A) \[97.5\text{ }\Omega \]
B) \[105\text{ }\Omega \]
C) \[150\text{ }\Omega \]
D) \[110\text{ }\Omega \]
Correct Answer: A
Solution :
(a) \[\text{97}\text{.5 }\Omega \] For null deflection in meter bridge, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{l}{(100-l)}\] \[\therefore \frac{65}{R}=\frac{40}{100-40}=\frac{40}{60}\] \[R=65\times \frac{60}{40}=97.5\,\Omega \]
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