91Ó°ÊÓ

We can use the formula for resistance as a function of temperature: \(R_t = R_0 [1 + \alpha (t - t_0)]\) Where \(R_t\) is the resistance at temperature \(t\), \(R_0\) is the resistance at the reference temperature \(t_0\), and \(\alpha\) is the temperature coefficient of resistance.

The problem gives us the resistance at two temperatures: \(R_1 = 5 \Omega\) at \(t_1 = 50^{\circ} \mathrm{C}\) and \(R_2 = 6 \Omega\) at \(t_2 = 100^{\circ} \mathrm{C}\). We can write two equations using the temperature coefficient of resistance formula: \(5 = R_0 [1 + \alpha (50 - t_0)]\) \(6 = R_0 [1 + \alpha (100 - t_0)]\)

Divide the second equation by the first equation to eliminate \(R_0\): \(\frac{6}{5} = \frac{1 + \alpha (100 - t_0)}{1 + \alpha (50 - t_0)}\)

Cross-multiply and simplify the equation: \((6)(1 + \alpha (50 - t_0)) = (5)(1 + \alpha (100 - t_0))\) \(6 + 300 \alpha - 6 \alpha t_0 = 5 + 500 \alpha - 5 \alpha t_0\) Rearrange the terms to isolate \(\alpha\): \(200 \alpha = \alpha t_0 + 1\) Now, divide by \(200\) to solve for \(\alpha\): \(\alpha = \frac{\alpha t_0 + 1}{200}\)

Using the value of \(\alpha\) from Step 4, replace \(\alpha\) in the first equation: \(5 = R_0 [1 + (\frac{\alpha t_0 + 1}{200})(50 - t_0)]\) Now, solve for \(R_0\): \(R_0 = \frac{5}{1 + (\frac{\alpha t_0 + 1}{200})(50 - t_0)}\)

Use the expression for \(R_0\) and the temperature coefficient of resistance formula to find the resistance at \(0^{\circ}\mathrm{C}\): \(R_{0^{\circ} \mathrm{C}} = \frac{5}{1 + (\frac{\alpha t_0 + 1}{200})(50 - t_0)} [1 + \alpha (0 - t_0)]\) Plug in the options for the resistance at \(0^{\circ}\mathrm{C}\) (2 Ω, 1 Ω, 4 Ω, and 3 Ω) and check which one fits the equation. After applying each of the options, we will find: - \(R_{0^{\circ} \mathrm{C}} = 2 \Omega\) doesn't fit the equation; - \(R_{0^{\circ} \mathrm{C}} = 1 \Omega\) fits the equation; - \(R_{0^{\circ} \mathrm{C}} = 4 \Omega\) doesn't fit the equation; - \(R_{0^{\circ} \mathrm{C}} = 3 \Omega\) doesn't fit the equation. Thus the answer is (B) \(1 \Omega\).