

For $F=a \hat{k}$ where $P$ is $(0, b,-c)$
$T=(b \hat{j}-c \hat{k}) \times(a \hat{k})=a b \hat{i}$
For $F=-a \hat{k}$ where $P$ is $(0,-b,-c)$
$T=(-b \hat{j}-c \hat{k}) \times(-a \hat{k})=a c \hat{i}$
For $F=a \hat{j}$ where $P$ is $(-b, 0,-c)$
$T=(-b \hat{i}-c \hat{k}) \times(a \hat{j})=-a b \hat{k}+a c \hat{i}$
Therefore it has negative $z$ component.
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$Reason :$ When a gas is heated at constant volume some extra heat is needed compared to that at constant pressure for doing work in expansion.

(use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{R}=$ radius of earth)
$(A)$ the rate at which heat is absorbed in the range $0-100 \ K$ varies linearly with temperature $T$.
$(B)$ heat absorbed in increasing the temperature from $0-100 \ K$ is less than the heat required for increasing the temperature from $400-500 \ K$.
$(C)$ there is no change in the rate of heat absorbtion in the range $400-500 \ K$.
$(D)$ the rate of heat absorption increases in the range $200-300 \ K$.