Question

Answer 2nd question pls

Answer 1

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♣️ GiveN:

• Diameter of the wire = 0.5 mm
• Resistivity of the wire = 1.6 ×10^-8 Ω m
• Resistance = 12 Ω

♣️ To FinD:

• Length of the wire
• New resistance when diameter is halved but length is same.

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The above question is based on a numerical based on resistance and resistivity. So, it's important to know about these first of all.

• Resistance:

It is the opposition to the flow of current in the circuit. Higher resistance means lower conductivity. Resistance is calculated by the ohm's law i.e.

$\large{ \boxed{ \sf{ \purple{V = IR}}}}$

✍ Note:

Symbols have their usual meanings.

• Resistivity

It is the measure of resistance or resisting power of any conductor i.e. specific material at constant temperature. The relationship between Resistance and Resistivity(ρ):

$\large{ \boxed{ \purple{ \sf{R = \rho \: \frac{L}{A} }}}}$

✍ Note:

Symbols have their usual meanings.

By using these two formuals, let's solve the question.

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i) According to question, we are provided with the resistance, resistivity and diameter.

• ρ = 1.6 × 10^-8 Ω m
• R = 12 Ω
• diameter = 0.5 mm

First we all need to convert the diameter into m, as the resistivity has unit "Ω m", We know, 1 mm = 1 × 10^-3 m

Then, Diameter = 0.5 × 10^-3 m

Radius(r) = 0.25 × 10^-3 m (to find the area of cross section)

Finding the area as we need it later,

$\large{ \sf{ \mapsto \: Area \: of \: crossection = \pi \: {r}^{2}}} \\ \\ \large{ \sf{ \mapsto \: Area \: of \: crossection = \frac{22}{7} \times 0.0625 \times {10}^{ - 6} \: {m}^{2} }} \\ \\ \large{ \sf{ \mapsto \: Area \: of \: crossection \approx \: \boxed{ \red{0.1964 \times {10}^{ - 6} {m}^{2} }}}}$

By using formula, finding the length

$\large{ \sf{ \mapsto \: R = \rho \: \frac{l}{A}}} \\ \\ \large{ \sf{ \mapsto \: L = \frac{RA}{ \rho} }} \\ \\ \large{ \sf{ \mapsto \: L = \frac{12 \Omega \: \times 0.1964 \times {10}^{ - 6} {m}^{2} }{1.6 \times {10}^{ - 8} \Omega \: m }}} \\ \\ \large{ \sf{ \mapsto \: L = \frac{12 \times 0.1964}{1.6} \times {10}^{2} \: m}} \\ \\ \large{ \sf{ \mapsto \: \boxed{ \sf{ \red{L = 1.473\times {10}^{2} \: m}}}}}$

✒Thus, the required length(L) of wire = 1.473 ×10^2 m(approx.)

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b) Let the new resistance be R'

And the new area be A'

New diameter = d/2 , then new radius = r/2

Length and resistivity is constant.

$\large{ \sf{\mapsto{ then \: new \:area(A') = \pi \: ( \frac{r}{2} ) {}^{2} }}} \\ \\ \large{ \sf{\mapsto{ new\: area(A') = \frac{\pi \: {r}^{2} }{4}}}} \\ \\ \large{ \sf{\mapsto{we \: have \: A = \pi \: {r}^{2} }}} \\ \\ \large{ \sf{\mapsto{ then \: new \:area(A') = \frac{A}{4}}}}$

So, Now by using formula,

$\large{ \sf{ \mapsto \: R' = \rho \: \frac{L}{A'}}} \\ \\ \large{ \sf{ \mapsto \: R'= \rho \: \frac{L}{ \frac{A}{4} }}} \\ \\ \large{ \sf{ \mapsto \: R' = 4 \: \rho \: \frac{L}{A} }} \\ \\ \large{ \sf{ \mapsto \: R' = 4R}} \\ \\ \large{ \sf{we \: have \: R = 12 \: \Omega \: }} \\ \\ \large{ \sf{ \mapsto \: R' = 4 \times 12 \Omega}} \\ \\ \large{ \sf{ \mapsto \: \boxed{ \red{ \sf{R' = 48 \Omega}}}}}$

✒ Thus, the new resistance = 48 Ω

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Answer 2

Answer:

a) 1.44 × 10 m²

b) 192 ohm

Explanation:

Given:

Diameter of wire = 0.5 mm

Resistivity of wire = $1.6 \times {10}^{-8}$ ohm m.

To find:

a) Length of wire when resistance is 12 Ohm

b) Resistance of another wire of same length but half diameter.

Solution(a):

1st) We have to find the Cross sectional Area of wire.

2nd) We will use the Resistivity formula.

As we know, 1 mm =$1 \times {10}^{-3}$

Diameter of wire = $0.5 \:\:\: mm$

Radius (r) of wire =

$\dfrac{0.5}{2} \times \:\:\: m {10}^{-3} \implies 0.25 \times {10}^{-3}\:\:\: m$

Cross Sectional Area(A) = πr²

$= 3.14 \times {0.25 \times {10}^{-3}}^{2}\\ \\= 3.14 \times 0.0625 \times {10}^{-6} m^2\\\\= 0.19625 \times {10}^{-6} m^2$

According to question,

$r = p \times \dfrac{l}{a} \\ \\ l = \frac{ra}{p} \\ \\ = \frac{12 \times 0.19625 \times {10}^{ - 6} }{1.6 \times {10}^{ - 8} } \: \: m \\ \\ = 7.5 \times 0.19625 \times {10}^{2} \: \: m\\ \\ = 1.44\times {10}^{2} \: \: m \: (approx)$

The new length be 1.44 × 10²m (approx).

Solution(b):

Length and resistivity will remain unchanged.

Let, The orginal Diameter be d .

The new Diameter be = d/2

So, Radius = d/4

Therefore, Cross sectional Area = πr²

= (A)1/16

Now, Using Formula.

$\frac{R_1}{R_2} = \dfrac{p \frac{l1}{A_1} }{ p\frac{l_2}{A_2} } \\ \\ \frac{R_1}{R_2} = \dfrac{\cancel{p}\frac{\cancel{l_1}}{A_1} }{ \cancel{p}\frac{\cancel{l_2}}{A_2} } \\ \\ 12 = \frac{A_2}{A_1} \times R_2 \\ \\ 12 = \frac{A_1}{A_1} \times \frac{1}{16} \times R_2 \\ \\ 12 \times 16 = R_2 \\ \\ 192 = R_2$

Therefore, The new resistance is 192 ohm

Concept Used

Resistivity : The resistance of a conductor which have unit length and unit cross sectional area is known as resistivity.

Resistivity is denoted by (ρ) .

$ρ = R \dfrac{A}{l}$

Here, R = Resistance of wire

A = Cross Sectional Area of wire

l = length of wire