Question

A 1.70 m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 13.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.2 A , while at 92.0 ∘C it reads 17.3 A . You can ignore any thermal expansion of the rod.

1) Find the resistivity and for the material of the rod at 20 ∘C.

2) Find the temperature coefficient of resistivity at 20 ∘C for the material of the rod.

Answer 1

Given, I l = 1.70m diameter , d = 0.500 cm = 5x15 3m. a sadices Y= 0= 2.5 x103 m potential difference, 1 - 130 v cat T = 20 c current I = 18.2 A at T = 92.0c current, I = 17 3A I 27 Let, Alow, s be the resistivity s at Tagore I got 273 = 2930 k r euroint, I = 18.2 A desistance, R = V = 13 = 0.414 an Nou, - 18.2 Alous, Area of crossection , Aarr² = 3-14X (2.5 x103)² = 3-14X6-250156 m2 A = 1.9625x165 m2 we know that, I Now, 5) pe RA - 017714 x 19625 X 109. Sex 162m! 1.70 at 20°c is - Hence, resistivity 8.24x102 m 27 Now, for for T = 20°C -8.24x100 om To = 92°C l = Rate = 13.0V I 2 17 3 A and R u a 13 0.751 2 In 143 les Rg A = 0.451 x 7.9625x165 - 8167 X10-2m Nous flow, 12 = 2 0 1170 Now, set, & be the coefficient of resistivity .

we know that, ds Se afi fi (T2-T) gaa 8.67X16 6 -8.24x106 _ - 0:43 x 10 Szuxio & (92-20) 8.24x156x72 ga 0:43 = 7.25 Xiók oë 0.43 8.24 x 72 593.28 is f. 24x10 c - coefficient of resistivity at goc