A) 1
B) 3/2
C) 2/3
D) 2
Correct Answer: D
Solution :
[d] Using \[\frac{{{\mu }_{2}}}{v}-\frac{{{\mu }_{1}}}{u}=\frac{{{\mu }_{2}}-{{\mu }_{1}}}{R}\] \[\frac{\mu }{2R}-\frac{1}{\infty }=\frac{\mu -1}{R}\Rightarrow \frac{\mu }{2}=\frac{\mu -1}{1}\] \[\Rightarrow \] \[\mu =2\]
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