Question

A wire having a mass of 0.45 gm posses a resistance of 0.014 if the restivity of material of wire is 1.78×10^-7 ohm m calculate its length and radius
Given, that the density bif material of the wire is 8.93 × 10^3 kg m^-3

Please give me correct answer ☺️

Answer 1

$\boxed{ Answer = 2.84 mm }$

Explanation:

Here, m = 0.45 kg , R = 0.014 $\Omega,$

$\rho = 1.78 \times {10}^{ - 8} ohm \: m \: \\ and \: density \: of \: the \: material \: of \: wire \:d = 8.93 \times {10}^{3}kg \: m^{ - 3} \\ let \: l \: be \: the \: length \: and \: r \: the \: radius \: of \: wire \:$

Now, R = $\rho \frac{l}{A} \: = \rho\frac{l}{\pi {r}^{2}} \: \: \: \: \: .....(i) \\ Also ,\: the \: mass \: of \: wire, m= volume \times density \\ or \: \: \: m = \pi {r}^{2}l \times d \: \: \: \: \: \: \: \: \: \: .....(ii) \\ Multiplying \: equation \: (i) \: and \: (ii), \: we \: have \\ R \times M = \rho \frac{l}{\pi {r}^{2}} \times \pi{r}^{2}l \times d \\ or \: \: l = \sqrt{ \frac{R \: m}{ \rho \: d} } = \sqrt{ \frac{0.014 \times 0.45}{1.78 \times {10}^{ - 7} \times 8.93 \times 10^{3} } } = 1.99 \: m \\ from \: equation \: (ii) \: we \: have \: \\ r = \sqrt{ \frac{m}{\pi \: ld} } = \sqrt{ \frac{0.45}{\pi \times 1.99 \times 8.93 \times {10}^{3} } } = 2.84 \times {10}^{ - 3} m = 2.84 \:mm$

Answer 2

Answer:

$\large \bold\red{ Length= 0.2 \: m}\\\\\large \bold\red{ Radius = 2.8 \times {10}^{ - 4}\:m }$

Explanation:

Given,

A wire having,

Mass,

• $\bold{m = 0.45\: gm = 4.5\times{10}^{-4}\:Kg}$

Resistance,

• $\bold{R= 0.014\:\Omega = 1.4\times{10}^{-2}\:\Omega}$

Density,

• $\bold{d=8.93\times{10}^{3}\:Kg{m}^{-3}}$

To find :

• Radius = Let denoted by (r)
• Length = Let demoted by (l)
• Area of cross section = Let denoted by (a)
• Volume = Let denoted by (v)

Now,

We know that,

$\large \boxed{\bold\purple{v = \pi {r}^{2} l}}$

And,

$\large \boxed{\bold\purple{v = \frac{m}{d} }}$

Therefore,

We get,

$= > \frac{m}{d} = \pi {r}^{2} l$

Substituting the respective values,

We get,

$= > \frac{4.5 \times {10}^{ - 4} }{8.93 \times {10}^{3} } = \frac{22}{7} \times {r}^{2} l \\ \\ = > {r}^{2} l = 0.5 \times {10}^{ - 4 - 3} \times \frac{7}{22} \\ \\ = > {r}^{2} l = 5 \times {10}^{ - 8} \times 0.32 \\ \\ = > {r}^{2} l = 1.6 \times {10}^{ - 8}$

Also,

We know that,

$\large \boxed{\bold\purple{R = \rho \frac{l}{a} }}$

Therefore,

Substituting the values,

We get,

$= > 1.4 \times {10}^{ - 2} = 1.78 \times {10}^{ - 7} \times \frac{l}{\pi {r}^{2} } \\ \\ = > \frac{ {l}^{2} }{ {r}^{2}l } = \cancel\frac{14 \times 22}{178 \times 7} \times \frac{ {10}^{ - 3} }{ {10}^{ - 9} } \\ \\ = > \frac{ {l}^{2} }{ {r}^{2}l } = \frac{22}{89} \times {10}^{6}$

Further,

We get,

$= > \frac{ {l}^{2} }{1.6 \times {10}^{ - 8} } = 2.5 \times {10}^{6} \\ \\ = > {l}^{2} = 2.5 \times 1.6 \times {10}^{6 - 8} \\ \\ = > {l}^{2} = 25 \times 16 \times {10}^{ - 2 - 2} \\ \\ = > {l}^{2} = {(5 \times 4 \times {10}^{ - 2}) }^{2} \\ \\ = > l = 20 \times {10}^{ - 2} \\ \\ = > \large \bold{ l = 0.2 \: m}$

Therefore,

Substituting the values ,

We get,

$= > {r}^{2} = \frac{1.6 \times {10}^{ - 8} }{0.2} \\ \\ = > {r}^{2} = 8 \times {10}^{ - 8} \\ \\ = > r = 2 \sqrt{2} \times {10}^{ - 4} \\ \\ = > r = 2 \times 1.4 \times {10}^{ - 4} \\ \\ = > \large \bold{ r = 2.8 \times {10}^{ - 4}\:m }$