Question

The resistance of a wire is \(5\,ohm\) at \(25^\circ C\) and \(7\,ohm\) at \(100^\circ C\). The resistance of the wire at \(0^\circ C\) is

• A. \( \frac{13}{3}\,ohm \)
• B. \( \frac{5}{3}\,ohm \)
• C. \( \frac{2}{3}\,ohm \)
• D. \(0.1\,ohm\)

Hint:

For resistance-temperature questions, use \(R_t = R_0(1+\alpha t)\) and form two equations to eliminate \(\alpha\).

Answer 1

The Correct Option is A

Step 1: Use resistance-temperature relation.
\[ R_t = R_0(1+\alpha t) \]

Step 2: Write equation for \(25^\circ C\).
\[ 5 = R_0(1+25\alpha) \]

Step 3: Write equation for \(100^\circ C\).
\[ 7 = R_0(1+100\alpha) \]

Step 4: Divide the equations.
\[ \frac{7}{5} = \frac{1+100\alpha}{1+25\alpha} \]

Step 5: Solve for \(\alpha\).
\[ 7(1+25\alpha)=5(1+100\alpha) \]
\[ 7+175\alpha=5+500\alpha \]
\[ 2=325\alpha \]
\[ \alpha=\frac{2}{325} \]

Step 6: Find \(R_0\).
\[ 5 = R_0\left(1+25 \times \frac{2}{325}\right) \]
\[ 5 = R_0\left(1+\frac{50}{325}\right) \]
\[ 5 = R_0\left(\frac{375}{325}\right) \]
\[ R_0 = 5 \times \frac{325}{375} \]
\[ R_0 = \frac{13}{3}\,ohm \]

Step 7: Final conclusion.
\[ \boxed{\frac{13}{3}\,ohm} \] Hence, correct answer is option (A).