A wire has a resistance of 2.5 Ω at 28 °C and a resistance of 2.9 Ω at 100 °C. The temperature coefficient of resistivity of material of the wire is

Question

A wire has a resistance of 2.5 Ω at 28 °C and a resistance of 2.9 Ω at 100 °C. The temperature coefficient of resistivity of material of the wire is (A) 1.06×10-3 °C-1 (B) 3.5×10-2 °C-1 (C) 2.22 × 10-3 °C-1 (D) 3.95×10-2 °C-1

Tardigrade Admin · Accepted Answer

Here, R1=2.5 Ω, T1 = 28 °C R2 = 2.9 Ω and T2 = 100 °C As R2= R1[1+α(T2-T1)] ∴ 2.9 = 2.5[1+α(100-28)] (2.9/2.5)-1 = α[72] or α = (1/72)×(2.9-2.5/2.5) = (1/72)×(0.4/2.5) = 2.22×10-3 °C-1

Q. A wire has a resistance of $2.5 Ω$ at $28^{\circ} C$ and a resistance of $2.9 Ω$ at $100^{\circ} C$ . The temperature coefficient of resistivity of material of the wire is

$1.06 \times 1 0^{- 3}^{\circ} C^{- 1}$

$3.5 \times 1 0^{- 2}^{\circ} C^{- 1}$

$2.22 \times 1 0^{- 3}^{\circ} C^{- 1}$

$3.95 \times 1 0^{- 2}^{\circ} C^{- 1}$

Q. A wire has a resistance of 2.5Ω at 28∘C and a resistance of 2.9Ω at 100∘C. The temperature coefficient of resistivity of material of the wire is

1.06×10−3∘C−1

3.5×10−2∘C−1

2.22×10−3∘C−1

3.95×10−2∘C−1

Here, R1​=2.5Ω,T1​=28∘C R2​=2.9Ω and T2​=100∘C As R2​=R1​[1+α(T2​−T1​)] ∴2.9=2.5[1+α(100−28)] 2.52.9​−1=α[72] or α=721​×2.52.9−2.5​ =721​×2.50.4​ =2.22×10−3∘C−1

Q. A wire has a resistance of $2.5 Ω$ at $28^{\circ} C$ and a resistance of $2.9 Ω$ at $100^{\circ} C$ . The temperature coefficient of resistivity of material of the wire is

$1.06 \times 1 0^{- 3}^{\circ} C^{- 1}$

$3.5 \times 1 0^{- 2}^{\circ} C^{- 1}$

$2.22 \times 1 0^{- 3}^{\circ} C^{- 1}$

$3.95 \times 1 0^{- 2}^{\circ} C^{- 1}$