An electric bulb of volume $$300 \mathrm{~cm}^3$$ was sealed off during manufacture at a pressure of $$1 \mathrm{~mm}$$ of mercury at $$27{ }^{\circ} \mathrm{C}$$. The number of air molecules contained in the bulb is, $$(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$$ and $$N_A=6.02 \times 10^{23})$$
Find the binding energy of the tritium nucleus: [Given: mass of $$1 \mathrm{H}^3=3.01605 \mathrm{~u} ; \mathrm{~m}_{\mathrm{p}}=1.00782 \mathrm{~u} ; \mathrm{~m}_{\mathrm{n}}=1.00866 \mathrm{~u}$$.]
In a single slit diffraction experiment, for slit width '$$\alpha$$' the width of the central maxima is '$$\beta$$'. If we double the slit width then the corresponding width of the central maxima will be:
Two charges '$$-q$$' each are fixed, separated by distance '$$2 d$$'. A third charge '$$q$$' of mass '$$m$$' placed at the mid-point is displaced slightly by '$$x$$' $$(x< < d)$$ perpendicular to the line joining the two fixed charges as shown in Fig. The time period of oscillation of '$$q$$' will be