A) 7.50\[\Omega \]
B) 9\[\Omega \]
C) 9.5\[\Omega \]
D) 10\[\Omega \]
Correct Answer: B
Solution :
The resistance of a wire of length\[l,\]area of cross-section A, specific resistance\[\rho \]is given by \[R=\rho \frac{l}{A}\] when r is radius of wire, then \[A=\pi {{r}^{2}}\] \[\therefore \] \[R=\rho \frac{l}{\pi {{r}^{2}}}\] \[\Rightarrow \] \[\rho =\frac{R\pi {{r}^{2}}}{l}\] Given, \[R=10\text{ }\Omega ,\text{ }r=9\text{ }mm=9\times {{10}^{-3}}m\] \[\rho =\frac{10\times \pi \times {{(9\times {{10}^{-3}})}^{2}}}{l}\] When \[r=3\text{ }mm=3\times {{10}^{-3}}m,\] \[R=\frac{\rho l}{\pi {{(3\times {{10}^{-3}})}^{2}}}=\frac{10\times \pi \times {{(9\times {{10}^{-3}})}^{2}}l}{l\pi {{(3\times {{10}^{-3}})}^{2}}}\] \[=90\,\,\Omega \] New resistance has 10 such wires in parallel, hence resultant resistance is \[\frac{1}{R}=\frac{10}{R}=\frac{10}{90}=\frac{1}{9}\] \[\therefore \] \[R=9\,\Omega \] Note: - (Given that density of \[Cu=9\text{ }g/c{{m}^{3}}\] and atomic weight of \[Cu=63\] and that one free electron is contributed by each atom)
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