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, \[\operatorname{dV} = I dR = \frac{2\,V}{{{L}^{2}}} x dx\] \[E(x)=\frac{dV}{dx}=\frac{2V}{{{L}^{2}}}x\]You need to login to perform this action.
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