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 meter bridge \[S=\left( \frac{100-l}{l} \right)R\] \[=\left( \frac{100-20}{20} \right)R=4R\] Again, \[S=\left( \frac{100-l}{l} \right)(R+15)\] \[=\left( \frac{100-40}{40} \right)(R+15)\] \[4R=\frac{3}{2}(R+15)\] \[8R-3R=45\] \[5R=45\] \[R=9\Omega \].You need to login to perform this action.
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