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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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Similar questions

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.

Q. The temperature coefficient of resistance of platinum is

α = 3.92 \times 10^{- 1} K^{- 1}

20^{o} C

. Find the temperature at which the increase in the resistance of platinum wire is

10

of its value at

20^{o} C

Q. Temperature coefficient of resistance of platinum is

4 \times 10^{- 3} / K

20^{\circ} C

. Temperature at which increase in resistance of platinum is

10

% is value at

20^{\circ} C

Q. A platinum wire has a resistance of

10 Ω

0^{\circ} C

and

20 Ω

273^{\circ} C

. Find the value of temperature coefficient of platinum.

Q. The electrical resistance of pure platinum increases linearly with increasing temperature over a small range of temperature. This property is used in a Platinum resistance thermometer. The relation between

R_{θ}

(Resistance at

θ K

) and

R_{0}

(Resistance at

θ_{0} K)

is given by

R_{θ} = R_{0} [1 + α (θ - θ_{0})]

, where

α =

temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads

0^{\circ} C

when its resistance is

80 Ω

and

100^{\circ} C

when its resistance is

90 Ω

, find the temperature at which its resistance is

86 Ω

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