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
Two metal spheres, one of radius $$\frac{R}{2}$$ and the other of radius $$2 \mathrm{R}$$ respectively have the same surface charge density They are brought in contact and separated. The ratio of their new surface charge densities is