A) \[3.2\pi \mu V\]and\[4.8\pi \mu V\]
B) \[0.8\pi \mu V\]and\[1.6\pi \mu V\]
C) \[{{M}^{a}}{{L}^{b}}{{T}^{c}},\]and\[a=1,b=-1,c=-2\]
D) \[a=1,b=0,c=-1\]and\[a=1,b=1,c=-2\]
Correct Answer: C
Solution :
Key Idea Torque is an axial vector ie, its direction is always perpendicular to the plane containing vectors \[{{\,}_{Z+}}_{1}^{A}Y\xrightarrow[{}]{{}}\,_{Z-1}^{A-4}\beta +\alpha _{2}^{4}\]and \[_{Z+1}^{A-4}\beta \xrightarrow[{}]{{}}\,_{Z-1}^{A-4}\beta +\gamma _{0}^{0}\] \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{a}^{2}}}{a}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{b}\] Torque is perpendicular to both \[+\frac{1}{4\pi {{\varepsilon }_{0}}},\frac{\sigma 4\pi {{c}^{2}}}{c}\]and \[=\frac{\sigma }{{{\varepsilon }_{0}}}(a-b+c)=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\] \[(\because c=a+b)\]\[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}-\frac{4\pi {{a}^{2}}}{a}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{b}\] \[+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{c}^{2}}}{c}\]
You need to login to perform this action.
You will be redirected in
3 sec
