Two long conductors separated by a distance '$$\mathrm{d}$$' carry currents $$I_1$$ and $$I_2$$ in the same direction. They exert a force '$$F$$' on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance between them is also increased to $$3 \mathrm{~d}$$. The new value of force between them is

The a.c. source is connected to series LCR circuit. If voltage across $$R$$ is $$40 \mathrm{~V}$$, that across $$\mathrm{L}$$ is $$80 \mathrm{~V}$$ and that across $$\mathrm{C}$$ is $$40 \mathrm{~V}$$, then the e.m.f. '$$e$$' of a.c. source is

In the study of transistor as an amplifier if $$\alpha=\frac{I_C}{I_E}=0.98$$ and $$\beta=\frac{I_C}{I_B}=49$$, where $$I_C, I_B$$ and $$\mathrm{I}_{\mathrm{E}}$$ are collector, base and emitter current respectively then $$\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)$$ is equal to

A liquid drop of radius '$$R$$' is broken into '$$n$$' identical small droplets. The work done is [T = surface tension of the liquid]