MCQ
During the thermodynamic process shown in figure for an ideal gas


- A$\Delta T=0$
- B$\Delta Q=0$
- C$W < 0$
- ✓$\Delta U > 0$

For a straight $P-V$ graph line $P \propto V$
If pressure increases, volume increases then $T$ also increases $[P V \propto T]$
So $\Delta T \neq 0$
Volume increasing so work is positive, $W > 0$
and temperature also increasing so $\Delta Q > 0$
$\because \Delta Q=\Delta U+\Delta W$
So $\Delta U > 0$
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$A$: standing on the horizontal surface
$B$: standing on the block To an observer
$A$, the work done by the normal reaction $N$ between the block and the spring on the block is
(Given : permeability of free space $\mu_{0}=4 \pi \times 10^{-7}\;NA ^{-2}$, speed of light in vacuum $c =3 \times 10^{8} \;ms ^{-1}$ )