1. Check Statement–1

We are given the linear relation $R = R_0(1+\alpha \,\Delta t).$

Here,

• $R_0 = 100\,\Omega$ at $27^\circ C$
• $R = 150\,\Omega$ at $227^\circ C$
• so $\Delta t = 227-27 = 200^\circ C$

Using $150 = 100(1+\alpha \cdot 200),$ we get $1.5 = 1 + 200\alpha$ $200\alpha = 0.5$ $\alpha = \frac{0.5}{200} = 2.5\times 10^{-3}\,/^\circ C.$

So Statement–1 is true.

2. Check Statement–2

The formula $R = R_0(1+\alpha\,\Delta t)$ is an approximate linear relation, valid when the temperature coefficient can be treated as constant over the range of temperature.

The statement says it is valid only when change in temperature is small and then incorrectly writes something like $\Delta T = (R-R_0) \ll R_0.$

This is dimensionally wrong because:

• $\Delta T$ has unit of temperature,
• $(R-R_0)$ and $R_0$ have unit of resistance.

So the written condition is physically meaningless. Hence Statement–2 is false.

3. Choose the correct option

• Statement–1: True
• Statement–2: False

Therefore, the correct option is: $\boxed{\text{D}}$

4. Comparison with stored answer

Stored correct answer: C

My derived answer: D

I disagree with the stored answer because Statement–1 is clearly true from direct calculation, while Statement–2 is false due to the dimensionally incorrect condition given in it.