A) 7.43
B) 10.56
C) 14.38
D) 16.48
Correct Answer: A
Solution :
The resistance \[(R)\]wire of length \[l\]and area\[A\] is \[R=\rho \frac{l}{A}\] Given, \[l=5\,cm=5\times {{10}^{-2}}m,\,\rho =3.5\times {{10}^{-5}}\Omega -m\] \[A=\pi (r_{1}^{2}-r_{2}^{2})=\pi (1\times {{10}^{-4}}-0.25\times {{10}^{-4}})\] \[\Rightarrow \]\[A=0.75\pi \times {{10}^{-4}}m\] \[\Rightarrow \]\[A=7.5\pi \times {{10}^{-5}}m\] \[\therefore \] \[R=\frac{3.5\times {{10}^{-5}}\times 5\times {{10}^{-2}}}{7.5\pi \times {{10}^{-5}}}\] \[=7.43\times {{10}^{-3}}\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
