A) 16
B) 8
C) 32
D) 64
Correct Answer: C
Solution :
As, \[{{V}_{rms}}=\sqrt{\frac{3RT}{M}}=\frac{{{({{V}_{rms}})}_{1}}}{{{({{V}_{rms}})}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[{{T}_{1}}V_{1}^{\gamma -1}=T_{2}^{\cdot }V_{2}^{\gamma -1}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}\] \[\frac{{{({{V}_{rms}})}_{1}}}{{{({{V}_{rms}})}_{2}}}={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\frac{\gamma -1}{2}}}\] \[=\left[ \frac{{{V}_{2}}}{{{V}_{1}}} \right]\frac{\frac{7}{5}-1}{2}\] \[={{\left[ \frac{{{V}_{2}}}{{{V}_{1}}} \right]}^{\frac{2}{5}\times \frac{1}{2}}}={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\frac{1}{5}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{5}}\] \[{{(2)}^{5}}=32\]
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