The temperature coefficient of resistance of platinum is -3-92-10-1 K-1 at 20oC- Find the temperature at which the increase in the resistance of platinum wire is 10

The correct option is B 45.5^oC Given, Resistance at 20 ^oC is 1 Ω #160; #10; #10;Increase in temperature, Δ R=10 Ω #10; #10; Δ ...

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Byju's Answer

Standard XII

Physics

Dependence of Resistance on Temperature

The temperatu...

Question

The temperature coefficient of resistance of platinum is $α = 3.92 \times 10^{- 1} K^{- 1}$ at $20^{o} C$ . Find the temperature at which the increase in the resistance of platinum wire is $10$ of its value at $20^{o} C$

{40.5}^{o} C

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{45.5}^{o} C

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{48.5}^{o} C

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{43.5}^{o} C

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Solution

The correct option is B ${45.5}^{o} C$
Given,
Resistance at $20^{o} C i s 1 Ω$

Increase in temperature, $Δ R = 10 Ω$

$Δ R = α R_{20} Δ T =$

$\Rightarrow 10 = 3.92 \times 10^{- 1} \times 1 \times (T - 20)$

$\Rightarrow T = 45.5^{o} C$
Hence, at $45.5^{o} C$ resistance increased by $10 Ω$

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The temperature coefficient of resistance of platinum is α=3.92×10−1 K−1 at 20oC. Find the temperature at which the increase in the resistance of platinum wire is 10 of its value at 20oC

The correct option is B 45.5oCGiven, Resistance at 20oCis1Ω Increase in temperature, ΔR=10Ω ΔR=αR20ΔT= ⇒10=3.92×10−1×1×(T−20) ⇒T=45.5oC Hence, at 45.5oC resistance increased by 10Ω

The temperature coefficient of resistance of platinum is $α = 3.92 \times 10^{- 1} K^{- 1}$ at $20^{o} C$ . Find the temperature at which the increase in the resistance of platinum wire is $10$ of its value at $20^{o} C$

The correct option is B ${45.5}^{o} C$
Given,
Resistance at $20^{o} C i s 1 Ω$

Increase in temperature, $Δ R = 10 Ω$

$Δ R = α R_{20} Δ T =$

$\Rightarrow 10 = 3.92 \times 10^{- 1} \times 1 \times (T - 20)$

$\Rightarrow T = 45.5^{o} C$
Hence, at $45.5^{o} C$ resistance increased by $10 Ω$