A) \[270\,\Omega \]
B) \[90\,\Omega \]
C) \[45\,\Omega \]
D) \[7.5\,\Omega \]
Correct Answer: D
Solution :
The resistance of a wire of length \[l\] and area of cross-section \[A\] is given by \[R=\rho \frac{l}{A}\] where \[\rho \] is 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, \[{{r}_{1}}=9mm,\,\,{{R}_{1}}=5\Omega ,\,\,{{r}_{2}}=3\,\,mm\] \[\therefore \] \[\frac{5}{{{R}_{2}}}=\frac{{{3}^{2}}}{{{9}^{2}}}\] Equivalent resistance of 6 wires each of resistance \[{{R}_{2}}\] connected in parallel is \[R=\frac{{{R}_{2}}}{6}=\frac{45}{6}=7.5\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
