This question contains Statement I and Statement II. Of the four choices given after the statements,

Q: But the math in Statement I is perfectly correct! Why is it false?

The arithmetic is correct, but the physics is wrong! The formula $R = R_0(1+\alpha\Delta t)$ is an approximation. Applying it outside its valid range (where $\Delta R \ll R_0$) gives a physically meaningless result.

Q: How do we know when $\Delta R$ is 'small enough'?

Generally, if $\Delta R$ is within $5\%$ to $10\%$ of $R_0$, the linear approximation holds well. Here, it's $50\%$, which is way too large to ignore higher-order terms.

Q: What is the exact formula if the linear one fails?

The actual relationship is often modeled as a polynomial: $R = R_0(1 + \alpha\Delta T + \beta\Delta T^2 + \dots)$ or an exponential function depending on the material.

Q: Why does the linear approximation exist if it's not always true?

Because for most everyday applications (like a wire heating up by $10^\circ\text{C}$), the linear formula is incredibly accurate and much easier to calculate than a complex polynomial.

Q: Is Statement II always true for all materials?

Yes, for standard conductors, the linear formula is universally taught as a first-order approximation. Statement II correctly defines the mathematical boundary of this approximation.

Question

This question contains Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement I The temperature dependence of resistance is usually given as $R = R_0(1+\alpha\Delta t)$. The resistance of a wire changes from $100\Omega$ to $150\Omega$ when its temperature is increased from $27^\circ	ext{C}$ to $227^\circ	ext{C}$. This implies that $\alpha = 2.5 	imes 10^{-3} /^\circ	ext{C}$.
Statement II $R = R_i(1+\alpha\Delta T)$ is valid only when the change in the temperature $\Delta T$ is small and $\Delta R = (R - R_0) << R_0$.

Options:

(A) Statement I is true, Statement II is false

(B) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I

(C) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I

(D) Statement I is false, Statement II is true

Sigma Percentile · Accepted Answer

Imagine you are taking an exam, and you see a question that gives you an initial resistance, a final resistance, and a temperature change. It asks you to verify the temperature coefficient $\alpha$. You quickly plug the numbers into the trusty formula $R = R_0(1+\alpha\Delta T)$, do the math, and boom! The answer matches perfectly. You confidently mark 'Statement I is true' and move on.

But wait! You just fell into one of the most beautiful traps in physics.

### The Origin of the Formula
The formula $R = R_0(1+\alpha\Delta T)$ is not a fundamental law of the universe like Newton's laws. It is a **linear approximation**.

In reality, the resistance of a material changes with temperature in a much more complex way. If we use a Taylor series expansion, the true relationship looks something like this:
$$R(T) = R_0 [1 + \alpha \Delta T + \beta (\Delta T)^2 + \gamma (\Delta T)^3 + \dots]$$

For very small changes in temperature ($\Delta T$), the higher-order terms like $(\Delta T)^2$ and $(\Delta T)^3$ become incredibly tiny. We can safely ignore them, leaving us with the simple, linear equation we all know and love.

### The Boundary Condition
Because we threw away the higher-order terms, our linear formula comes with a strict boundary condition: it is only valid when $\Delta T$ is small. Mathematically, this means the change in resistance $\Delta R$ must be much, much less than the initial resistance $R_0$.

This is exactly what **Statement II** tells us. It correctly identifies the limitation of the formula. Therefore, Statement II is absolutely true.

### The Trap in Statement I
Now, let's look at the numbers given in Statement I.
The initial resistance is $R_0 = 100\Omega$.
The final resistance is $R = 150\Omega$.

Let's calculate the change in resistance:
$$\Delta R = 150\Omega - 100\Omega = 50\Omega$$

Here is the catch: $50\Omega$ is exactly $50\%$ of the initial resistance $100\Omega$. Is $50\%$ "much less" than the original value? Absolutely not!

Because $\Delta R$ is so large, the curve of the actual resistance has bent significantly away from our straight-line approximation. The linear formula has completely broken down.

### The Final Verdict
Even though plugging the numbers into the formula gives $\alpha = 2.5 	imes 10^{-3} /^\circ	ext{C}$, the calculation is physically meaningless because the formula itself is invalid for this scenario. The arithmetic is flawless, but the physics is fundamentally flawed.

Therefore, **Statement I is false**. The correct choice is option (d). Always remember to check the boundary conditions of your formulas!

Select Answer:

Visualized Solution

$R = R_{0} (1 + α Δ T)$

Validity Condition

Analyzing Statement I

Checking the Condition

Conclusion

Final Answer

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Origin of the Formula The formula $R = R_{0} (1 + α Δ T)$ is not a fundamental law of the universe like Newton's laws

The Boundary Condition Because we threw away the higher-order terms, our linear formula comes with a strict boundary condition: it is only valid when $Δ T$ is small

The Trap in Statement I Now, let's look at the numbers given in Statement I

The Final Verdict Even though plugging the numbers into the formula gives $α = 2.5 \times 1 0^{- 3} /^{\circ} C$ , the calculation is physically meaningless because the formula itself is invalid for this scenario

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