The length of a certain wire is doubled and at the same time its radius is also doubled. What is the new resistance of this wire

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

Explanation:

Consider a wire with:

Resistance = R

Length = L

Area = A = πr²

where, r = radius

ρ = resistivity

Then:

R = ρL/A

R = ρL/πr²   equation 1

Now, the new wire has:

Resistance = R'

Resistivity = ρ

Length = L' = 2 L

Radius = r' = 2r

Area = πr'² = π(2r)² = 4πr²

Therefore,

R' = ρL'/πr'²

R' = ρ(2 L)/4πr²

R' = (1/2)(ρL/πr²)

using equation 1:

R' = R/2

Therefore, the new resistance of the wire is twice the value of the initial resistance.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j