Space between two concentric spheres of radii r1 and r2- such that r1 - r2- is filled with a material of resistivity - Find the resistance between inner and outer surface of the material-

The correct option is B r_2 r_1r_1r_2ρ4π Given, #160;Radii of the concentric sphere s, r_1 and r_2 #160; #160; #160; #160; #16 ...

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Standard XII

Physics

Conductivity and Resistivity

Space between...

Question

Space between two concentric spheres of radii $r_{1}$ and $r_{2}$ , such that $r_{1} < r_{2}$ , is filled with a material of resistivity $ρ$ . Find the resistance between inner and outer surface of the material.

\frac{r_{1}}{r_{2}} \frac{ρ}{2}

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\frac{r_{2} - r_{1}}{r_{1} r_{2}} \frac{ρ}{4 π}

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\frac{r_{1} r_{2}}{r_{2} - r_{1}} \frac{ρ}{4 π}

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None of these

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Solution

The correct option is B $\frac{r_{2} - r_{1}}{r_{1} r_{2}} \frac{ρ}{4 π}$
Given,
Radii of the concentric sphere s, $r_{1}$ and $r_{2}$ [ $r_{1} < r_{2}$ ]
Resistivity of the material $= ρ$
As we know,
Resistance can be given by the formula,
$R = ρ \frac{l}{A}$
In this case,
length of the resistor here it's radius, $l = r$
and Area of the spherical resistor, $A = 4 π r^{2}$
$∴ R = ρ \frac{d r}{4 π r^{2}}$
[where $r$ is any radius and $d r$ is small element]
Total resistance,
$R = \frac{ρ}{4 π} \int_{r_{1}}^{r_{2}} \frac{d r}{r^{2}}$
$\Rightarrow R = \frac{ρ}{4 π} {[- \frac{1}{r}]}_{r_{1}}^{r_{2}}$
$\Rightarrow R = \frac{ρ}{4 π} [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$

$\Rightarrow R = [\frac{r_{2} - r_{1}}{r_{1} r_{2}}] \frac{ρ}{4 π}$
Hence, the correct option is $(B)$

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Space between two concentric spheres of radii r1 and r2, such that r1<r2, is filled with a material of resistivity ρ. Find the resistance between inner and outer surface of the material.

Space between two concentric spheres of radii $r_{1}$ and $r_{2}$ , such that $r_{1} < r_{2}$ , is filled with a material of resistivity $ρ$ . Find the resistance between inner and outer surface of the material.