A wire with a resistance R is lengthened to 7.34 times its original length by pulling it through a small hole. Find the new resistance of the wire after it was stretched. Answer in units of R.

The resistance of a wire is given as

R=ρl/A.

So the first wire has resistance R and length l and resistivity ρ and area A.

Now, the length of the wire is now stretched to 7.34 times it initial

L2=7.34l

Since the length ha increase by 7.34, the cross sectional area is reduced by 7.34

Therefore, A2 = A/7.34

The resistivity of the wire did not change

Therefore

Resistivity 1 is equal to resistivity 2

ρ. = ρ

R1A1/L1=R2A2/L2

Since R=R, A1=A and L1 = l

Also R2=?, A2 = A/7.34 and L2=7.34l

RA/l=R2A/7.34l

R2=RA/l*7.34l/A/7.34

R2=7.34*7.34R

R2=53.88R

The new resistance is 53.88 times the old resistance