A) \[1.8\times {{10}^{5}}N/{{m}^{2}}\]
B) \[1.5\times {{10}^{5}}N/{{m}^{2}}\]
C) \[1.4\times {{10}^{5}}N/{{m}^{2}}\]
D) \[1.2\times {{10}^{5}}N/{{m}^{2}}\]
Correct Answer: B
Solution :
\[\frac{\text{Adiabatic elasticicity}\ ({{E}_{\varphi }})}{\text{Isothermal}\ \text{elasticicity}\ ({{E}_{\theta }})}=\gamma \]Þ\[{{E}_{\theta }}=\frac{{{E}_{\varphi }}}{\gamma }\] Þ \[{{E}_{\theta }}=\frac{2.1\times {{10}^{5}}}{1.4}\]\[=1.5\times {{10}^{5}}N/{{m}^{2}}\]
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