The pressure in the tyre of a car is four times the atmospheric pressure at $300 K$. If this tyre suddenly bursts, its new temperature will be $(\gamma = 1.4)$
  • A$300\,{(4)^{1.4/0.4}}$
  • B$300\,{\left( {\frac{1}{4}} \right)^{ - 0.4/1.4}}$
  • C$300\,{(2)^{ - 0.4/1.4}}$
  • D$300\,{(4)^{ - 0.4/1.4}}$
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