The temperature coefficient of resistance of a wire of 0.00125/° C . At 300 K, its resistance is 1 Ω . The resistance of the wire will be 2 Ω , at following temperature

Question

The temperature coefficient of resistance of a wire of 0.00125/° C . At 300 K, its resistance is 1 Ω . The resistance of the wire will be 2 Ω , at following temperature (A) 1127 K (B) 1128 K (C) 600 K (D) 1400 K

Tardigrade Admin · Accepted Answer

According to definition of thermal resistance coefficient α =(R2-R1/R1t2-R2t1) 0.00125=(2-1/1× t2-2× 27°) 0.00125=(1/t2 2 - 54°) or t2=854°C Hence, t2=854+273 =1127 K

Q. The temperature coefficient of resistance of a wire of $0.00125 /^{\circ} C$ . At $300 K$ , its resistance is $1 Ω$ . The resistance of the wire will be $2 Ω$ , at following temperature

1127 K

1128 K

600 K

1400 K

According to definition of thermal resistance coefficient
$α = \frac{R _{2} - R _{1}}{R _{1} t _{2} - R _{2} t _{1}}$
$0.00125 = \frac{2 - 1}{1 \times t _{2} - 2 \times 2 7 ^{\circ}}$
$0.00125 = \frac{1}{t _{2} 2 - 5 4 ^{\circ}}$
or $t_{2} = 85 4^{\circ} C$
Hence, $t_{2} = 854 + 273$
$= 1127 K$

Q. The temperature coefficient of resistance of a wire of 0.00125/∘C . At 300K, its resistance is 1Ω . The resistance of the wire will be 2Ω , at following temperature