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  • A portion of length is cut out of a conical solid wire. The two ends of this portion have circular cross section of radii <im

A portion of length l is cut out of a conical solid wire. The two ends of this portion have circular cross section of radii \mathrm{r_{1}} and \mathrm{r_{2}\left(r_{2}>r_{1}\right)}.It is connected length wise to a circuit  and a current \mathrm{i} is flowing in it. resistivity of the material is \mathrm{\rho}. Then the resistance of the given portion is given as:

Option: 1

\mathrm{\frac{\rho L}{\pi r_{1} r_{2}}}


Option: 2

\mathrm{\frac{\rho L}{\pi\left(r_{2}{ }^{2}-r_{1}{ }^{2}\right)}}


Option: 3

\mathrm{\rho L \frac{r_{1}+r_{2}}{r_{1}{ }^{3}+r_{2}{ }^{3}}}


Option: 4

\mathrm{\frac{\rho}{\pi\left(r_{2}-r_{1}\right)}}


Answers (1)

best_answer

From figure.
\mathrm{\tan \theta=\frac{r_{2}-r_{1}}{L}}
radius at a distance \mathrm{x} from left end is \mathrm{r=r_{1}+x \tan \theta=r_{1}+x \frac{\left(r_{2}-r_{1}\right)}{L}}
\mathrm{\Rightarrow r =\frac{r_{1} L+x\left(r_{2}-r_{1}\right)}{L} }.
\mathrm{\Rightarrow A =\pi r^{2}=\frac{\pi}{L^{2}} \cdot\left[r_{1} L+\left(r_{2}-r_{1}\right) \cdot x\right]^{2} }
\mathrm{d R =\frac{\rho d x}{\pi r^{2}}=\frac{\rho d x \cdot L^{2}}{\pi\left[r_{1} L+\left(r_{2}-r_{1}\right) \cdot x\right]^{2}} }

\mathrm{\Rightarrow R =\int d R=\frac{\rho L^{2}}{\pi} \cdot \int_{0}^{L} \frac{d x}{\left[r_{1} L+\left(r_{2}-r_{1}\right) \cdot x\right]^{2}} }
          \mathrm{ =\left.\frac{\rho L^{2}}{\pi}\left[r_{1} L+\left(r_{2}-r_{1}\right) \cdot x\right]^{-1}\right|_{0} ^{L} \cdot\left(\frac{1}{r_{2}-r_{1}}\right) .}
          \mathrm{ =\frac{-\rho L}{\pi\left(r_{2}-r_{1}\right)}\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)=\frac{\rho L}{\pi r_{1} r_{2}}}

Posted by

Deependra Verma

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