A) \[270\,\Omega \]
B) \[9.0\,\Omega \]
C) \[45\,\Omega \]
D) \[7.5\,\Omega \]
Correct Answer: B
Solution :
The relation between resistance of a wire of length \[l\] and area of cross-section A s given by \[R=\rho \frac{l}{A}\] Where \[\rho \] is the specific resistance Also, \[A=\pi {{r}^{2}},\] r being radius of wire \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{r_{2}^{2}}{r_{1}^{2}}\] Given, \[{{r}_{1}}=9mm,\,{{R}_{1}}=5\Omega ,\,{{r}_{2}}=3mm.\] \[\therefore \] \[\frac{5}{{{R}_{2}}}=\frac{{{3}^{2}}}{{{9}^{2}}}\] \[{{R}_{2}}=5\times 9\] \[=45\Omega \] Equivalent resistance of 6 wires each of resistances \[{{R}_{2}}\]. Connected in parallel is \[R'=\frac{{{R}_{2}}}{6}=\frac{45}{6}=7.5\Omega \] Since the value of the equivalent resistance of the resistances connecting in parallel is less than the value of the smallest resistance among those resistances.
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