B.

0.544 N/m

Given,

Radius of capillary tube = 0.5 mm

                                   = 0.5 × 10^-3 m

Level inside tube = 0.8 cm

                         = 0.8 × 10^-2 m

Angle of contract, θ = 120°

Mass density of mercury, ρ = 13.6 × 10³ kg/m³

Acceleration due to gravity, g = 10 m/s²

             $$\begin{gathered} \mathrm{h} = \frac{2\mathrm{T} \cos \mathrm{\theta }}{\mathrm{r\rho g}} \\ \mathrm{T} = \frac{\mathrm{hr\rho g}}{2 \cos \mathrm{\theta }} \\ = \frac{0.8 \times {10}^{-2} \times 0.5 \times {10}^{-3} \times 13.6 \times {10}^3 \times 10}{2 \times \cos 120} \\ = \frac{0.8 \times {10}^{-2} \times 0.5 \times {10}^{-3} \times 13.6 \times {10}^3 \times 10}{2 \times {\displaystyle \frac{1}{2}}} \\ = 0.8 \times 0.5 \times 13.6 \times {10}^{-1} \\ = 0.544 \mathrm{N}/\mathrm{m} \end{gathered}$$

A.

2 m

Given, radius of first capillary tube = r
Height of rised liquid in first capillary tube = h
Mass of liquid in first capillary tube = m

We know that,

$$\begin{gathered} \mathrm{m} \propto {\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{or} \mathrm{m} \propto {\mathrm{r}}^2\mathrm{h} \\ \mathrm{rh} = \mathrm{constant} \\ \mathrm{So}, \mathrm{h} \propto \frac{1}{\mathrm{r}} \Rightarrow \mathrm{m} \propto \mathrm{r} \\ \mathrm{Now}, \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1} = \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1} \\ \Rightarrow \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1} = \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1} = 2 \\ {\mathrm{m}}_2 = 2{\mathrm{m}}_1 \mathrm{or} {\mathrm{m}}_2 = 2\mathrm{m} \left[\because {\mathrm{m}}_1 = \mathrm{m}\right] \end{gathered}$$

B.

directly proportional to both radius R and velocity v.

Retarding force acting on a ball falling into a viscous fluid
F = 6πηRv
where R = radius of the ball
v = velocity of ball and
η = coefficient of viscosity
∴ F ∝ R and F ∝ v
Or in words, retarding force is proportional to both R and v

D.

4 cm

Water will rise to the full length of capillary tube

C.

air flows from the smaller bubble to the bigger.

The excess pressure inside the soap bubble is inversely proportional to the radius of soap bubble i.e. P ∝1/r, r being the radius of the bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.