Question

A wire is given material having length l and area of Cross section A has a resistance of 4 Ω . What would be the resistance of another wire of the Same material having length l/2 and area of cross section 2A ??


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Answer 1

Hey here is your answer

Since R= rho l/A
Hope you got

Answer 2

Hey friend, Harish here.

Here is your answer.

Let the data of the original wire be :

R₁ - Resistance = 4 Ω

L₁ = l - Length

A₁ = A  - Cross sectional Area

ρ₁ = ρ - Resistivity  

And, the data of the new wire be :

R₂ - Resistance (Unknown)

L₂ = l/2 - Length

A₂ = 2A  - Cross sectional Area

ρ₂ = ρ - Resistivity

To Find :

The resistance of the new wire.

Solution :

We know that :

$\mathrm{R = \rho \times \frac{L}{A}}$

Then,

$\mathrm{R_1 = \rho_1 \times \frac{L_1}{A_1}}\\\\\implies 4\ \Omega = \mathrm{\rho \times \frac{l}{A}}$

Let this be equation (i)

And,

$\mathrm{R_2 = \rho_2 \times \frac{L_2}{A_2}} \\ \\\implies \mathrm{R_2= \rho \times \frac{\frac{l}{2}}{2A}} \\\\\implies \mathrm{R_2= \rho \times \frac{l}{4A} = \frac{1}{4} \times \Big(\rho \times \frac{l}{A}\Big)}$

and this be equation (ii)

Now substituting eq (i) in (ii) we get :

$\mathrm{R_2 = \frac{1}{4} \times R_1=\frac{1}{4} \times \big( 4 \Omega \big) = 1 \Omega }$

∴ R₂ (New Resistance ) = 1 Ω

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Hope my answer is helpful to you.