The volume of this specimen is constant, the volume is the cross-sectional area A times length L :
V=A_1L_1=A_2L_2,\\\space\\
\dfrac{\pi d_1^2}{4}L_1=\dfrac{\pi d_2^2}{4}L_2→d_1^2L_1=d_2^2L_2,\\\space\\
[L_2]=L_1\dfrac{d_1^2}{d_2^2}.
The resistance:
R 2 = 100 R 1 , R = ρ L A : 100 ρ L 1 A 1 = ρ L 2 A 2 → 100 L 1 A 1 = L 2 A 2 → 100 L 1 d 1 2 = L 2 d 2 2 , d 2 = d 1 10 [ L 2 ] L 1 , d 2 = d 1 10 [ L 1 d 1 2 d 2 2 ] L 1 = d 1 2 10 d 2 , d 2 = d 1 10 = 0.79 cm . R_2=100R_1,\\\space\\
R=\dfrac {\rho L}A:\\\space\\
100\dfrac {\rho L_1}{A_1}=\dfrac {\rho L_2}{A_2}→100\dfrac { L_1}{A_1}=\dfrac {L_2}{A_2}→100\dfrac {L_1}{d_1^2}=\dfrac {L_2}{d^2_2},\\\space\\
d_2=\dfrac{d_1}{10}\sqrt{\dfrac {[L_2]}{L_1}},\\\space\\
d_2=\dfrac{d_1}{10}\sqrt{\dfrac {\bigg[L_1\dfrac{d_1^2}{d_2^2}\bigg]}{L_1}}=\dfrac{d_1^2}{10d_2},\\\space\\
d_2=\dfrac {d_1}{\sqrt{10}}=0.79\text{ cm}. R 2 = 100 R 1 , R = A ρ L : 100 A 1 ρ L 1 = A 2 ρ L 2 → 100 A 1 L 1 = A 2 L 2 → 100 d 1 2 L 1 = d 2 2 L 2 , d 2 = 10 d 1 L 1 [ L 2 ] , d 2 = 10 d 1 L 1 [ L 1 d 2 2 d 1 2 ] = 10 d 2 d 1 2 , d 2 = 10 d 1 = 0.79 cm .
The new diameter is 7.9 mm.
