A) 3
B) 6
C) 9
D) 12
Correct Answer: C
Solution :
Let S be larger and R be smaller resistance connected in two gaps of meter bridge. \ \[S=\left( \frac{100-l}{l} \right)R=\frac{100-20}{20}R=4R\] .....(i) When \[15\Omega \] resistance is added to resistance R, then \[S=\left( \frac{100-40}{40} \right)(R+15)=\frac{6}{4}(R+15)\] .... (ii) From equations (i) and (ii) \[R=9\Omega \]You need to login to perform this action.
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