Temperature coefficient of resistance of platinum is 4- 10-3K-1 at 0-C- The temperature at which the increase in the resistance of platinum wire is 10

R_t=R_0[1+α t] R_t=R_0+10/100R_0=1.1R_0 1.1R_0=R_0[1+α t] 1.1 1=α t 0.1=4× 10^ 3t t=100/4=25^∘C ...

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Byju's Answer

Standard XII

Physics

Dependence of Resistance on Temperature

Temperature c...

Question

Temperature coefficient of resistance of platinum is $4 \times 10^{- 3} K^{- 1}$ at $0^{\circ}$ C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at $0^{\circ}$ C.

250^{\circ} C

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500^{\circ} C

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50^{\circ} C

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25^{\circ} C

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Solution

The correct option is D $25^{\circ} C$
$R_{t} = R_{0} [1 + α t] R_{t} = R_{0} + \frac{10}{100} R_{0} = 1.1 R_{0} 1.1 R_{0} = R_{0} [1 + α t] 1.1 - 1 = α t 0.1 = 4 \times 10^{- 3} t t = \frac{100}{4} = 25^{\circ} C$

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Temperature coefficient of resistance of platinum is $4 \times 10^{- 3} K^{- 1}$ at $0^{\circ}$ C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at $0^{\circ}$ C.

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Temperature coefficient of resistance of platinum is 4×10−3K−1 at 0∘C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at 0∘C.

The correct option is D 25∘C Rt=R0[1+αt]Rt=R0+10100R0=1.1R01.1R0=R0[1+αt]1.1−1=αt0.1=4×10−3tt=1004=25∘C

Temperature coefficient of resistance of platinum is $4 \times 10^{- 3} K^{- 1}$ at $0^{\circ}$ C. The temperature at which the increase in the resistance of platinum wire is 10% of its value at $0^{\circ}$ C.

The correct option is D $25^{\circ} C$
$R_{t} = R_{0} [1 + α t] R_{t} = R_{0} + \frac{10}{100} R_{0} = 1.1 R_{0} 1.1 R_{0} = R_{0} [1 + α t] 1.1 - 1 = α t 0.1 = 4 \times 10^{- 3} t t = \frac{100}{4} = 25^{\circ} C$