A) 3
B) 6
C) 9
D) 12
Correct Answer: C
Solution :
Let S be the large and R be the smaller resistance. From formula for metre bridge \[S=\left( \frac{100-l}{l} \right)R\] \[=\frac{100-20}{20}R=4R\] Again, \[S=\left( \frac{100-l}{100} \right)(R+15)\] \[=\frac{100-40}{40}(R+15)\] \[=\frac{3}{2}(R+15)\] \[\therefore \] \[4R=\frac{3}{2}(R+15)\] \[\frac{8R}{3}-R=15\Rightarrow \frac{5R}{3}=15\] \[R=9\Omega \]You need to login to perform this action.
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