If the terminal speed of a sphere A [density $$\rho_{\mathrm{A}}=7.5 \mathrm{~kg} \mathrm{~m}^{-3}$$ ] is $$0.4 \mathrm{~ms}^{-1}$$, in a viscous liquid [density $$\rho_{\mathrm{L}}=1.5 \mathrm{~kg} \mathrm{~m}^{-3}$$ ], the terminal speed of sphere B [density $$\rho_B=3 \mathrm{~kg} \mathrm{~m}^{-3}$$ ] of the same size in the same liquid is

A needle is $$7 \mathrm{~cm}$$ long. Assuming that the needle is not wetted by water, what is the weight of the needle, so that it floats on water?

$$\left[\mathrm{T}=\right.$$ surface tension of water $$\left.=70 \frac{\mathrm{dyne}}{\mathrm{cm}}\right]$$

[acceleration due to gravity $$=980 \mathrm{~cm} \mathrm{~s}^{-2}$$]

Water rises in a capillary tube of radius '$$r$$' up to a height '$$\mathrm{h}$$'. The mass of water in a capillary is '$$\mathrm{m}$$'. The mass of water that will rise in a capillary tube of radius $$\frac{'r'}{3}$$ will be

A drop of liquid of density '$$\rho$$' is floating half immersed in a liquid of density '$$d$$'. If '$$T$$' is the surface tension, then the diameter of the drop of the liquid is