A) 9.2 A
B) 4.5 A
C) 6.2 A
D) 12.2 A
Correct Answer: A
Solution :
For a fuse wire maximum safe current i is related to radius r-through \[{{i}^{2}}\propto {{r}^{3}}\] or \[i\propto {{r}^{3/2}}\] So, \[\frac{{{i}_{2}}}{{{i}_{1}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{3/2}}={{\left( \frac{0.3}{0.2} \right)}^{3/2}}={{\left( \frac{3}{2} \right)}^{3/2}}\] or \[{{i}_{2}}={{\left( \frac{27}{8} \right)}^{\frac{1}{2}}}\times {{i}_{1}}=\sqrt{\frac{27}{8}}\times 5\] So, \[{{i}_{2}}=9.18\approx 9.2A\]
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