
- A$5, 10,15$
- B$5, 15, 30$
- ✓$5, 25, 35$
- D$5, 15, 25$


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Reason : Electric field at all points on Gaussian surface depends only on charges $q_1$ and $q_2$ .
It may be helpful to use the following: Boltzmann constant $k=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}$; $\frac{e^2}{4 \pi \varepsilon_0}=1.44 \times 10^9 \mathrm{eVm}$.
$1.$ In the core of nuclear fusion reactor, the gas becomes plasma because of
$(A)$ strong nuclear force acting between the deuterons
$(B)$ Coulomb force acting between the deuterons
$(C)$ Coulomb force acting between deuteron-electron pairs
$(D)$ the high temperature maintained inside the reactor core
$2.$ Assume that two deuteron nuclei in the core of fusion reactor at temperature $T$ are moving towards each other, each with kinetic energy $1.5 \mathrm{kT}$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $T$ required for them to reach a separation of $4 \times 10^{-15} \mathrm{~m}$ is in the range
$(A)$ $1.0 \times 10^9 \mathrm{~K}$ $(B)$ $2.0 \times 10^9 \mathrm{~K}$ $(C)$ $3.0 \times 10^9 \mathrm{~K}$ $(D)$ $4.0 \times 10^9 \mathrm{~K}$
$3.$ Results of calculations for four different designs of a fusion reactor using $D-D$ reaction are given below. Which of these is most promising based on Lawson criterion?
$(A)$ deuteron density $=2.0 \times 10^{12} \mathrm{~cm}^{-3}$, confinement time $=5.0 \times 10^{-3} \mathrm{~s}$
$(B)$ deuteron density $=8.0 \times 10^{14} \mathrm{~cm}^{-3}$, confinement time $=9.0 \times 10^{-1} \mathrm{~s}$
$(C)$ deuteron density $=4.0 \times 10^{23} \mathrm{~cm}^{-3}$, confinement time $=1.0 \times 10^{-11} \mathrm{~s}$
$(D)$ deuteron density $=1.0 \times 10^{24} \mathrm{~cm}^{-3}$, confinement time $=4.0 \times 10^{-12} \mathrm{~s}$
Give the answer question $1,2,$ and $3.$

