Two drops of same radius are falling through air with steady velocity of $v $ $cm/s$. If the two drops coalesce, what would be the terminal velocity?

A spherical body of radius $R$ consists of a fluid of constant density and is in equilibrium under its own gravity. If $P ( r )$ is the pressure at $r ( r < R )$, then the correct option$(s)$ is(are)

$(A)$ $P ( I =0)=0$ $(B)$ $\frac{ P ( r =3 R / 4)}{ P ( r =2 R / 3)}=\frac{63}{80}$

$(C)$ $\frac{ P ( r =3 R / 5)}{ P ( r =2 R / 5)}=\frac{16}{21}$ $(D)$ $\frac{ P ( r = R / 2)}{ P ( r = R / 3)}=\frac{20}{27}$

A sphere is dropped under gravity through a fluid of viscosity $\eta$ . If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is ($\rho$ = density of sphere ; $r$ = radius)

Give two uses of Stoke’s law.

Two uniform solid balls of same density and of radii $r$ and $2r$ are dropped in air and fall vertically downwards. The terminal velocity of the ball with radius $r$ is $1\,cm\,s^{-1}$ , then find the terminal velocity of the ball of radius $2r$ (neglect bouyant force on the balls.) ........... $cm\,s^{-1}$

As shown schematically in the figure, two vessels contain water solutions (at temperature $T$ ) of potassium permanganate $\left( KMnO _4\right)$ of different concentrations $n_1$ and $n_2\left(n_1>n_2\right)$ molecules per unit volume with $\Delta n=\left(n_1-n_2\right) \ll n_1$. When they are connected by a tube of small length $\ell$ and cross-sectional area $S , KMnO _4$ starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the diffusion. The speed $v$ of the molecules is limited by the viscous force $-\beta v$ on each molecule, where $\beta$ is a constant. Neglecting all terms of the order $(\Delta n)^2$, which of the following is/are correct? ( $k_B$ is the Boltzmann constant)-

$(A)$ the force causing the molecules to move across the tube is $\Delta n k_B T S$

$(B)$ force balance implies $n_1 \beta v \ell=\Delta n k_B T$

$(C)$ total number of molecules going across the tube per sec is $\left(\frac{\Delta n}{\ell}\right)\left(\frac{k_B T}{\beta}\right) S$

$(D)$ rate of molecules getting transferred through the tube does not change with time