A) True
B) False
C) Insufficient data
D) None of the above
Correct Answer: A
Solution :
We know that, \[R=\rho \frac{l}{a}\] \[R=\rho \frac{l}{\pi {{r}^{2}}}\] \[\therefore \]Heat produced per second \[={{I}^{2}}\times \rho \frac{l}{\pi {{r}^{2}}}\] Rate of loss of heat per unit surface area, \[H=\frac{{{I}^{2}}\rho \frac{l}{\pi {{r}^{2}}}}{2\pi rl}=\frac{{{I}^{2}}\rho }{2{{\pi }^{2}}{{r}^{3}}}\] As the heat is independent of\[l\]so both the wires will fuse at the same value of current because they have same\[\rho \]and r.
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