| A potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire, is k volt/cm and the ammeter, present in the circuit, reads 1.0 A when two way key is switched off. The balance points, when the key between the terminals (i) 1 and 2 (ii) 1 and 3, is plugged in, are found to be at lengths \[{{l}_{1}}cm\] and \[{{l}_{2}}cm\] respectively. The magnitudes, of the resistors R and X, in ohm, are then, equal, respectively, to [AIPMT (S) 2010] |
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A) \[k({{l}_{2}}-{{l}_{1}})\] and \[k{{l}_{2}}\]
B) \[k{{l}_{2}}\] and \[k\,({{l}_{2}}-{{l}_{1}})\]
C) \[k({{l}_{2}}-{{l}_{1}})\] and \[k\,{{l}_{1}}\]
D) \[k{{l}_{2}}\] and \[k{{l}_{2}}\]
Correct Answer: B
Solution :
| [b] The balancing length for R (when 1, 2 are connected) be is \[{{l}_{1}}\] and balancing length for \[R+X\] (when 1, 3 is connected is \[{{l}_{2}}\]) |
| 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 |
| \[X=k({{l}_{2}}-{{l}_{1}})\] |
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