Question

two wires of equal length made up of material of resistivity ratio 1 is to 2 and area ratio 3 is to 2 have the potential drop across them in the ratio X is to why in series the X to Y ratio is

a. 3:1
b. 2:5
c. 5:2
d. 1:3

plz give the answer with explanation ​

Answer 1

Given: Two wires of the same material having lengths in the ratio of 1:2 and diameters in the ratio of 2:3 are connected in series with an accumulator.

To find the ratio of the potential difference across the two wires.

Solution:

Let ρ be the resistivity of the material of the wire.

First wire: 

Length = L

Diameter = 2d

Area of Cross-section, A = π(22d)2=πd2

Resistance, R=ρAL=ρπd2L

If current I flows through the combination then the potential difference across R is, V=IR

Second wire: 

Length = 2L

Diameter = 3d

Area of Cross-section, A′=π(23d)2=49×πd2=49×A

Resistance, R′=ρA2L′=ρ49A2L=ρ9A8L=98ρAL=98R

If current I flows through the combination then the potential difference across R′ is, V′=IR′=98IR

So, the ratio of potential difference across the wires in series is, 

V:V′=IR:98IR=9:8

Answer 2

Explanation:

Given: Two wires of the same material having lengths in the ratio of 1:2 and diameters in the ratio of 2:3 are connected in series with an accumulator.

To find the ratio of the potential difference across the two wires.

Solution:

Let ρ be the resistivity of the material of the wire.

First wire:  

Length = L

Diameter = 2d

Area of Cross-section A= π(  2d/2)square=πdsquare

Resistance, R=ρ L/A= ρ L/πd square

If current I flows through the combination then the potential difference across R is, V=IR

Second wire:

Length = 2L

Diameter = 3d

Area of Cross-section, A'=π(3d/2)square=9/4xπd square=9/4xA

Resistance, R'=p 2L'/A=p 2L/9/4A=p 8L/9A=8/9p L/A=8/9R

If current I flows through the combination then the potential difference across R' is, V' =1R'=8/9IR

So, the ratio of potential difference across the wires in series is,

V:V' = IR:8/9IR=9:8

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