A) 17.7 min
B) 14.4 min
C) 20.9mm
D) 13.9 min
Correct Answer: B
Solution :
If N be the initial number of turns in the coil and r be the radius of coil, then its resistance, \[R=\rho .\frac{L}{A}=\rho .\frac{N\times 2\pi r}{A}\] \[\frac{{{V}^{2}}tA}{4.2\rho N\,2\pi r}=Q=msd\theta \] \[\frac{t}{N}=\frac{msd\theta \times 4.2\rho 2\pi r}{{{V}^{2}}A}\]= constant \[\frac{{{t}_{1}}/{{N}_{1}}}{{{t}_{2}}/{{N}_{2}}}=1\] \[{{t}_{2}}=\frac{{{N}_{2}}}{{{N}_{1}}}\times {{t}_{1}}=\left( \frac{\frac{9}{10}{{N}_{1}}}{{{N}_{1}}} \right)\times {{t}_{1}}=\frac{9}{10}\times 16\min \] \[=14.4\min \]
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