One mole of an ideal gas having initial volume $V$,pressure $2P$ and temperature $T$ undergoes a cyclic process $ABCDA$ as shown below. The net work done in the complete cycle is:

One mole of an ideal gas goes from an initial state $A$ to a final state $B$ via two processes: It first undergoes isothermal expansion from volume $V$ to $3V$ and then its volume is reduced from $3V$ to $V$ at constant pressure. The correct $P-V$ diagram representing the two processes is:

An ideal gas goes through a reversible cycle $a \to b \to c \to d$ and has the $V - T$ diagram shown below. Processes $d \to a$ and $b \to c$ are adiabatic. The corresponding $P - V$ diagram for the process is (all figures are schematic and not drawn to scale):

Starting with the same initial conditions,an ideal gas expands from volume $V_{i}$ to $V_{f}$ in three different ways. The work done by the gas is $W_{1}$ if the process is purely isothermal,$W_{2}$ if the process is purely adiabatic,and $W_{3}$ if the process is purely isobaric. Then,choose the correct option.

Read the following statements:
$A.$ When the small temperature difference between a liquid and its surroundings is doubled,the rate of loss of heat of the liquid becomes twice.
$B.$ Two bodies $P$ and $Q$ having equal surface areas are maintained at temperatures $10^{\circ}C$ and $20^{\circ}C$. The thermal radiation emitted in a given time by $P$ and $Q$ are in the ratio $1:1.15$.
$C.$ $A$ Carnot engine working between $100 K$ and $400 K$ has an efficiency of $75\%$.
$D.$ When the small temperature difference between a liquid and its surroundings is quadrupled,the rate of loss of heat of the liquid becomes twice.
Choose the correct answer from the options given below:

In Column-$I$ processes and in Column-$II$ the first law of thermodynamics are given. Match them appropriately:

Column-$I$	Column-$II$
$(a)$ Adiabatic	$(i)$ $\Delta Q = \Delta U$
$(b)$ Isothermal	$(ii)$ $\Delta Q = \Delta W$
	$(iii)$ $\Delta U = -\Delta W$