Question

The length of a certain wire is doubled and at the same time its radius is increased by a factor of 4. What is the change in the resistance of this wire? 

• it is reduced by a factor or 8 

• it increased by a factor of 8 

• it is reduced by a factor of 4 

• it increases by a factor of 4

Answer 1

2. Sol::

We know that

Resistance(R)= ρ*L/A

ρ= resistivity of material
L= length
And
A = cross sectional area = πr²
So

R= ρ*L/πr².................(1)

Now as given
Length of wire is doubled
So ,
let new length be L1,

Then
L1= 2L
And radius is increased by a factor of 4
So
r1= 4r

Now the resistance will be

R1= ρL1/π*(r1)²
= ρ(2L)/ π*(4r)²
= 2ρl/16πr²........................(2)

Now from equations (1)&(2)

R1/R= (2ρl/16πr²)/ρ*L/πr²
R1 = (1/8)R

Hence resistance will be REDUCED BY A FACTOR of 8