A copper - -1-6- 10-8- m wire and an Aluminium - -2-5- 10-8- m wire have equal cross sectional area and equal resistances- The ratio of their lengths is

The correct option is C 25:16 We know, R= ρ× L A Here, A_Cu = A_Al R_Cu = R_Al Hence, #160; #160;ρ^_Cu #160;×L^_Cu = #160; #160;ρ ...

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Standard X

Physics

Factors Affecting Resistance

A copper ρ ...

Question

A copper $(ρ = 1.6 \times 10^{- 8} Ω m)$ wire and an Aluminium $(ρ = 2.5 \times 10^{- 8} Ω m)$ wire have equal cross sectional area and equal resistances. The ratio of their lengths is

5 : 4

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4 : 5

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16 : 25

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25 : 16

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Solution

The correct option is C $25 : 16$
We know,
$R = \frac{ρ \times L}{A}$
Here,
$A_{C u} = A_{A l}$
$R_{C u} = R_{A l}$
Hence,
$ρ_{C u} \times L_{C u} = ρ_{A l} \times L_{A l}$
or, $\frac{L_{C u}}{L_{A l}} = \frac{ρ_{A l}}{ρ_{C u}} = \frac{2.5 \times 10^{- 8}}{1.6 \times 10^{- 8}} = \frac{25}{16}$

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A copper (ρ=1.6×10−8Ωm) wire and an Aluminium (ρ=2.5×10−8Ωm) wire have equal cross sectional area and equal resistances. The ratio of their lengths is

The correct option is C 25:16We know,R=ρ×LAHere,ACu=AAlRCu=RAlHence,ρCu×LCu=ρAl×LAlor, LCuLAl=ρAlρCu=2.5×10−81.6×10−8=2516

A copper $(ρ = 1.6 \times 10^{- 8} Ω m)$ wire and an Aluminium $(ρ = 2.5 \times 10^{- 8} Ω m)$ wire have equal cross sectional area and equal resistances. The ratio of their lengths is

The correct option is C $25 : 16$
We know,
$R = \frac{ρ \times L}{A}$
Here,
$A_{C u} = A_{A l}$
$R_{C u} = R_{A l}$
Hence,
$ρ_{C u} \times L_{C u} = ρ_{A l} \times L_{A l}$
or, $\frac{L_{C u}}{L_{A l}} = \frac{ρ_{A l}}{ρ_{C u}} = \frac{2.5 \times 10^{- 8}}{1.6 \times 10^{- 8}} = \frac{25}{16}$