The temperature coefficient of resistance for a wire is 0.00125 $/^{\circ} C$ . At 300K its resistance is 1 ohm. The temperature at which the resistance becomes 2 ohm is

1154 K

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1100 K

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1400 K

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1127 K

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Solution

The correct option is D 1127 K
$\frac{ρ_{1}}{ρ_{2}} = \frac{(1 + α t_{1})}{(1 + α t_{2})} = \frac{1}{2} \Rightarrow \frac{(1 + 0.00125 \times 27)}{(1 + 0.00125 \times t)}$

$\Rightarrow t = 854^{\circ} C \Rightarrow T = 1127 K$

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Similar questions

Q. The temperature coefficient of resistance for a wire is

0.00125 C^{- 1}

. At 300 K its resistance is 1 ohm. The temperature at which the resistance becomes 2 ohm is

Q. The temperature coefficient of resistance of a wire is

0.00125 /^{\circ} C

. At

300 K

its resistance is

1 ohm

. The temperature at which the resistance becomes

2 ohm

is:

The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1 ohm. This resistance of the wire will be 2 ohm at
(IIT-JEE 1980)

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The temperature coefficient of resistance for a wire is 0.00125/∘C. At 300K its resistance is 1 ohm. The temperature at which the resistance becomes 2 ohm is

The correct option is D 1127 Kρ1ρ2=(1+α t1)(1+α t2)=12⇒(1+0.00125× 27)(1+0.00125× t) ⇒ t =854∘C⇒ T =1127 K

The temperature coefficient of resistance for a wire is 0.00125 $/^{\circ} C$ . At 300K its resistance is 1 ohm. The temperature at which the resistance becomes 2 ohm is

The correct option is D 1127 K
$\frac{ρ_{1}}{ρ_{2}} = \frac{(1 + α t_{1})}{(1 + α t_{2})} = \frac{1}{2} \Rightarrow \frac{(1 + 0.00125 \times 27)}{(1 + 0.00125 \times t)}$

$\Rightarrow t = 854^{\circ} C \Rightarrow T = 1127 K$