A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The viscous force of the liquid on the rising ball is greater than the weight of the ball by a factor of

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 $$=980 \mathrm{~cm} \mathrm{~s}^{-2}$$]

Under isothermal conditions, two soap bubbles of radii '$$r_1$$' and '$$r_2$$' combine to form a single soap bubble of radius '$$R$$'. The surface tension of soap solution is ( $$P=$$ outside pressure)