When the temperature difference between the source and sink increases,the efficiency of the heat engine

Explain: Carnot cycle and Carnot engine.

Match the temperatures of the source and sink ($T_1$ and $T_2$ respectively) of a Carnot heat engine given in List-$I$ with the corresponding efficiencies given in List-$II$.

List-$I$	List-$II$
$A$. $T_1 = 500 \text{ K}, T_2 = 300 \text{ K}$	$i$. $0.2$
$B$. $T_1 = 500 \text{ K}, T_2 = 350 \text{ K}$	$ii$. $0.3$
$C$. $T_1 = 800 \text{ K}, T_2 = 400 \text{ K}$	$iii$. $0.4$
$D$. $T_1 = 450 \text{ K}, T_2 = 360 \text{ K}$	$iv$. $0.5$

$A$ Carnot engine,whose efficiency is $40 \%$,takes heat from a source maintained at a temperature of $600 \ K$. If it is desired to have an efficiency of $60 \%$,then the intake temperature for the same exhaust (sink) temperature should be: (in $K$)

$A$ Carnot heat engine works with an ideal diatomic gas and an adiabatic volume expansion ratio of $32$. Then its efficiency is ....... $\%$

$A$ diatomic gas is used in a Carnot heat engine. If the efficiency of the given Carnot heat engine is $80\%$,then find the ratio of the initial volume to the final volume of the gas during adiabatic expansion.