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Chapter 4: Q6E (page 842)

You want to produce three 1.00-mm-diameter cylindrical wires,

each with a resistance of 1.00 鈩� at room temperature. One wire is gold, one

is copper, and one is aluminum. Refer to Table 25.1 for the resistivity

values. (a) What will be the length of each wire? (b) Gold has a density of1.93脳10-4kgm3.

What will be the mass of the gold wire? If you consider the current price of gold, is

this wire very expensive?

Short Answer

Expert verified

(a) The length of each wire is,

Length of copper wire 鈥� 45.66m

Length of gold wire 鈥� 32.19

Length of aluminum wire 鈥� 28.35m

(b)The mass of the gold wire is 0.488 kg and the wire is very expensive.

Step by step solution

01

Formula of resistance and resistivity of wires.

Consider the formula for the resistance as:

R=蚁LAL=RA蚁

Here, 蚁is the resistivity, L is the length and A is the area of cross section.

Consider the resistivity values- as:

Copper - 1.72脳10-8惟.m.

Gold - 2.44脳10-8惟.m.

Aluminum - 2.77脳10-8惟.m.

Consider the formula for the density as:

M=V蚁M=(AL)蚁

02

Calculate the length of each wire.

(a)

Solve for the length of the copper wire.

L=1惟蟺41mm21.72脳10-8惟.m=45.66m

Solve for the length of the gold wire.

L=1惟蟺41mm22.44脳10-8惟.m=32.19m

Solve for the length of the aluminum wire.

L=1惟蟺41mm22.77脳10-8惟.m=28.35m

03

Calculate the mass of the gold wire.

(b)

The density of gold is given as1.93脳104kgm3

Therefore mass of gold wire is,

M=1.93脳104kgm37.854脳10-7m232.19m=0.488kg

Therefore, the mass of the gold wire is 0.488 kg and the wire is very expensive.

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