Question

# As part of a class project you are given 0.900 g of nichrome and asked to...

As part of a class project you are given 0.900 g of nichrome and asked to fabricate a wire with uniform cross-section. You use up 95% of the nichrome and make a wire with a resistance of 0.673 Ω. The resistivity of nichrome is 1.00 10-6 Ω · m and its density is 8.31 103 kg/m3. (a) What length of wire do you end up with? m (b) What is the diameter of the wire? mm

Resistance,R=r* l/A ,   r=resistivity,l=length,A=area

l/A=R/r=0.673 ohm/1.00*10^-6 ohm .m =0.673*10^6 m^-1

or, l/A=0.673*10^6 m^-1…..(1)

mass of Nichrome used=95% of 0.9 g=0.95*0.9 g=0.855 g=0.855*10^-3 kg

density = 8.31* 10^3 kg/m3

Volume =mass/density=0.855*10^-3 kg/8.31* 10^3 kg/m3=0.1029*10^-6 m3

a) l/A*l/l=l^2/V

or, 0.673*10^6 m^-1=l^2/0.1029*10^-6 m3

or, 0.673*10^6 m^-1 *0.1029*10^-6 m3=l^2

or,0.0692 m^2=l^2

l=0.263 m

b) diameter of wire=?

Area=A=l/0.673*10^6 m^-1 (from eqn 1)

A=0.263m/0.673*10^6 m^-1 =0.391*10^-6 m^2

Area of cross –section of Cylindrical wire=∏r^2=0.391*10^-6 m^2

∏r^2=0.391*10^-6 m^2

r^2=0.391*10^-6 m^2/3.14=0.124*10^-6 m^2

r=0.352 *10^-6 m=0.352*10^-3 mm

r=0.352*10^-3 mm

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