$|\overrightarrow{\mathrm{C}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos 120^{\circ}}$
$|\overrightarrow{\mathrm{C}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-\mathrm{AB}} \quad\left[\text { Ascos } 120^{\circ}=-\frac{1}{2}\right]$
similarly, $|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos 120^{\circ}}$
$=\sqrt{A^{2}+B^{2}+A B}$
$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|>\mathrm{C}$
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$2\,\frac{{{d^2}x}}{{d{t^2}}} + 32x = 0$
where $x$ is the displacement from the mean position of rest. The period of its oscillation (in seconds) is
Step $1$ It is first compressed adiabatically from volume $V_{1}$ to $1 \;m ^{3}$.
Step $2$ Then expanded isothermally to volume $10 \;m ^{3}$.
Step $3$ Then expanded adiabatically to volume $V _{3}$.
Step $4$ Then compressed isothermally to volume $V_{1}$. If the efficiency of the above cycle is $3 / 4$, then $V_{1}$ is ............ $m^3$
| List-$i$ | List-$2$ |
| $(A)$Kinetic energy of plant | $(1)$ $-\frac{\mathrm{GMm}}{\mathrm{a}}$ |
| $(B)$ Gravaitatioin potentiyal energy of sun -plant system | $(2)$ $\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
| $(C)$Total mecaniacal energy of palnt | $(3)$ $\frac{\mathrm{Gm}}{\mathrm{r}}$ |
| $(D)$Escap energyat the surface of plant for unit mass object | $(4)$ $-\frac{\mathrm{GMm}}{2 \mathrm{a}}$ |
(Where $\mathrm{a}=$ radius of planet orbit, $\mathrm{r}=$ radius of planet, $M=$ mass of Sun, $m=$ mass of planet)
Choose the correct answer from the options given below: