_{2}}=\frac{W{{\ell }^{3}}}{3Y(\pi {{r}^{4}}/4)}=\frac{4W{{\ell }^{3}}}{3Y(\pi {{r}^{4}})}\] (fix circular cross-section) Now \[\frac{{{\delta }_{1}}}{{{\delta }_{2}}}=\frac{3\pi {{r}^{4}}}{{{b}^{4}}}=\frac{3\pi {{r}^{4}}}{{{(\pi {{r}^{2}})}^{2}}}=\frac{3}{\pi }\] \[(\because \,\,{{b}^{2}}=\pi {{r}^{2}}\]i.e., they have same cross- sectional area)
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