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3 years ago

15

A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be

to have a resistance of 0.001 Ω?

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, $A=1\ cm^2=0.0001\ m^2$

Resistivity of gold wire, $\rho=2.8\times 10^{-8}\ \Omega-m$

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

$R=\rho \dfrac{l}{A}$

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}$

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

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${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

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${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

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${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

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$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

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☯ As we know that,

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

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$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

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$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

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$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

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$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

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$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

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$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

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$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

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$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

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$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

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$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

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☯ For left penetration (s₂)

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$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

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$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

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$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

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$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

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$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

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