A) 0.25 \[\Omega \]
B) 0.95\[\Omega \]
C) 0.5 \[\Omega \]
D) 0.75 \[\Omega \]
Correct Answer: C
Solution :
Given, e = 2V and \[l\]= 4m Potential drop per unit length \[\phi =\frac{e}{l}=\frac{2}{4}=0.5V/m\] For the first case, Þ \[e'=\phi {{l}_{1}}\,\,\,\,\,\,\,\,\,....(i)\] (\[e'\]® emf of the given cell) For the second case, \[V=\phi {{l}_{2}}\,\,\,\,\,\,\,\,\,....(ii)\] From Eqs. (i) and (ii), \[e'/V={{l}_{1}}/{{l}_{2}}\] \[e'=l(r+R)\] and \[V=lR\] for the second case So, \[r=R\left( \frac{{{l}_{1}}}{{{l}_{2}}}-1 \right)=9.5\left( \frac{3}{2.85}-1 \right)=9.5(1.05-1)\] \[=9.5\times 0.05=0.475\simeq 0.5\Omega \]You need to login to perform this action.
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