A) 3
B) 6
C) 9
D) 12
E) None of these
Correct Answer: C
Solution :
Let S be the large and R be the smaller resistance. From formula for meter bridge \[S=\left( \frac{100-l}{l} \right)R\] \[=\left( \frac{100-20}{20} \right)R=4R\] Again, \[S=\left( \frac{100-l}{l} \right)(R+15)\] \[=\left( \frac{100-40}{40} \right)(R+15)\] \[=\frac{3}{2}(R+15)\] \[\therefore \] \[4R=\frac{3}{2}(R+15)\] or \[8R-3R=45\] or \[5R=45\] \[\therefore \] \[R=9\,\Omega \]You need to login to perform this action.
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