NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    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 \[{{\text{l}}_{\text{1}}}\text{cm}\] and \[{{\text{l}}_{2}}\text{cm}\] respectively. The magnitudes, of the resistors R and X, in ohm, are then, equal, respectively, to

    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.
You will be redirected in 3 sec spinner