Reasoning: According to the principle of floatation, the weight of a floating body equals the weight of the fluid displaced by it. By comparing the immersed volume and the total volume, the density or relative density of the body can be determined.

Reasoning: Archimedes’ principle states that the upward buoyant force acting on an immersed body is equal to the weight of the fluid displaced by that body.

Reasoning: Stokes’ law gives the viscous drag on a sphere: $\mathbf{F =}\mathbf{6\pi\eta rv\ }$
It is commonly applied when the sphere has reached terminal velocity, where its velocity becomes constant and its acceleration is zero.

Reasoning: When two identical drops combine, their total volume doubles:
$\mathbf{R}^{\mathbf{3}}\mathbf{= 2}\mathbf{r}^{\mathbf{3}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R =}\mathbf{2}^{\mathbf{1/3}}\mathbf{\ }\mathbf{r}$
Terminal velocity is proportional to the square of the radius:
$\mathbf{v}_{\mathbf{t}}\mathbf{\propto}$ $\mathbf{r}^{\mathbf{2}}$
Therefore, $\mathbf{\ }\frac{\mathbf{V}}{\mathbf{v}}\mathbf{\ =}\left( \frac{\mathbf{R}}{\mathbf{r}} \right)^{\mathbf{2}}\mathbf{=}\mathbf{2}^{\mathbf{\ 2/3}}$
$\mathbf{V = 20}\mathbf{\ }\mathbf{(}\mathbf{2}^{\mathbf{2/3}}\mathbf{) \approx 20}\mathbf{\ }\mathbf{(1.59) \approx 32\, cm\ }\mathbf{s}^{\mathbf{- 1}}$

Reasoning: According to Stokes’ law: $\mathbf{F = 6\pi\eta rv}$
Thus, viscous force is directly proportional to the radius of the sphere and its velocity: $\mathbf{F \propto rv}$

Reasoning: Terminal velocity is proportional to the square of the radius:
$\mathbf{v}_{\mathbf{t}}\mathbf{\propto}$ $\mathbf{r}^{\mathbf{2}}$
Therefore, $\frac{\mathbf{v}_{\mathbf{1}}}{\mathbf{v}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{r}_{\mathbf{1}}^{\mathbf{2}}}{\mathbf{r}_{\mathbf{2}}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ $\frac{\mathbf{4}}{\mathbf{9}}\mathbf{=}\frac{\mathbf{r}_{\mathbf{1}}^{\mathbf{2}}}{\mathbf{r}_{\mathbf{2}}^{\mathbf{2}}}$ Taking the square root. $\mathbf{\Longrightarrow}$ $\frac{\mathbf{r}_{\mathbf{1}}}{\mathbf{r}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}$

Reasoning: According to the equation of continuity: $\mathbf{A}_{\mathbf{1}\mathbf{\ }}\mathbf{v}_{\mathbf{1}}\mathbf{=}\mathbf{A}_{\mathbf{2}\mathbf{\ }}\mathbf{v}_{\mathbf{2}}$
Since area is proportional to the square of diameter:
$\mathbf{d}_{\mathbf{1}}^{\mathbf{2}}{\mathbf{\ }\mathbf{v}}_{\mathbf{1}}\mathbf{=}\mathbf{d}_{\mathbf{2}}^{\mathbf{2}}{\mathbf{\ }\mathbf{v}}_{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ ${\mathbf{(10}\mathbf{)}}^{\mathbf{2}}\mathbf{\ }\mathbf{(9) =}{\mathbf{(6}\mathbf{)}}^{\mathbf{2}}\mathbf{\ }\mathbf{v}_{\mathbf{2}}$
$\mathbf{v}_{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{100}\mathbf{\ }\mathbf{\times}\mathbf{\ }\mathbf{9}}{\mathbf{36}}\mathbf{= 25}\mathbf{\ }\mathbf{cm}\mathbf{s}^{\mathbf{- 1}}$

Reasoning: Using the equation of continuity: $\mathbf{A}_{\mathbf{1}\mathbf{\ }}\mathbf{v}_{\mathbf{1}}\mathbf{=}\mathbf{A}_{\mathbf{2}\mathbf{\ }}\mathbf{v}_{\mathbf{2}}$
Given, $\mathbf{A}_{\mathbf{2}}\mathbf{= 9}\mathbf{\ }\mathbf{A}_{\mathbf{1}}$ Therefore, $\mathbf{A}_{\mathbf{1}\mathbf{\ }}\mathbf{v}_{\mathbf{1}}\mathbf{= 9}\mathbf{\ }\mathbf{A}_{\mathbf{1}\mathbf{\ }}\mathbf{v}_{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ $\mathbf{v}_{\mathbf{1}}\mathbf{=}\mathbf{9}\mathbf{\ }\mathbf{v}_{\mathbf{2}}$ Thus, the speed at point $\mathbf{1}$ is nine times the speed at point $\mathbf{2}$.

Reasoning: In liquids, increasing temperature weakens intermolecular forces, so viscosity decreases. In gases, increasing temperature increases molecular motion and momentum transfer, so viscosity increases.

Reasoning: According to the equation of continuity: $\mathbf{Av = constant}$
As the speed ‘$\mathbf{v}$’ of falling water increases, its cross-sectional area ‘$\mathbf{A}$’ must decrease to maintain a constant flow rate.

Reasoning: Fast-moving air between the balloons produces lower pressure according to Bernoulli’s principle. The higher air pressure outside the balloons pushes them towards each other.

Reasoning: For laminar flow under a constant pressure difference, the average flow velocity is proportional to the square of the tube's radius:
$\mathbf{v \propto}$ $\frac{\mathbf{r}^{\mathbf{2}}}{\mathbf{\eta}}$
If the radius is doubled. $\mathbf{\Longrightarrow}$ $\mathbf{v}^{\mathbf{'}}\mathbf{\propto}{\mathbf{(2r}\mathbf{)}}^{\mathbf{2}}\mathbf{= 4}\mathbf{r}^{\mathbf{2}}$ Therefore, the flow velocity becomes four times greater.

Reasoning: The Venturi effect occurs in the narrow section of a pipe. As the cross-sectional area decreases, the fluid speed increases and its pressure decreases.

Reasoning: According to the equation of continuity: $\mathbf{A}_{\mathbf{1}\mathbf{\ }}\mathbf{v}_{\mathbf{1}}\mathbf{=}\mathbf{A}_{\mathbf{2}\mathbf{\ }}\mathbf{v}_{\mathbf{2}}$
Given, $\mathbf{v}_{\mathbf{2}}\mathbf{= 4}\mathbf{\ }\mathbf{v}_{\mathbf{1}}$ $\mathbf{\Longrightarrow}$ Therefore, $\mathbf{A}_{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{A}_{\mathbf{1}}}{\mathbf{4}}$ Since area is proportional to the square of diameter:
$\mathbf{d}_{\mathbf{2}}^{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{d}_{\mathbf{1}}^{\mathbf{2}}}{\mathbf{4}}$ $\mathbf{\Longrightarrow}$ $\mathbf{d}_{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{d}_{\mathbf{1}}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{4}}{\mathbf{2}}\mathbf{= 2}\mathbf{\ }\mathbf{cm}$

Reasoning: Bernoulli’s equation in its basic form applies to steady, non-viscous, incompressible flow, typically along a streamline. Unsteady, rotational, and especially turbulent flows violate the simple assumptions used in the standard Bernoulli equation.

Reasoning: In laminar flow, fluid moves smoothly in parallel layers with very little mixing between adjacent layers. Eddies and irregular swirling motions are characteristic of turbulent flow, not laminar flow.