1. 3×10⁻¹ ^°C⁻¹ 2. 3×10⁻⁴ ^°C⁻¹ 3. 3×10⁻³ ^°C⁻¹ 4. 3×10⁻² ^°C⁻¹

Answer: 4. 3×10⁻² ^°C⁻¹ R_t=R₀(1+α t) , so α=R_t-R₀/R₀ t . α=6.8-2/2×80=4.8/160=3×10⁻² ^°C⁻¹ . The answer is 3×10⁻² ^°C⁻¹ .

The resistance of platinum wire at 0^°C is 2 Ω and 6.8 Ω at 80^°C. The temperature coefficient of resistance of the wire is :

1. \(3\times10^{-1}\,^\circ\mathrm{C}^{-1}\)

2. \(3\times10^{-4}\,^\circ\mathrm{C}^{-1}\)

3. \(3\times10^{-3}\,^\circ\mathrm{C}^{-1}\)

4. \(3\times10^{-2}\,^\circ\mathrm{C}^{-1}\)

Doubtly Asked 30 Oct 2026 NEET UG 2023 Practice test neet Save WhatsApp Report

Best answer

Answer: 4. \(3\times10^{-2}\,^\circ\mathrm{C}^{-1}\)

\(R_t=R_0(1+\alpha t)\), so \(\alpha=\dfrac{R_t-R_0}{R_0\,t}\).

\(\alpha=\dfrac{6.8-2}{2\times80}=\dfrac{4.8}{160}=3\times10^{-2}\,^\circ\mathrm{C}^{-1}\).

The answer is \(3\times10^{-2}\,^\circ\mathrm{C}^{-1}\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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