A) 20
B) 10
C) 60
D) 40
Correct Answer: A
Solution :
The balance condition of a metre bridge experiment \[\frac{R}{S}=\frac{{{l}_{1}}}{(100-{{l}_{1}})}\] Here : \[R={{R}_{1}},S={{R}_{2}}.\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{l}_{1}}}{(100-{{l}_{1}})}\] Ist case:\[\frac{{{R}_{1}}+10}{{{R}_{2}}}=\frac{50}{50}\] \[\Rightarrow \] \[{{R}_{1}}+10={{R}_{2}}\] IInd case;\[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{40}{60}\] \[\Rightarrow \] \[{{R}_{2}}=\frac{60}{40}{{R}_{1}}\] ?(2) So, equations (1) and (2) give \[{{R}_{1}}+10=\frac{60}{40}{{R}_{1}}\] \[\Rightarrow \] \[\frac{60}{40}{{R}_{1}}-{{R}_{1}}=10\] \[\Rightarrow \] \[\frac{20}{40}{{R}_{1}}=10\] \[\Rightarrow \] \[{{R}_{1}}=\frac{10\times 40}{20}\] \[\therefore \] \[{{R}_{1}}=20\Omega \]You need to login to perform this action.
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