The temperature coefficient of resistance for a wire is 0.00125° C -1. At 300 K its resistance is 1 Ω. The temperature at which the resistance becomes 1.5 Ω is?

Question

The temperature coefficient of resistance for a wire is 0.00125° C -1. At 300 K its resistance is 1 Ω. The temperature at which the resistance becomes 1.5 Ω is? (A) 450 K (B) 727 K (C) 454 K (D) 900 K

Tardigrade Admin · Accepted Answer

Rt2=R0(1+α t2) and Rt1=R0(1+α t1) ∴ (Rt2/Rt1)=(1+α t2/1+α t1) or (1.5/1)=(1+0.00125 × t2/1+0.00125 × 27) On solving we get t2=454° C =454+273=727 K

Q. The temperature coefficient of resistance for a wire is $0.0012 5^{\circ} C^{- 1}$ . At $300 K$ its resistance is $1 Ω$ . The temperature at which the resistance becomes $1.5 Ω$ is?

450 K

727 K

454 K

900 K

$R_{t_{2}} = R_{0} (1 + α t_{2})$ and $R_{t_{1}} = R_{0} (1 + α t_{1})$
$∴ \frac{R _{t_{2}}}{R _{t_{1}}} = \frac{1 + α t _{2}}{1 + α t _{1}}$
or $\frac{1.5}{1} = \frac{1 + 0.00125 \times t _{2}}{1 + 0.00125 \times 27}$
On solving we get
$t_{2} = 45 4^{\circ} C = 454 + 273 = 727 K$

Q. The temperature coefficient of resistance for a wire is 0.00125∘C−1. At 300K its resistance is 1Ω. The temperature at which the resistance becomes 1.5Ω is?