$(a)\;2 \pi \rho_{0} V_{0}\qquad(b)\;-2 \pi \rho_{0} V_{0}\qquad(c)\;\large\frac{ \pi} {2} \rho_{0} V_{0}\qquad(d)\;-\large\frac{ \pi} {2} \rho_{0} V_{0}$

Thermodynamics

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Answer : $\large\frac{ \pi} {2} \rho_{0} V_{0}$

Explanation :

Since $\;\bigtriangleup Q = \bigtriangleup W\;$ for a cycle $\;(\bigtriangleup u=0)$

Therefore , $\;\bigtriangleup Q =\;$ area under the graph

=area of the ellipse with semi - axes length as $\;\rho_{0}\;$ & $\;\large\frac{V_{0}}{2}$

$= \pi (\rho) (\large\frac{V_{0}}{2})$

Since the direction of the process is clockwise with $\;\rho \;$ & V on y - axis & x - axis respectively , $\;\bigtriangleup Q\;$ is +ve

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