A) \[\text{k(}{{\text{l}}_{\text{2}}}-{{\text{l}}_{\text{1}}}\text{)}\,\text{and}\,\text{k}{{\text{l}}_{\text{2}}}\]
B) \[\text{k}{{\text{l}}_{\text{2}}}\,\text{and}\,\text{k}\,\text{(}{{\text{l}}_{\text{2}}}-{{\text{l}}_{\text{1}}})\]
C) \[\text{k(}{{\text{l}}_{\text{2}}}-{{\text{l}}_{\text{1}}})\,\,\text{and}\,\text{k}\,{{\text{l}}_{\text{1}}}\]
D) \[\text{k}{{\text{l}}_{\text{2}}}\,\text{and}\,\text{k}{{\text{l}}_{2}}\]
Correct Answer: B
Solution :
The balancing length for R (when 1, 2 are connected) be is \[{{\text{l}}_{\text{1}}}\]and balancing length for R + X (when 1, 3 is connected is\[{{\text{l}}_{2}}\]) Then \[iR=k{{l}_{1}}\] and \[i(R+X)=k{{l}_{2}}\] Given \[i=1A\] \[\therefore \] \[R=k{{l}_{1}}\] ..(i) \[R+X=k{{l}_{2}}\] ...(ii) Subtracting Eq. (i) from Eq. (ii), we get \[X=k({{l}_{2}}-{{l}_{1}})\]You need to login to perform this action.
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