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Ans.122.7 m and 1/4 times​

Answer 1

$\huge\bigstar\underline\mathfrak\red{Explanation}$

We know that,

Area of cross-section of the wire (A)= πr²

Given :

Diameter = 0.5mm = 0.0005 m

Resistance(R)= 10 ohm

________________________

Also,

R = rho × (l/A)

l= RA/rho

=[ 10× 3.14× (0.0005/2)² ]/(1.6×10^-8)

= (10×3.14×25)/(4×1.6)

= 122.72 m

Hence, length of the wire is 122.72 m.

_________________________

If we doubled the diameter,

New diameter = 2×0.5 = 1 mm = 0.001 m

____________________________

Now, let new resistance be R',

R' = rho×(l/A)

= (1.6×10^-8 × 122.72 )/[π(1/2×10^-3)]

= 250.2 × 10^-2

= 2.5 ohm

Hence, new resistance is 2.5 ohm.

______________________________

Answer 2

Answer:

D= 0.5×10^-3m

therefore the ar of cross section

A= πD^2/4

= 3.14× (0.5×10^-3)^2/4

resistivity= 1.6×10^-8 Ohm-m

resistance= 10 Ohm

we know that, R= resistivity× length/area

l= 10×3.14×0.5×10^-3×0.5×10^-3 / 4×1.6×10^-8

l= 122.7m

now when diameter is doubled

A' = π(2D)^2 / 4

= 4×πD^2 / 4

= 4A

therefore l'= RA' / resistivity

= R(4A) / resistivity

= 4× RA / resistivity

=4(122.7)

= 490.8 m