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(A) In the given $T-S$ cycle:$b-c$ and $a-d$ are isothermal processes. Since internal energy $U$ of an ideal gas depends only on temperature $(U = nC_

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(A) In the given $T-S$ cycle:$b-c$ and $a-d$ are isothermal processes. Since internal energy $U$ of an ideal gas depends only on temperature $(U = nC_vT)$, $U$ remains constant during these processes. Thus, on a $U-V$ diagram, these appear as horizontal lines.$a-b$ and $c-d$ are isentropic (adiabatic) processes $(S = 	ext{constant})$. For an adiabatic process, $TV^{\gamma-1} = 	ext{constant}$. Since $U \propto T$, we have $UV^{\gamma-1} = 	ext{constant}$, which implies $U \propto V^{1-\gamma}$. This represents a curve on the $U-V$ diagram.Comparing the processes: In $b-c$, $T$ is high, so $U$ is high. In $a-d$, $T$ is low, so $U$ is low. Thus, the cycle on the $U-V$ diagram consists of two horizontal lines at different $U$ levels, connected by two curves. This matches the representation in option $A$.

Column $I$	Column $II$
$(A)$ Process $A \rightarrow B$	$(p)$ Internal energy decreases.
$(B)$ Process $B \rightarrow C$	$(q)$ Internal energy increases.
$(C)$ Process $C \rightarrow D$	$(r)$ Heat is lost.
$(D)$ Process $D \rightarrow A$	$(s)$ Heat is gained.
	$(t)$ Work is done on the gas.

An ideal gas is taken reversibly around the cycle $a-b-c-d-a$ as shown on the temperature $T$ - entropy $S$ diagram. The most appropriate representation of the above cycle on an internal energy $U$ - volume $V$ diagram is

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