The resistance of a wire at room temperature 30° C is found to be 10 Ω . Now to increase the resistance by 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is 0.002/° C ]

Question

The resistance of a wire at room temperature 30° C is found to be 10 Ω . Now to increase the resistance by 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is 0.002/° C ] (A)  36° C  (B)  83° C  (C)  63° C  (D)  33° C

Tardigrade Admin · Accepted Answer

R=R0(1+α t) ∴ R0(1+30α )=10 Ω and R0(1+α )=11 Ω Therefore, (11/10)=(1+α t/1+30α ) Or 11+330α =10+10α t Or 11+330× 0.002=10+10× 0.002t Or 11.66=0.02t+10 Or 0.02t=1.66 Or t=83° C

Q. The resistance of a wire at room temperature $30^{\circ} C$ is found to be $10 Ω$ . Now to increase the resistance by 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002/^{\circ} C$ ]

$36^{\circ} C$

$83^{\circ} C$

$63^{\circ} C$

$33^{\circ} C$

$66^{\circ} C$

$R = R_{0} (1 + α t)$
$∴$ $R_{0} (1 + 30 α) = 10 Ω$
and $R_{0} (1 + α) = 11 Ω$
Therefore, $\frac{11}{10} = \frac{1 + α t}{1 + 30 α}$
Or $11 + 330 α = 10 + 10 α t$
Or $11 + 330 \times 0.002 = 10 + 10 \times 0.002 t$
Or $11.66 = 0.02 t + 10$ Or $0.02 t = 1.66$
Or $t = 83^{\circ} C$

Q. The resistance of a wire at room temperature 30∘C is found to be 10Ω . Now to increase the resistance by 10%, the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is 0.002/∘C ]

36∘C

83∘C

63∘C

33∘C

66∘C

R=R0​(1+αt) ∴ R0​(1+30α)=10Ω and R0​(1+α)=11Ω Therefore, 1011​=1+30α1+αt​ Or 11+330α=10+10αt Or 11+330×0.002=10+10×0.002t Or 11.66=0.02t+10 Or 0.02t=1.66 Or t=83∘C