question_answer

A cylindrical solid of length L and radius a is having varying resistivity given by \[\rho  ={{\rho }_{0}}x\], where \[{{\rho }_{0}}\] is a positive constant and x is measured from left end of solid. The cell shown in the figure is having emf V and negligible internal resistance. The electric field as a function of x is best described by  

A) \[\frac{2V}{{{L}^{2}}}x\]       

B)        \[\frac{2V}{{{\rho }_{0}}{{L}^{2}}}x\]

C) \[\frac{V}{{{L}^{2}}}x\]                    

D)        None of these

Correct Answer: A

Solution :

 Consider an element part of solid at a distance x from left end of width dx.  Resistance of this elemental part is, \[dR=\frac{\rho dx}{\pi {{a}^{2}}}=\frac{{{\rho }_{0}}xdx}{\pi {{a}^{2}}}\] \[R=\,\,\int{dR}=\,\,\int\limits_{0}^{L}{\frac{{{\rho }_{0}}xdx}{\pi {{a}^{2}}}}=\frac{{{\rho }_{0}}{{L}_{2}}}{2\pi {{a}^{2}}}\] Current through cylinder is, \[I=\frac{V}{R}=\frac{V\times 2\pi {{a}^{2}}}{{{\rho }_{0}}{{L}^{2}}}\] Potential drop across element is, \[\operatorname{dV} = I dR = \frac{2\,V}{{{L}^{2}}} x dx\] \[E(x)=\frac{dV}{dx}=\frac{2V}{{{L}^{2}}}x\]