NEET (UG) 2023 — Question 26

Given $T_1 = -50^\circ\mathrm{C} = 223\ \mathrm{K}$. RMS speed $v \propto \sqrt{T}$. Let initial speed be $v$. The speed becomes $4v$ (increased by 3 times means final = initial + 3 × initial). Then $\frac{v_2}{v_1} = \frac{4v}{v} = 4$. Since $v \propto \sqrt{T}$, $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$. So $4 = \sqrt{\frac{T_2}{223}}$. Squaring: $16 = \frac{T_2}{223}$ ⇒ $T_2 = 223 \times 16 = 3568\ \mathrm{K}$. Convert to Celsius: $T_2 = 3568 - 273 = 3295^\circ\mathrm{C}$. Thus the answer is option A.