A) \[40\,\Omega .\]
B) \[80\,\Omega .\]
C) \[60\,\Omega .\]
D) \[180\,\Omega .\]
Correct Answer: D
Solution :
Key Idea: When length is increased volume remians same and specific resistance constant. The resistance of a wire of length I is given by \[R=\rho \frac{l}{A}\] ?(i) When wire is stretched, volume remains constant, hence, \[V=Al=\text{constant}\] ?(ii) Multiply and divide Eq. (i) by\[l,\] we get \[R=\frac{\rho {{l}^{2}}}{Al}\] \[\therefore \] \[\frac{{{R}_{2}}}{{{R}_{1}}}={{\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)}^{2}}\] Given, \[{{l}_{2}}=3{{l}_{1}}\] \[\frac{R{{ & }_{2}}}{20}={{\left( \frac{3{{l}_{1}}}{{{l}_{1}}} \right)}^{2}}=9\] \[\Rightarrow \] \[{{R}_{2}}=20\times 9=180\,\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
