2. Solved Papers
  3. JAMIA MILLIA ISLAMIA

JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2007

  • question_answer
        In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance\[X\]is balanced against another resistance Y. If\[X<Y,\]then where will be the new position of the null point from the same end, if one decides to balance a resistance of\[4X\]against\[Y\]?

    A)  50 cm                                  

    B)  80 cm

    C)  40cm                                   

    D)  70cm

    Correct Answer: A

    Solution :

                    Meter bridge is an arrangement which works on Wheatstones principle, so the balancing condition is \[\frac{R}{S}=\frac{{{l}_{1}}}{{{l}_{2}}}\] Where \[{{l}_{2}}=100-{{l}_{1}}\] 1st case: \[R=X,S=Y,{{l}_{1}}=20\,cm,\] \[{{l}_{2}}=100-20=80cm\] \[\therefore \]  \[\frac{X}{Y}=\frac{20}{80}\]                       ?.(i) IInd case: Let the position of null point is obtained at a distance\[l\]from same end. \[\therefore \] \[R=4X,S=Y,{{l}_{1}}=l,{{l}_{2}}=100-l\] So, from Eq. (i)                 \[\frac{4X}{Y}=\frac{l}{100-l}\] \[\Rightarrow \]               \[\frac{X}{Y}=\frac{l}{4(400-l)}\]                                                ?(ii) Therefore, from Eqs. (i) and (ii)                                 \[\frac{l}{4(100-l)}=\frac{20}{80}\] \[\Rightarrow \]               \[\frac{l}{4(100-l)}=\frac{1}{4}\] \[\Rightarrow \]               \[l=100-l\] \[\Rightarrow \]               \[2l=100\] Hence, \[l=50\text{ }cm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner