A) \[\frac{2}{15}l\]
B) \[\frac{3}{15}l\]
C) \[\frac{3}{10}l\]
D) \[\frac{4}{10}l\]
Correct Answer: C
Solution :
[c] : In first case, potential gradient, \[K=\frac{{{\varepsilon }_{0}}}{l}\]where \[{{\varepsilon }_{0}}\] is the emf of the battery in potentiometer circuit. As per question\[\varepsilon =\frac{Kl}{5}=\frac{{{\varepsilon }_{0}}}{l}\times \frac{l}{5}=\frac{{{\varepsilon }_{0}}}{5}\] In second case, length of potentiometer wire \[=l+\frac{l}{2}=\frac{3l}{2}\] Potential gradient, \[K'=\frac{{{\varepsilon }_{0}}}{3l/2}=\frac{2{{\varepsilon }_{0}}}{3l}\] If \[l'\] is the new balancing length, then \[\varepsilon =\frac{{{\varepsilon }_{0}}}{5}=\frac{2{{\varepsilon }_{0}}}{3l}\times l'\]or\[l'=\frac{3}{10}l\]
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