Sankalp NEET Full Test-2 Question-45 Solution

Question 45: The resistance of platinum wire at 0^{\circ} \mathrm{C} is 2 \Omega and 6.8 \Omega at 80^{\circ} \mathrm{C}. The temperature coefficient of resistance of the wire is :

(1) 3 \times 10^{-2}{ }^{\circ} \mathrm{C}^{-1}

(2) 3 \times 10^{-1}{ }^{\circ} \mathrm{C}^{-1}

(3) 3 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}

(4) 3 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}

Answer: Option (1)

Explanation:

The resistance of a conductor at temperature T is given by

R_T = R_0 \left(1 + \alpha T \right),

where R_0 is the resistance at 0^{\circ} \mathrm{C} and \alpha

is the temperature coefficient of resistance.

Given R_0 = 2 \Omega and R_{80} = 6.8 \Omega.

Substituting the values,

6.8 = 2 \left(1 + 80\alpha \right).

Dividing both sides by 2,

3.4 = 1 + 80\alpha.

80\alpha = 2.4.

\alpha = \frac{2.4}{80} = 0.03.

Thus,

\alpha = 3 \times 10^{-2} { }^{\circ} \mathrm{C}^{-1}.

Hence, the correct answer is Option (1).

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