A) 0.04 A
B) 0.03 A
C) 0.02 A
D) 0.01 A
Correct Answer: C
Solution :
Key Idea: Potential difference acre galvanometer and shunt is same. Let G be the resistance of galvanometer and \[({{k}_{1}}+{{k}_{2}})F=-{{k}_{1}}{{k}_{2}}y\] the current which on passing through the galvanometer produces full scale deflection. Since G and 5 are in parallel, the potential difference across them will be \[\Rightarrow \] \[F=-\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}y\] Hence, \[\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\] ??(1) \[n=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\] ??.(2) Equating Eqs. (1) and (2), we get \[\frac{1}{k}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}\] \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\] \[n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\] \[=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\] \[=\frac{Force}{Length}=\frac{[ML{{T}^{-2}}]}{[L]}=[M{{T}^{-2}}]\] [from Eq. (1)]You need to login to perform this action.
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