A) \[\frac{3c}{2a{{b}^{2}}}\]
B) \[\frac{2{{a}^{2}}c}{b}\]
C) \[\frac{3a}{2{{b}^{2}}c}\]
D) \[\frac{2ac}{{{b}^{2}}}\]
Correct Answer: C
Solution :
[c] According to questions, \[\frac{{{\ell }_{s}}}{{{\ell }_{b}}}=a,\frac{{{r}_{s}}}{{{r}_{b}}}=b,\,\frac{{{y}_{s}}}{{{y}_{b}}}=c,\frac{\Delta {{\ell }_{s}}}{\Delta {{\ell }_{b}}}=?\] As, \[y=\frac{F\ell }{A\Delta \ell }\Rightarrow \Delta \ell =\frac{F\ell }{Ay}\] \[\Delta {{\ell }_{s}}=\frac{3mg{{\ell }_{s}}}{\pi r_{s}^{2}.{{y}_{s}}}\,\,\left[ \therefore \,{{F}_{s}}=(M+2M)g \right]\] \[\Delta {{\ell }_{b}}=\frac{2Mg{{\ell }_{b}}}{\pi r_{s}^{2}.{{y}_{b}}}\,\,\,[\because {{F}_{b}}=2Mg]\] \[\therefore \,\,\,\,\frac{\Delta {{\ell }_{s}}}{\Delta {{\ell }_{b}}}=\frac{\frac{3Mg{{\ell }_{s}}}{\pi r_{s}^{2}.{{y}_{s}}}}{\frac{2Mg{{\ell }_{b}}}{\pi r_{b}^{2}.{{y}_{b}}}}=\frac{3a}{2{{b}^{2}}C}\]
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