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NEET PHYSICSEasy

The resistance of platinum wire at 0∘C0^{\circ}\text{C} is 2 Ω2 \text{ } \Omega and 6.8 Ω6.8 \text{ } \Omega at 80∘C80^{\circ}\text{C}. The temperature coefficient of resistance of the wire is

1

3×10−4 ∘C−13 \times 10^{-4} \text{ }^{\circ}\text{C}^{-1}

2

3×10−3 ∘C−13 \times 10^{-3} \text{ }^{\circ}\text{C}^{-1}

3

3×10−2 ∘C−13 \times 10^{-2} \text{ }^{\circ}\text{C}^{-1}

4

3×10−1 ∘C−13 \times 10^{-1} \text{ }^{\circ}\text{C}^{-1}

Step-by-Step Solution

The formula for temperature dependence of resistance is Rt=R0(1+αΔt)R_t = R_0(1 + \alpha \Delta t). Here R0=2 ΩR_0 = 2 \text{ } \Omega, Rt=6.8 ΩR_t = 6.8 \text{ } \Omega, and Δt=80∘C\Delta t = 80^{\circ}\text{C}. So, 6.8=2(1+α⋅80)6.8 = 2(1 + \alpha \cdot 80). Dividing by 2, 3.4=1+80α3.4 = 1 + 80\alpha, which gives 2.4=80α2.4 = 80\alpha. Thus, α=2.480=24800=0.03=3×10−2 ∘C−1\alpha = \frac{2.4}{80} = \frac{24}{800} = 0.03 = 3 \times 10^{-2} \text{ }^{\circ}\text{C}^{-1}.

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