A) \[17{}^\circ C\]
B) \[887{}^\circ C\]
C) \[16{}^\circ C\]
D) none of these
Correct Answer: B
Solution :
From the formula \[T{{V}^{\gamma -1}}=\text{constant}\] \[\left[ Here:\,\,{{V}_{2}}=\frac{{{V}_{1}}}{8},\,\,\gamma =\frac{5}{3} \right]\] Hence, \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\] \[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}={{\left( \frac{1}{1/8} \right)}^{5/3-1}}\] \[{{T}_{2}}={{(8)}^{2/3}}\times {{T}_{1}}\] \[=4\times 290=1160\,K=887{}^\circ C\]
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