Question

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Answer 1

Answer:

Length of wire = 122.72m

The new resistance is =  2.5 Ω

Explanation:

Given Problem:

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8 Ω m.

1)What will be the length of this wire to make its resistance 10 Ω?

2)How much does the resistance change if the diameter is doubled?

Solution:

To Find:

1)What will be the length of this wire to make its resistance 10 Ω?

2)How much does the resistance change if the diameter is doubled?

--------------------

Method:

Area of cross-section of the wire, A =π (d/2) 2

Diameter= 0.5 mm = 0.0005 m

Resistance, R = 10 Ω

We know that,

$R = P\frac{1}{A}$

$l = \frac{RA}{p}$

$=>\frac{10*3.14*[\frac{0.0005}{2} ]}{1.6*10^-8}$

$=>\frac{10*.14*25}{4*1.6} =1.22.72m$

∴ Length of the wire = 122.72m

If the diameter of the wire is doubled, new diameter=2×0.5=1mm=0.001m

Let new resistance be Rʹ

$R = P=\frac{1}{A}$

$=>\frac{1.6*10^-8*122.72}{pi(\frac{1}{2}*10^-3)^2 }$

$=>\frac{1.6*10^-8*122.72*4}{3.14*10^-6}$

$=>250.2*10^-2 = 2.5$

∴ The new resistance is 2.5 Ω

Answer 2

(a)The diameter of copper wire is 0.5mm,

radius 0.5/2=0.25mm.

=0.25/1000m or 0.25×10^-3m

area of cross section of wire ,A=pie×r^2

22/7×(0.25×10-3)2

=0.1964×10^6m^2

resistivity,p=1.6×10^-8 ohm meter

resistance, R=10 ohm meter

and length, L=?

now putting the values in the formula

p=R×A/L

we get : 1.6×10^8=10×0.1964×10^-6/L

l=10×0.1964×10^6/1.6×10^-8

l=1964/16

l=122.7m

(b) the resistance of wire is inversely propertional to the square of diameter of the wire is double, then the resistance will will become (1/2)^2=1/4.

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