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.You need to login to perform this action.
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