A) 19.6 min
B) 12.7 min
C) 14.4 min
D) 15.0 min
Correct Answer: C
Solution :
We know,\[\frac{{{v}^{2}}t}{4.2R}=Q=msd\theta \] Let N = initial number of turns in the coil R = resistance of the coil Now, \[R=\rho \frac{l}{A}=\frac{\rho \times N\times 2\pi r}{A}\] Also,\[\frac{{{v}^{2}}tA}{4.2\times \rho \times N\times 2\pi r}=Q=msd\theta \] \[\Rightarrow \]\[\frac{t}{N}=\] constant\[\Rightarrow \]\[\frac{{{t}_{1}}}{{{N}_{1}}}=\frac{{{t}_{2}}}{{{N}_{2}}}\] \[\Rightarrow \]\[{{t}_{2}}=\frac{{{N}_{2}}}{{{N}_{1}}}\times {{t}_{1}}=\frac{9}{10}\times 16=14.4\min \]
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