A) 1.75
B) 2.0
C) 2:3
D) 5.0
Correct Answer: C
Solution :
The elonfation in wire \[\Delta l=\frac{F.L}{\pi {{r}^{2}}Y}\] elonfation in both the wires is same wires one of same length and under the same weight (load) So \[\frac{1}{r_{1}^{2}{{Y}_{1}}}=\frac{1}{r_{2}^{2}{{Y}_{2}}}\] \[{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=\frac{{{Y}_{2}}}{{{Y}_{2}}}\]or \[{{\left( \frac{{{D}_{1}}}{{{D}_{2}}} \right)}^{2}}\] \[={{\left( \frac{3}{{{D}_{2}}} \right)}^{2}}=\frac{12\times {{10}^{10}}}{7\times {{10}^{10}}}\] \[\frac{3}{{{D}_{2}}}=\sqrt{\frac{12}{7}}\Rightarrow 3\sqrt{\frac{7}{12}}\] \[\approx 2.3\,mm\]
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