1. Find the resistance of the toaster at the final temperature

Given:

• Initial resistance at $27^\circ \mathrm{C}$: $R_0 = 60\,\Omega$
• Supply voltage: $V = 220\,\mathrm{V}$
• Current at heated condition: $I = 2.75\,\mathrm{A}$

Using Ohm’s law, $R = \frac{V}{I} = \frac{220}{2.75} = 80\,\Omega$

So, at the attained temperature, the resistance becomes $R = 80\,\Omega$

2. Use the linear temperature dependence of resistance

For small/moderate temperature ranges, resistance varies as $R = R_0\left[1+\alpha (T-T_0)\right]$

Here,

• $R_0 = 60\,\Omega$
• $T_0 = 27^\circ \mathrm{C}$
• $R = 80\,\Omega$
• $\alpha = 2\times 10^{-4}/^\circ\mathrm{C}$

Substitute: $80 = 60\left[1 + 2\times 10^{-4}(T-27)\right]$

Divide by $60$: $\frac{80}{60} = 1 + 2\times 10^{-4}(T-27)$ $\frac{4}{3} = 1 + 2\times 10^{-4}(T-27)$

So, $\frac{1}{3} = 2\times 10^{-4}(T-27)$

Hence, $(T-27)=\frac{1/3}{2\times 10^{-4}}=\frac{1}{6\times 10^{-4}}=1666.67$

Therefore, $T = 1666.67 + 27 = 1693.67^\circ \mathrm{C}$

So the temperature attained is approximately $\boxed{1694^\circ \mathrm{C}}$

3. Check options

• A: $1235^\circ \mathrm{C}$ — incorrect
• B: $1667^\circ \mathrm{C}$ — this is only the temperature rise, not the final temperature
• C: $694^\circ \mathrm{C}$ — incorrect
• D: $1694^\circ \mathrm{C}$ — correct

Thus, the correct option is D.