Question

Plz solve question no. 6?

Answer 1

HEY , HERE'S THE ANSWER ⏬


$radius = \frac{0.5}{2} = 0.5mm = 0.025cm \\ p = 1.6 \times 10 {}^{ - 6} ohm - cm \\ r = p \times \frac{l}{a} = p \times \frac{l}{r {}^{2} \pi } \\ l = \frac{3.14 \times (0.025) { }^{2} \times 10}{1.6 \times 10 {}^{ - 6} } \\ l = 12265.625 \: cm \\ l = 122.6 \: m \\ again \: \: \: r = p \times \frac{l}{a}$
Thus , when the diameter of the wire is doubled , the resistance becomes one-fourth of the original value.
New resistance
$= \frac{10}{4} = 2.5 \: ohm \\ decrease \: in \: resistance \: = 10 - 2.5 = 7.5 \: ohm.$
HOPE IT HELPS YOU ✌️!!!