A) \[{{50}^{o}}C\]
B) \[{{100}^{o}}C\]
C) \[{{273}^{o}}C\]
D) \[{{546}^{o}}C\]
Correct Answer: C
Solution :
Acc. to Charles law, \[V\propto t\] \[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\] \[\frac{2}{273}=\frac{4}{{{T}_{2}}}\] \[{{T}_{2}}=\frac{4\times 273}{2}\] \[{{T}_{2}}=546K\] or \[{{T}_{2}}=546-273\] \[{{T}_{2}}={{273}^{o}}C\]
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