A) \[{{3}^{1/4}}mm\]
B) \[{{4}^{1/3}}mm\]
C) \[{{3}^{1/2}}mm\]
D) \[{{2}^{1/3}}mm\]
Correct Answer: B
Solution :
The temperature of the wire increase to such a value at which the heat produced per second equals heat lost per second due to radiation. \[{{i}^{2}}\left( \frac{\rho l}{\pi {{r}^{2}}} \right)=H\times 2\pi rl\]. where H is heat lost per second per unit area due to radiation Hence, \[{{i}^{2}}\propto {{r}^{3}}\] So, \[\frac{i_{1}^{2}}{i_{2}^{2}}=\frac{r_{1}^{3}}{r_{2}^{3}}\] or \[{{r}_{2}}={{r}_{1}}{{\left( \frac{{{i}_{2}}}{{{i}_{1}}} \right)}^{2/3}}\] [Here: \[{{r}_{1}}=1mm,\,\,{{i}_{1}}=1.5A,\,\,{{i}_{2}}=3A\]] \[{{r}_{2}}=1\times {{\left( \frac{3}{1.5} \right)}^{2/3}}={{4}^{1/3}}mm\]
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