A) 2
B) 3
C) 1
D) 4
Correct Answer: B
Solution :
According to analysis \[d\propto {{\rho }^{x}}{{s}^{y}}{{f}^{z}}\Rightarrow \,d=k{{\rho }^{x}}{{s}^{y}}{{f}^{z}}\] equating dimensions, \[[L]\,={{[M{{L}^{-3}}]}^{x}}\,{{\left[ \frac{M{{L}^{2}}{{T}^{-3}}}{{{L}^{2}}} \right]}^{y}}\,{{[{{T}^{-1}}]}^{z}}\] \[[L]\,=[{{M}^{x+y}}{{L}^{-3x}}{{T}^{-3y-z}}]\] Equating power on both side \[x+y\,=0,\,\,1=-3x,\,\,0=-3y-z\] \[x=\frac{-1}{3}\,,\,\,y=\frac{1}{3},\,z=-1\Rightarrow \,d=k\,{{\rho }^{-1/3}}\,{{s}^{1/3}}\,{{f}^{-1}}\] So, according to question \[\frac{1}{n}=\frac{1}{3}\Rightarrow \,n=3\]You need to login to perform this action.
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