Mechanical Properties of Fluids MCQs for NEET — Physics Questions with Answers

$$\begin{gathered} \mathrm{Volume} \mathrm{of} \mathrm{n} \mathrm{small} \mathrm{drops} = \mathrm{Volume} \mathrm{of} \mathrm{large} \mathrm{drop}. \\ \mathrm{n}\frac{4}{3}{\mathrm{\pi r}}^3 = \frac{4}{3}{\mathrm{\pi R}}^3 \\ {\mathrm{nr}}^3={\mathrm{R}}^3 \\ \mathrm{Change} \mathrm{in} \mathrm{surface} \mathrm{area} \\ ∆\mathrm{A}=\mathrm{n}4{\mathrm{\pi r}}^2-4{\mathrm{\pi R}}^2 \\ =4\mathrm{\pi }\left({\mathrm{nr}}^2-{\mathrm{R}}^2\right) \\ \mathrm{Reduction} \mathrm{in} \mathrm{surface} \mathrm{energy} = \mathrm{T}·∆\mathrm{A} \\ =\mathrm{T}4\mathrm{\pi }\left({\mathrm{nr}}^2-{\mathrm{R}}^2\right) \\ =\mathrm{T}4\mathrm{\pi }\left(\frac{{\mathrm{R}}^3}{\mathrm{r}}-{\mathrm{R}}^2\right) \\ =\mathrm{T}4{\mathrm{\pi R}}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \mathrm{Heat} \Rightarrow \mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}=\frac{\mathrm{T}·4{\mathrm{\pi R}}^3}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\ \mathrm{Q}=\mathrm{ms}∆\mathrm{T} \\ \frac{\mathrm{T}·4{\mathrm{\pi R}}^3}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]=1\times \frac{4}{3}{\mathrm{\pi R}}^3\times 1\times ∆\mathrm{T} \\ ∆\mathrm{T}=\frac{3\mathrm{T}}{\mathrm{J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \end{gathered}$$

1. Pascal’s law

According to the provided text, 'A number of devices, such as hydraulic lift and hydraulic brakes, are based on the Pascal’s law. In these devices, fluids are used for transmitting pressure.'

The NCERT text explicitly states, 'whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is another form of the Pascal’s law.'

The text explains, 'The pressure $P = F_1/A_1$ is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area $A_2$, which results in an upward force of $P \times A_2$. Therefore, $F_2 = PA_2 = (F_1/A_1)A_2$'. This implies $F_1/A_1 = F_2/A_2$.

The NCERT text states that 'the applied force has been increased by a factor of $A_2/A_1$ and this factor is the mechanical advantage of the device.'

From Example 9.5: $F_1 = 10 N$. $D_1 = 1.0 cm$, $D_2 = 3.0 cm$. Areas are proportional to $D^2$. So, $A_2/A_1 = (D_2/D_1)^2 = (3.0/1.0)^2 = 9$. Since $F_2/A_2 = F_1/A_1$, then $F_2 = F_1 (A_2/A_1) = 10 N \times 9 = 90 N$.

As stated in Example 9.5 (b), 'Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston.' This means $V_1 = V_2$, or $A_1 L_1 = A_2 L_2$.

From Example 9.6: Force to be lifted ($F_2$) = mass $\times g = 1350 \text{ kg} \times 9.8 \text{ m/s}^2 = 13230 \text{ N}$. Radii are $r_1 = 5.0 \text{ cm}$ and $r_2 = 15 \text{ cm}$. We know $F_1/A_1 = F_2/A_2$, so $F_1 = F_2 (A_1/A_2) = F_2 (\pi r_1^2 / \pi r_2^2) = F_2 (r_1/r_2)^2$. $F_1 = 13230 \text{ N} \times (5/15)^2 = 13230 \text{ N} \times (1/3)^2 = 13230 \text{ N} \times (1/9) = 1470 \text{ N}$.