Question

Chapter :- Electricity =_=


#Kindly Solve it with a proper explanation :)

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Answer 1

Resistance of wire A is RA, and RB, lengths is lA and lB, area is AA and AB and resistivity is p which is same for both the wires as the material is same.

Volume of the 1st wire is AAlA and of the second wire is ABlB

Now given that:

lA = 1m

lB = 2m

AA = 1cm2 = 10-4m2

AB = 100mm2 = 10-4m2

Now to find p , a 3rd wire of same material is given

Thickness of this wire = 2cm which is diameter

Hence radius = 1cm = 10-2m

Area of this wire will be ∏*10-4m2

Resistance R of this wire is given as 5 ohm

Length is also 5m

Hence from the formula R = pl/A

We get p = ∏*10-4ohm m

We will use this value of p for the final case

Here we are given that a new wire is formed by melting A and B hence the volume of the new wire = Al

This is equal to:

Al = AAlA +ABlB

= 1*10-4 + 2 *10-4

Al= 3*10-4

Now it is given in the question that length l of this wire = 5m

Hence A = 3/5 * 10-4m2

To find the resistance R, we are given:

R = pl/A

R = ∏*10-4 * 5*5/3 *10-4

= 25∏/3 ohms.

Thank you and now is it understandable.

Answer 2

25Ω

Given :

Two conducting wires A and B (made of the same material ) of lengths 1 m and 2 m and area of cross sec-tions 1cm² and 100mm²

The resistance of a wire of length 5 m and thickness 2 cm made of the same material as A and B is 5ω

To Find :

The resistance of the new wire of length 5 m formed by melting A and B.

Solution :

• Step 1 :

The volume of new wire (C) =$\bf V_C=V_A+V_B$

So,

Area of the cross-section of a new circle (C),

$\implies \sf A_C=\dfrac{V_A+V_B}{l_C}$

$\implies \sf R = \rho' \dfrac{l}{A}$

• Step 2 :

Let's find the value of ρ,

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

$\sf \implies \dfrac{50 \Omega \times227\times1cm^2}{500m}$

We know that :

$\bigstar\;\bf R_C=\rho'\dfrac{lC}{AC}$

After Subsituing the value,

We get the answer as - 25Ω.