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Resistance

A copper tube...

Question

A copper tube is of internal radius 4 mm and outer radius 5 mm. Its resistance is R $_{1}$ . The tube is filled with a suitable copper wire. The resistance of the arrangement is R $_{2}$ .Then R $_{2}$ /R $_{1}$ is

25/9

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1/2

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9/25

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Solution

The correct option is D 9/25
$R_{1} = \frac{ρ L}{A} = \frac{ρ L}{π (r_{2}^{2} - r_{1}^{2})} = \frac{ρ L}{π (25 - 16)} = \frac{ρ L}{9 π}$
$R_{2} = \frac{ρ L}{π r_{2}^{2}} = \frac{ρ L}{π 25}$
$\frac{R_{2}}{R_{1}} = \frac{9}{25}$

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A copper tube is of internal radius 4 mm and outer radius 5 mm. Its resistance is R1. The tube is filled with a suitable copper wire. The resistance of the arrangement is R2.Then R2/R1 is

The correct option is D 9/25R1=ρLA=ρLπ(r22−r21)=ρLπ(25−16)=ρL9πR2=ρLπr22=ρLπ25R2R1=925

A copper tube is of internal radius 4 mm and outer radius 5 mm. Its resistance is R $_{1}$ . The tube is filled with a suitable copper wire. The resistance of the arrangement is R $_{2}$ .Then R $_{2}$ /R $_{1}$ is

The correct option is D 9/25
$R_{1} = \frac{ρ L}{A} = \frac{ρ L}{π (r_{2}^{2} - r_{1}^{2})} = \frac{ρ L}{π (25 - 16)} = \frac{ρ L}{9 π}$
$R_{2} = \frac{ρ L}{π r_{2}^{2}} = \frac{ρ L}{π 25}$
$\frac{R_{2}}{R_{1}} = \frac{9}{25}$