MCQ
On the Celsius scale the absolute zero of temperature is at
- A$0^{\circ} \mathrm{C}$
- B$-32^{\circ} \mathrm{C}$
- C$100^{\circ} \mathrm{C}$
- ✓$-273.15^{\circ} \mathrm{C}$
✓
Answer
Correct option: D.
$-273.15^{\circ} \mathrm{C}$
$ T=273.15+t^{\circ} \mathrm{C} \Rightarrow 0=273.15+t^{\circ} \mathrm{C} $
$ \Rightarrow t=-273.15^{\circ} \mathrm{C}$
$ \Rightarrow t=-273.15^{\circ} \mathrm{C}$
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