A) 1.25 \[{{f}_{G}}<{{f}_{R}}<{{f}_{V}}\]
B) 2.5 \[{{f}_{G}}<{{f}_{V}}<{{f}_{R}}\]
C) 50 \[\mu \]
D) 10.0\[6.6\times {{10}^{-34}}\]
Correct Answer: A
Solution :
Key Idea: Potential difference across galvanometer and shunt is same. For G being resistance of galvanometer and \[\,\omega \] the current across it the current across S is \[I=\frac{1}{2}M{{R}^{2}}\]. Since, G and S are in parallel, potential across them is same \[\therefore \] \[J=\frac{1}{2}M{{R}^{2}}\omega \] \[I={{I}_{4}}+m{{R}^{2}}\] Given, \[=\frac{1}{4}m{{R}^{2}}+m{{R}^{2}}=\frac{5}{4}m{{R}^{2}}\] \[g=g\left( 1-\frac{h}{R} \right)\] \[\Rightarrow \] \[w=w\left( 1-\frac{h}{R} \right)\]You need to login to perform this action.
You will be redirected in
3 sec