for the second part,

The resistivity of a material increases with the increasing temperature and it can be linearly approximated as follows.

rho = rho_0(1+alpha(T-T_0))

or simple we can say Rho is directly proportionate to T

So rho = kT

T in kelvins,

therefore

R = kT * L /A

assuming all the copper wires have the same cross sectioanl area, A then R = C * T * L

Therefore the wire with the greatest ressitance would be one with the greatest T*L value,

The wire with greates T*L value is wire (1) (Which has 305*10)

Therefore the wire with greatest resistance is wire (1)

**Further Reading**

This question can be approached as below.

The resistance (R) is related to resistivity (rho), length (L) and cross sectional area (A) by the following equation,

R = rho * L /A

first part -- least resistance, since these are same material (Al) we can ignore it and conside about L/A ratio. Therefore the least resistance wire would be the one with the lowest L/A ratio.

We can calculate the L/A ratio as follows for each wire, We can take the diameter*diameter as the area since (A = pi * d^2 /4)

(i) L/A = 5/25 = 0.2

(ii) L/A = 10/9 = 1.11

(iii) L/A = 10/25 = 0.4

(iv)L/A = 5/9 = 0.555

Therefore the least resistance wire will be (i) with 5 cm length and 5 cm diameter.

You are only supposed to ask 1 question at a time. So I will answer only question 1 here and 2nd one I will add my answer.

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**Further Reading**