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JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Current Electricity

Resistance of a wire at

0^{\circ} C

,

10 0^{\circ} C

and

t^{\circ} C

is found to be

10Ω

,

10.2Ω

and

10.95Ω

respectively. The temperature t in Kelvin scale is ____.

Answer: 748

Solution

R = R_{0} (1 + α Δ T) \Rightarrow \frac{Δ R}{R _{0}} = α Δ T

.

0 \to 10 0^{\circ}

C:

\frac{10.2 - 10}{10} = α (100) \Rightarrow α = 2 \times 1 0^{- 4} /^{\circ}

C.

0 \to t

:

\frac{10.95 - 10}{10} = α t \Rightarrow 0.095 = 2 \times 1 0^{- 4} t \Rightarrow t = 47 5^{\circ}

C.

T = 475 + 273 = 748 K

.

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Integer answer

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