Properties of Matter

A soap bubble has two surfaces (inner and outer). The total surface energy $(U)$ is given by :

$ U=2 \times S \times A=2 \times S \times\left(4 \pi R^2\right) $

where $S$ is the surface tension and $A$ is the surface area.

The work done ( $W$ ) to increase the size is equal to the change in surface energy :

$ \mathrm{W}=\Delta \mathrm{U}=8 \pi \mathrm{~S}\left(\mathrm{R}_2^2-\mathrm{R}_1^2\right) $

$ \mathrm{W}=8 \pi \mathrm{~S}\left[\left(\frac{\mathrm{D}_2}{2}\right)^2-\left(\frac{\mathrm{D}_1}{2}\right)^2\right]=2 \pi \mathrm{~S}\left(\mathrm{D}_2^2-\mathrm{D}_1^2\right) $

The given values are,

Surface tension $\mathrm{N}=0.04 \mathrm{~N} / \mathrm{m}$

Initial diameter $D_1=7 \mathrm{~cm}=0.07 \mathrm{~m}$

Final diameter $\mathrm{D}_2=14 \mathrm{~cm}=0.14 \mathrm{~m}$

So, the work done to increase the size of bubble is,

$ \mathrm{W}=2 \times \frac{22}{7} \times 0.04 \times\left[(0.14)^2-(0.07)^2\right] $

$\Rightarrow $ $\mathrm{W}=\frac{44}{7} \times 0.04 \times[0.0196-0.0049]$

$ \mathrm{W}=0.003696 \mathrm{~J} $

Now, as we know $1 \mathrm{~J}=10^6 \mu \mathrm{~J}$,

$ \mathrm{W}=0.003696 \times 10^6 \mu \mathrm{~J}=3696 \mu \mathrm{~J} $

$\Rightarrow 15000-x=3696$

$\Rightarrow $ $x=15000-3696$

$\Rightarrow $ $ \mathrm{x}=11304 $

Therefore, the value of x is 11304.

Hence, the correct answer is $\mathbf{1 1 3 0 4}$.

When a spherical drop falls through a viscous medium (like air), it reaches a constant speed called terminal velocity $\left(\mathrm{v}_{\mathrm{t}}\right)$ when the net force on it is zero.

The forces involved are :

Weight ( W ) is acting downwards, $\mathrm{W}=\mathrm{mg}=$ Volume $\times \rho \times \mathrm{g}$.

Buoyant force ( $\mathrm{F}_{\mathrm{b}}$ ) is acting upwards.

Viscous drag force ( $\mathrm{F}_{\mathrm{v}}$ ) is acting upwards

According to Stokes' Law $\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{r}$ v.

At equilibrium, the net force must be zero.

$ \mathrm{W}-\mathrm{F}_{\mathrm{b}}=\mathrm{F}_{\mathrm{v}} $

This leads to the standard formula for terminal velocity :

$ v_t=\frac{2 r^2 g(\rho-\sigma)}{9 \eta} $

Where :

• $\mathrm{r}=$ Radius of the drop

• $\rho=$ Density of the drop (water)

• $\sigma=$ Density of the medium (air)

• $\eta=$ Coefficient of viscosity

• $\mathrm{g}=$ Acceleration due to gravity

Since $\mathrm{g}, \rho, \sigma$, and $\eta$ are constants for both the small and big drops, the terminal velocity is directly proportional to the square of the radius.

$ v_t \propto r^2 $

When 64 small drops combine to form one big drop, the total volume remains conserved.

• Let r be the radius of a small drop.

• Let R be the radius of the big drop.

• Let $\mathrm{n}=64$ be the number of drops.

Volume of Big Drop $=\mathrm{n} \times$ (Volume of Small Drop)

$\Rightarrow $ $\frac{4}{3} \pi \mathrm{R}^3=\mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^3$

$\Rightarrow $ $\mathrm{R}^3=\mathrm{n} \cdot \mathrm{r}^3$

$\Rightarrow $ $ \mathrm{R}=\mathrm{n}^{1 / 3} \cdot \mathrm{r} $

Substituting $\mathrm{n}=64$ :

$ \mathrm{R}=64^{1 / 3} \cdot \mathrm{r} $

$\Rightarrow $ $ \mathrm{R}=4 \mathrm{r} $

So, the ratio of the terminal velocities for big and small drop using the fact that terminal velocity is proportional to square of radius,

$ \frac{\mathrm{v}_{\text {big }}}{\mathrm{v}_{\text {small }}}=\frac{\mathrm{R}^2}{\mathrm{r}^2}=\left(\frac{\mathrm{R}}{\mathrm{r}}\right)^2=\left(\frac{4 \mathrm{r}}{4}\right)^2=16 $

Given $\mathrm{v}_{\text {small }}=10 \mathrm{~cm} / \mathrm{s}$.

$ v_{\text {big }}=16 \times v_{\text {small }} $

$\Rightarrow $ $v_{\mathrm{big}}=16 \times 10 \mathrm{~cm} / \mathrm{s}$

$\Rightarrow $ $\mathrm{v}_{\mathrm{big}}=160 \mathrm{~cm} / \mathrm{s}$

Dimensions of Each Quantity

We first write the dimensions of every symbol in the formula.

Time $[\mathrm{t}]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]$

Density $[\rho]=[\sigma]=\left[\frac{\text { Mass }}{\text { Volume }}\right]=\left[\mathrm{ML}^{-3}\right]$

Radius $[\mathrm{r}]=[\mathrm{L}]$

From Stokes' Law ( $F=6 \pi \eta$ rv), we have $\eta=$ F/rv.

Using Stokes’ law, $\eta$ depends on force, radius, and velocity, so its dimensions come from $F$, $r$, and $v$.

Dimension $=\left[\mathrm{MLT}^{-2}\right][\mathrm{L^{-1}}]\left[\mathrm{L^{-1}T}\right]=\left[\mathrm{ML}^{-1}{ \mathrm{~T}^{-1}}\right]$

Constant (A) is dimensionless

According to the given formula $t=A \rho^a r^b \eta^c \sigma^d$ :

$ [\mathrm{T}]=\left[\mathrm{ML}^{-3}\right]^{\mathrm{a}}[\mathrm{~L}]^{\mathrm{b}}\left[\mathrm{ML}^{-1}{ \mathrm{~T}^{-1}}\right]^{\mathrm{c}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{d}} $

$\Rightarrow $ $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1\right]=\left[\mathrm{M}^{\mathrm{a}+\mathrm{c}+\mathrm{d}}\right]\left[\mathrm{L}^{-3 \mathrm{a}+\mathrm{b}-\mathrm{c}-3 \mathrm{~d}}\right]\left[\mathrm{T}^{-\mathrm{c}}\right]$

Comparing the exponents for $\mathrm{M}, \mathrm{L}$, and T on both sides :

For $\mathrm{T}:-\mathrm{c}=1 \Rightarrow \mathrm{c}=-1$

For $\mathrm{M}: \mathrm{a}+\mathrm{c}+\mathrm{d}=0 \Rightarrow \mathrm{a}-1+\mathrm{d}=0 \Rightarrow \mathrm{a}+\mathrm{d}=1$

For $\mathrm{L}:-3 \mathrm{a}+\mathrm{b}-\mathrm{c}-3 \mathrm{~d}=0$

Substituting $\mathrm{c}=-1$ :

$ -3 a+b+1-3 d=0 $

$\Rightarrow $ $ b+1-3(a+d)=0 $

Substituting $(a+d)=1$

$ b+1-3(1)=0 $

$\Rightarrow $ $ b-2=0 \Rightarrow b=2 $

We need to find the value of $\frac{b+c}{a+d}$ :

$ \frac{b+c}{a+d}=\frac{2+(-1)}{1}=1 $

Therefore, the correct answer is $\mathbf{1}$.

When a spherical ball falls through a viscous fluid, it eventually reaches a constant speed called terminal velocity. This occurs when the net force acting on the ball is zero.

The forces acting on the ball are :

• Weight (W) which acts downwards. $\mathrm{W}=\frac{4}{3} \pi \mathrm{r}^3 \sigma \mathrm{~g}$ (where $\sigma$ is the density of the ball).

• Up-thrust $\left(\mathrm{F}_{\mathrm{B}}\right)$ is Acting upwards. $F_B=\frac{4}{3} \pi r^3 \rho g$ (where $\rho$ is the density of the fluid).

• Viscous drag $\left(\mathrm{F}_{\mathrm{v}}\right)$ is acting upwards. According to Stokes' Law, $\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{r} \mathrm{v}$, where $\eta$ is the viscosity of fluid and v is the speed of the ball.

At equilibrium :

$ \mathrm{W}=\mathrm{F}_{\mathrm{B}}+\mathrm{F}_{\mathrm{v}} $

$\Rightarrow $ $\frac{4}{3} \pi r^3 \sigma g=\frac{4}{3} \pi r^3 \rho g+6 \pi \eta r v$

$\Rightarrow $ $6 \pi \eta r v=\frac{4}{3} \pi r^3 g(\sigma-\rho)$

$\Rightarrow $ $ \mathrm{v}=\frac{2 \mathrm{r}^2 \mathrm{~g}(\sigma-\rho)}{9 \eta} $

For balls of the same material in the same fluid :

$ \mathrm{v} \propto \mathrm{r}^2 $

Therefore :

$ \frac{\mathrm{v}_2}{\mathrm{v}_1}=\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right)^2 $

Initial radius of the ball is $r_1=6 \mathrm{~mm}$ and the initial terminal velocity is $v_1=20 \mathrm{~cm} / \mathrm{s}$

Final radius of the ball is $r_2=3 \mathrm{~mm}$

Substituting the values :

$\frac{v_2}{20}=\left(\frac{3}{6}\right)^2$

$\Rightarrow $ $ \frac{v_2}{20}=\frac{1}{4} \Rightarrow v_2=\frac{20}{4}=5 \mathrm{~cm} / \mathrm{s} $

Therefore, the terminal velocity of the second ball will be $5 \mathrm{~cm} / \mathrm{s}$.

Hence, the correct answer is 5.

Given, Initial pressure of liquid $\left(\mathrm{P}_{\mathrm{i}}\right)=1 \mathrm{~atm}$

Final pressure of liquid $\left(\mathrm{P}_{\mathrm{f}}\right)=5 \mathrm{~atm}$

Change in pressure $(\mathrm{dP})=\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}=4 \mathrm{~atm}$

$=4 \times 10^5 \mathrm{~Pa}$

Change in volume $(\mathrm{dV})=-0.8 \mathrm{~cm}^3$

Bulk modulus $(B)=2 \times 10^9 \mathrm{~Pa}$

Now, $B=\frac{-d P}{(d V / V)} \Rightarrow V=-B\left(\frac{d V}{d P}\right)$

$\begin{aligned} \Rightarrow \mathrm{V} & =-2 \times 10^9 \times \frac{\left(-0.8 \times 10^{-6}\right)}{4 \times 10^5} \\ & =4 \times 10^{-3} \mathrm{~m}^3=4 \text { litre } \end{aligned}$

Given, volume of block $=\left(10 \times 10^{-2}\right)^3=10^{-3} \mathrm{~m}^3$

Let density of block $=\rho \mathrm{kg} / \mathrm{m}^3$

mass of block $=\rho \times 10^{-3} \mathrm{~kg}$

Buoyant Force $\left(F_B\right)=1000 \times \frac{10^{-3}}{2} \times 10=5 \mathrm{~N}$

F.B.D. of blocks

Balancing torque about point O , we get

$\begin{aligned} & \operatorname{mg}\left(2 \times 10^{-2}\right)-\mathrm{F}_{\mathrm{B}}\left(2 \times 10^{-2}\right)=0.2 \mathrm{~g}\left(25 \times 10^{-2}\right) \\ & \rho \times 10^{-3} \times 10 \times 2-10=50 \\ & \rho=3000 \mathrm{~kg} / \mathrm{m}^3 \end{aligned}$

Hence, mass of block $=\rho \times 10^{-3}$

$=3000 \times 10^{-3}=3 \mathrm{~kg}$

$\begin{aligned} &\text { Deformation angle }\\ &\begin{aligned} & 2 \theta_1=\theta_2 \\ & \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2} \end{aligned} \end{aligned}$

$\begin{aligned} & \Rightarrow 2\left(\frac{F}{\ell d_1 \eta_1}\right)=\frac{F}{\ell d_2 \eta_2} \\ & \Rightarrow \eta_2=\frac{\eta_1}{4}=1 \times 10^9 \Rightarrow x=1 \end{aligned}$

The excess pressure inside a soap bubble is determined by the formula:

$ \Delta \mathrm{P} = \frac{4 \mathrm{~T}}{\mathrm{R}} $

where $ \Delta \mathrm{P} $ is the excess pressure, $ \mathrm{T} $ is the surface tension of the soap film, and $ \mathrm{R} $ is the radius of the bubble.

Given that the excess pressure inside bubble A is half that inside bubble B, we have:

$ \Delta \mathrm{P}_{\mathrm{A}} = \frac{1}{2} \Delta \mathrm{P}_{\mathrm{B}} $

This implies:

$ \frac{R_{\mathrm{A}}}{R_{\mathrm{B}}} = \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{A}}} = 2 $

Now, considering the volumes of the bubbles, the relationship between volume and radius for a sphere is given by:

$ V = \frac{4}{3} \pi R^3 $

Thus, the ratio of the volumes of bubbles A and B is:

$ \frac{V_{\mathrm{A}}}{V_{\mathrm{B}}} = \left( \frac{R_{\mathrm{A}}}{R_{\mathrm{B}}} \right)^3 = 2^3 = 8 $

Therefore, the volume of bubble A is $ n = 8 $ times the volume of bubble B.

To find the original length of the string, we use the relationship between tension, elasticity constant ($ K $), and the change in length of the string.

Given the equation for tension:

$ \mathrm{T} = \mathrm{K}(\ell - \ell_0) $

where $\ell$ is the length of the string under tension and $\ell_0$ is the original length.

When the tension is 5 N, the equation becomes:

$ 5 = \mathrm{K}(1.4 - \ell_0) $

When the tension increases to 7 N, the equation is:

$ 7 = \mathrm{K}(1.56 - \ell_0) $

By setting up a ratio from the two equations, we have:

$ \frac{5}{1.4 - \ell_0} = \frac{7}{1.56 - \ell_0} $

Solving this proportion gives us the original length $ \ell_0 $:

$ \ell_0 = 1 \, \mathrm{m} $

Thus, the original length of the string is 1 meter.

To find the elastic potential energy density of the steel wire, we need to use the given information and formulae for strain and energy density.

Given:

The length of the wire, $ \ell = 2 \, \text{m} $

Young's modulus, $ Y = 2.0 \times 10^{11} \, \text{N/m}^2 $

Poisson's ratio, $ \mu = 0.2 $

Transverse strain, $ \frac{\Delta r}{r} = 10^{-3} $

The formula for Poisson's ratio is:

$ \mu = -\frac{\left(\frac{\Delta r}{r}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} $

From this, we solve for the longitudinal strain $\frac{\Delta \ell}{\ell}$:

$ \frac{\Delta \ell}{\ell} = \frac{1}{\mu} \times \left(\frac{\Delta r}{r}\right) $

Substitute the given values:

$ \frac{\Delta \ell}{\ell} = \frac{1}{0.2} \times 10^{-3} = 5 \times 10^{-3} $

The elastic potential energy density $ u $ is given by:

$ u = \frac{1}{2} Y \varepsilon_{\ell}^2 $

where $ \varepsilon_{\ell} = \frac{\Delta \ell}{\ell} $. Plug in the values:

$ u = \frac{1}{2} \times 2 \times 10^{11} \times \left(5 \times 10^{-3}\right)^2 $

Simplify further:

$ u = \frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{-6} $

$ u = 25 \times 10^5 \, \text{(in SI units)} $

Thus, the elastic potential energy density of the wire is $ 25 \times 10^5 $ SI units.

Step 1: Find the Pressure on Each Side of Window

At the window's depth, the pressure on each side comes from the liquid above. On one side is water, on the other is a denser liquid.

Step 2: Pressures Acting on Both Sides

The total pressure at depth $h = 3~\mathrm{m}$ in each liquid is:

Pressure from water side: $P_w = P_0 + \rho_w g h$

Pressure from dense liquid side: $P_\ell = P_0 + \rho_\ell g h$

Where:
$P_0$ = atmospheric pressure (cancels out later)
$\rho_w = 1000~\mathrm{kg}/\mathrm{m}^3$
$\rho_\ell = 1500~\mathrm{kg}/\mathrm{m}^3$
$g = 10~\mathrm{m}/\mathrm{s}^2$
$h = 3~\mathrm{m}$

Step 3: Net Force on Window

The net force on the window is the difference in pressure between the two liquids, times the window's area:

$F_{ext} = (P_\ell - P_w) \cdot A$

Step 4: Simplifying the Force Expression

The atmospheric pressure, $P_0$, is on both sides, so it cancels out:

$F_{ext} = (\rho_\ell g h - \rho_w g h) \cdot A$
$= (\rho_\ell - \rho_w) g h A$

Step 5: Substitute Values

$(\rho_\ell - \rho_w) = 1500 - 1000 = 500~\mathrm{kg}/\mathrm{m}^3$
$g = 10~\mathrm{m}/\mathrm{s}^2$
$h = 3~\mathrm{m}$
Window area, $A = 100~\mathrm{cm}^2 = 100 \times 10^{-4}~\mathrm{m}^2 = 0.01~\mathrm{m}^2$

So, $F_{ext} = 500 \times 10 \times 3 \times 0.01 = 150~\mathrm{N}$

Answer: The required force to keep the window closed is $150~\mathrm{N}$.

To find the volume contraction of the copper cube, we need to use the formula for volumetric strain under pressure, which involves the bulk modulus (K). The volumetric strain is given by the formula:

$ \text{Volumetric strain} = \frac{\Delta V}{V} = -\frac{P}{K} $

Where:

$ \Delta V $ is the change in volume or volume contraction,

$ V $ is the original volume of the cube,

$ P $ is the hydraulic pressure applied,

$ K $ is the bulk modulus of the material.

Given:

Edge length of the cube, $ a = 10 \, \text{cm} = 0.1 \, \text{m} $

Hydraulic pressure, $ P = 7 \times 10^6 \, \text{Pa} $

Bulk modulus of copper, $ K = 1.4 \times 10^{11} \, \text{Pa} $

Calculate the original volume ($ V $) of the cube:

$ V = a^3 = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 $

Calculate the volumetric strain:

$ \text{Volumetric strain} = -\frac{P}{K} = -\frac{7 \times 10^6}{1.4 \times 10^{11}} = -5 \times 10^{-5} $

Calculate the volume contraction ($ \Delta V $):

$ \Delta V = V \times \text{Volumetric strain} = 0.001 \, \text{m}^3 \times (-5 \times 10^{-5}) = -5 \times 10^{-8} \, \text{m}^3 $

Convert $\Delta V$ from cubic meters to cubic millimeters:

1 cubic meter = $10^9$ cubic millimeters. Therefore,

$ \Delta V = -5 \times 10^{-8} \, \text{m}^3 \times 10^9 \, \text{mm}^3/\text{m}^3 = -50 \, \text{mm}^3 $

Thus, the volume contraction of the copper cube, when subjected to the given hydraulic pressure, is $50 \, \text{mm}^3$.

Given, measured quantity $ = {{Modulus\,of\,elasticity} \over {torque}}$

$ = {\sigma \over {\varepsilon \tau }} = {F \over {A\varepsilon \tau }}$ where, $\sigma$ = stress, $\varepsilon $ = strain, $\tau$ = torque

$ = {F \over {A\varepsilon FL}}$

So, dimension $ = {1 \over {[{L^2}][L]}}$ ($\varepsilon $ is dimensionless)

$ = [{L^{ - 3}}] = [{M^0}{L^{ - 3}}{T^0}] = [{M^a}{L^b}{T^c}]$

$ \Rightarrow c = 0$

The bulk modulus of a material is defined as:

$ B = \frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)} $

Given:

Bulk Modulus, $ B = 2.15 \times 10^9 \, \text{Nm}^{-2} $

Change in volume percentage, $ \frac{\Delta V}{V} = 0.2\% = \frac{0.2}{100} = 0.002 $

To find the required increase in pressure, $ \Delta P $, we rearrange the formula for the bulk modulus:

$ B = \frac{-\Delta P}{- \left(\frac{\Delta V}{V}\right)} $

Plug in the known values:

$ 2.15 \times 10^9 = \frac{\Delta P}{0.002} $

Solving for $ \Delta P $:

$ \Delta P = 2.15 \times 10^9 \times 0.002 $

$ \Delta P = 4.3 \times 10^6 \, \text{Nm}^{-2} $

Since the problem specifies $ \Delta P = P \times 10^5 \, \text{Nm}^{-2} $, we have:

$ 4.3 \times 10^6 = P \times 10^5 $

$ P = 43 $

The value of $ P $ is therefore 43.

To find the difference between the pressure inside the air bubble and the atmospheric pressure, we use the fact that at depth $h$ in a liquid of density $\rho$, the external (hydrostatic) pressure exceeds atmospheric by $\rho g h$. Additionally, for a spherical bubble in a liquid, the internal pressure exceeds the external pressure by $\frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble.

Hence, the internal pressure of the bubble is

$ P_\text{inside} = P_\text{atm} + \rho g h + \frac{2T}{r}. $

Therefore, the difference between the bubble's internal pressure and the atmospheric pressure is

$ \Delta P = P_\text{inside} - P_\text{atm} = \rho g h + \frac{2T}{r}. $

Substituting the given values:

$\rho = 10^3 \,\text{kg/m}^3$

$g = 10 \,\text{m/s}^2$

$h = 0.20 \,\text{m}$

$T = 0.095 \,\text{J/m}^2$

$r = 1.0 \times 10^{-3} \,\text{m}$

we compute each term:

Hydrostatic term:

$ \rho g h = (10^3 \,\text{kg/m}^3) \cdot (10 \,\text{m/s}^2) \cdot (0.20 \,\text{m}) = 2000 \,\text{N/m}^2. $

Surface tension term:

$ \frac{2T}{r} = \frac{2 \times 0.095 \,\text{J/m}^2}{1.0 \times 10^{-3} \,\text{m}} = \frac{2 \times 0.095}{10^{-3}} = 2 \times 95 = 190 \,\text{N/m}^2. $

Hence,

$ \Delta P = 2000 \,\text{N/m}^2 + 190 \,\text{N/m}^2 = 2190 \,\text{N/m}^2. $

Thus, the required pressure difference between the inside of the bubble and the atmospheric pressure is

$ \boxed{2190 \,\text{N/m}^2.} $

To find the radius of curvature of the common surface between two soap bubbles, we use the formula:

$ r = \left| \frac{{r_1 \cdot r_2}}{{r_1 - r_2}} \right| $

Given that $ r_1 = 2 \, \text{cm} $ and $ r_2 = 4 \, \text{cm} $, we substitute these values into the formula:

$ r = \left| \frac{2 \cdot 4}{2 - 4} \right| = \left| \frac{8}{-2} \right| = 4 \, \text{cm} $

Thus, the radius of curvature of the common surface is $ 4 \, \text{cm} $.

To determine the extension in the length of the wire, we can use the formula derived from Young's modulus:

$ \text{Young's Modulus (Y)} = \frac{\text{Stress}}{\text{Strain}} $

Where:

Stress ($\sigma$) is given by:

$ \sigma = \frac{F}{A} $

and Strain ($\epsilon$) is:

$ \epsilon = \frac{\Delta L}{L} $

Here,

• $F$ is the force exerted,
• $A$ is the cross-sectional area,
• $\Delta L$ is the change in length,
• $L$ is the original length.

We are given:

• $F = 200 \mathrm{~N}$ (each person pulls with this force, but the total force in the wire should be considered as the tension experienced by one side, so it still remains 200 N),
• $A = 2 \mathrm{~cm}^2 = 2 \times 10^{-4} \mathrm{~m}^2$,
• $L = 2 \mathrm{~m}$,
• $Y = 1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$.

First, calculate the stress ($\sigma$):

$ \sigma = \frac{200 \mathrm{~N}}{2 \times 10^{-4} \mathrm{~m}^2} = 10^6 \mathrm{~N} \mathrm{~m}^{-2} $

Using Young’s modulus formula:

$ Y = \frac{\text{Stress}}{\text{Strain}} \implies \text{Strain} = \frac{\text{Stress}}{Y} $

Thus, the strain ($\epsilon$) is:

$ \epsilon = \frac{10^6 \mathrm{~N} \mathrm{~m}^{-2}}{1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}} = 10^{-5} $

Now, relate the strain to the change in length:

$ \epsilon = \frac{\Delta L}{L} \implies \Delta L = \epsilon \times L = 10^{-5} \times 2 \mathrm{~m} = 2 \times 10^{-5} \mathrm{~m} $

Convert this change in length to micrometers:

$ 1 \mathrm{~m} = 10^6 \mu \mathrm{m} $

$ 2 \times 10^{-5} \mathrm{~m} = 2 \times 10^{-5} \times 10^6 \mu \mathrm{m} = 20 \mu \mathrm{m} $

Therefore, the wire will extend in length by $20 \mu \mathrm{m}$.

To find the new terminal velocity of the larger drop formed by the coalescence of 8 smaller droplets, we need to understand the relationship between the radius of the droplets and their terminal velocity.

The terminal velocity for a small spherical droplet falling through the air is given by the Stokes' law:

$ v_t = \frac{2}{9} \frac{r^2 (\rho - \rho_{\text{air}}) g}{\eta} $

where:

• $v_t$ is the terminal velocity.
• $r$ is the radius of the droplet.
• $\rho$ is the density of the droplet.
• $\rho_{\text{air}}$ is the density of the air.
• $g$ is the acceleration due to gravity.
• $\eta$ is the viscosity of the air.

Given that the radius of the small droplets is $0.01 \ \text{mm}$ and their terminal velocity is $10 \ \text{cm/s}$, we now need to determine the radius of the larger drop formed by the coalescence of 8 smaller droplets.

When droplets coalesce, the volume of the larger drop is equal to the sum of the volumes of the smaller droplets. The volume of a sphere is given by:

$ V = \frac{4}{3} \pi r^3 $

Therefore, the volume of the large drop (V_large) can be calculated by:

$ V_{\text{large}} = 8 \times V_{\text{small}} = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right) $

Let the radius of the larger drop be $R$. Then:

$ \frac{4}{3} \pi R^3 = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right) $

Simplifying, we get:

$ R^3 = 8 r_{\text{small}}^3 $

Taking the cube root on both sides:

$ R = 2 r_{\text{small}} $

Therefore, the radius of the larger drop is twice the radius of the smaller droplet:

$ R = 2 \times 0.01 \ \text{mm} = 0.02 \ \text{mm} $

The terminal velocity of a droplet is proportional to the square of its radius. Therefore:

$ v_{t_{\text{large}}} \propto R^2 $

Given that the terminal velocity of the smaller droplets is 10 cm/s, the terminal velocity of the larger drop (formed by coalescing 8 smaller droplets) is:

$ v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{R}{r_{\text{small}}} \right)^2 $

$ v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{0.02 \ \text{mm}}{0.01 \ \text{mm}} \right)^2 $

$ v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( 2 \right)^2 $

$ v_{t_{\text{large}}} = 10 \ \text{cm/s} \times 4 $

$ v_{t_{\text{large}}} = 40 \ \text{cm/s} $

Therefore, the new terminal velocity of the larger drop will be 40 cm/s.

Let's start by understanding the problem. We need to determine the diameter of the soap bubble, given certain properties of the liquid and the soap solution.

The excess pressure inside a soap bubble can be calculated using the formula:

$ \Delta P = \frac{4T}{r} $

where:

• $\Delta P$ is the excess pressure.
• $T$ is the surface tension of the soap solution.
• $r$ is the radius of the soap bubble.

The column of liquid balances this excess pressure, and the pressure exerted by the liquid column is given by:

$ P = \rho g h $

where:

• $\rho$ is the density of the liquid.
• $g$ is the acceleration due to gravity.
• $h$ is the height of the liquid column.

Thus, we have:

$ \rho g h = \frac{4T}{r} $

Given values:

• $ \rho = 8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3} $
• $ g = 10 \mathrm{~m} \mathrm{~s}^{-2} $
• $ h = 0.04 \mathrm{~cm} = 0.04 \times 10^{-2} \mathrm{~m} $
• $ T = 0.28 \mathrm{~Nm}^{-1} $

Substituting the values into the pressure balance equation:

$ 8 \times 10^3 \times 10 \times 0.04 \times 10^{-2} = \frac{4 \times 0.28}{r} $

Simplifying, we get:

$ 8 \times 10^3 \times 10 \times 0.04 \times 10^{-2} = 3200 \times 10^{-4} = 0.32 $

So,

$ 0.32 = \frac{1.12}{r} $

Solving for $r$, we get:

$ r = \frac{1.12}{0.32} $

$ r \approx 3.5 \mathrm{~cm} $

We need the diameter, which is twice the radius:

$ \text{Diameter} = 2r = 2 \times 3.5 = 7 \mathrm{~cm} $

Therefore, the diameter of the soap bubble is $ 7 \mathrm{~cm} $.

$\begin{aligned} \Delta I_{\text {spring }} & =\left(\sqrt{L^2+x^2}\right)-L \\ & =\frac{x^2}{2 L} \\ T=k \Delta I & =\frac{k n^2}{2 L} \end{aligned}$

$\begin{array}{rl} &\Rightarrow 2 T \theta=m g \\ & 2\left(\frac{Y A}{l}\right) \frac{x^2}{2 l} \theta=m g \\ & A =\frac{m g}{\theta^3 Y} \\ & =\frac{20}{10^{-6} \times 2 \times 10^{11}} \\ & =10 \times 10^{-5} \\ & =1 \times 10^{-4} \end{array}$

$R_{\text {big }}=10 R_{\text {small }}$

$ \Rightarrow {{{E_{1000}}} \over {{E_{big}}}} = {{1000 \times T \times 4\pi {{\left[ {{{{R_{big}}} \over {10}}} \right]}^2}} \over {T \times 4\pi R_{big}^2}} = {{10} \over 1}$

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\quad$ (By Pascal's law)

$\begin{aligned} \Rightarrow F_1 & =10\left(\frac{14^2}{1.4^2}\right) \\ & =10 \times 100 \\ & =1000 \mathrm{~N} \end{aligned}$

The breaking stress of a wire, denoted as $\sigma$, is the maximum tension per unit area it can withstand before it breaks. It is given as:

$ \sigma = \frac{F}{A} $

where $F$ is the breaking force and $A$ is the cross-sectional area of the wire. The weight of the wire, when it's on the verge of breaking, equals the maximum force $F$ it can sustain. The weight of an object is given by:

$ W = mg $

where $m$ is the mass of the object and $g$ is the acceleration due to gravity. Since the wire's mass, $m$, can also be expressed in terms of its density ($\rho$), length ($L$), and area ($A$) as:

$ m = \rho V = \rho A L $

We can substitute this expression in the equation for weight:

$ W = \rho A L g $

On a planet where the acceleration due to gravity is $\frac{1}{3}$rd of that on Earth, we substitute $g_{\text{planet}} = \frac{1}{3}g_{\text{Earth}} = \frac{1}{3} \times 10 \, \text{m/s}^2 = \frac{10}{3} \, \text{m/s}^2$. The breaking force $F$ which is equivalent to the weight at the breaking point, is therefore given by:

$ F = \rho A L g_{\text{planet}} $

Since the breaking stress $\sigma$ is also $F/A$, we can set the two expressions equal to find the maximum length $L$ of the wire before breaking:

$ \sigma = \frac{\rho A L g_{\text{planet}}}{A} $

$ \sigma = \rho L g_{\text{planet}} $

Solving for $L$:

$ L = \frac{\sigma}{\rho g_{\text{planet}}} $

Given that $\sigma = 1.2 \times 10^8 \, \text{N/m}^2$, $\rho = 6 \times 10^4 \, \text{kg/m}^3$, and $g_{\text{planet}} = \frac{10}{3} \, \text{m/s}^2$, we can substitute these values into the formula:

$ L = \frac{1.2 \times 10^8}{6 \times 10^4 \times \left( \frac{10}{3} \right) } $

$ L = \frac{1.2 \times 10^8}{2 \times 10^5} $

$ L = 600 \, \text{m} $

Therefore, the maximum length of the wire that can be suspended from a rigid support on this planet without breaking is $600$ meters.

$\begin{aligned} F & =\left(p_0+\rho g h\right) A \\ & =\left(10^5+1.36 \times 10^4 \times 10 \times \frac{3}{10}\right) \frac{22}{7}\left(\frac{2}{100}\right)^2 \\ & =177 \mathrm{~N} \end{aligned}$

$\begin{aligned} & \Delta W=8 \pi\left(R_2^2-R_1^2\right) T \\ & 36960=8 \times \frac{22}{7} \times 40\left(R_2^2-\frac{49}{4}\right) \\ & R_2=7 \mathrm{~cm} \end{aligned}$

To determine the tension for the elastic spring's length of $(3a - 2b)$, we can use Hooke's law which states that the force exerted by a spring is proportional to the extension or compression of the spring from its natural length. Mathematically, Hooke's law is given by:

$ F = k \cdot \Delta x $

where:

• $F$ is the force exerted by the spring.
• $k$ is the spring constant (a measure of the stiffness of the spring).
• $\Delta x$ is the displacement from the natural length.

Let's denote the natural (unstretched) length of the spring as $l_0$. Given that the spring's length is $a$ under a tension of $3 \mathrm{~N}$, and its length is $b$ under a tension of $2 \mathrm{~N}$, we can write the equations as:

$ 3 \mathrm{~N} = k \cdot (a - l_0) $

$ 2 \mathrm{~N} = k \cdot (b - l_0) $

Next, we need to find the length displacement $(3a - 2b)$ in terms of the natural length $l_0$. We express the displacements as:

$ x = (3a - 2b) - l_0 $

To solve for this, let's express $a - l_0$ and $b - l_0$ from the given conditions:

$ a - l_0 = \frac{3 \mathrm{~N}}{k} $

$ b - l_0 = \frac{2 \mathrm{~N}}{k} $

Substitute these into the displacement equation:

$ x = (3a - 2b) - l_0 = 3 \left( \frac{3 \mathrm{~N}}{k} + l_0 \right) - 2 \left( \frac{2 \mathrm{~N}}{k} + l_0 \right) - l_0 $

On simplifying, we get:

$ x = 3 \frac{3 \mathrm{~N}}{k} + 3 l_0 - 2 \frac{2 \mathrm{~N}}{k} - 2 l_0 - l_0 $

$ x = \frac{9 \mathrm{~N}}{k} + 3 l_0 - \frac{4 \mathrm{~N}}{k} - 3 l_0 $

$ x = \frac{5 \mathrm{~N}}{k} $

Finally, by Hooke's Law, the force corresponding to this displacement is:

$ F = k \cdot x = k \cdot \frac{5 \mathrm{~N}}{k} = 5 \mathrm{~N} $

Therefore, the tension required for the elastic spring's length $(3a - 2b)$ is 5 N.

To solve this problem, we use the fact that the longitudinal strain in a wire is given by the formula:

where $\Delta L$ is the change in length, $L$ is the original length, $F$ is the force applied, $A$ is the area of cross-section of the wire, and $Y$ is the Young's modulus of the material of the wire.

The force applied by each load due to gravity is calculated using $F = mg$, where $m$ is the mass of the load and $g$ is the acceleration due to gravity ($10 \text{ m/s}^2$ in this case).

For the upper wire, the total force applied is the weight of both masses (2 kg and 1 kg):

For the lower wire, the force applied is just the weight of the 1 kg mass:

Since the area $A$ and Young's modulus $Y$ are the same for both wires, these values cancel out when we calculate the ratio of the strains. Therefore, the ratio of the strains is directly proportional to the ratio of the forces:

Hence, the ratio of longitudinal strain of the upper wire to that of the lower wire is 3.

To solve this problem, we need to employ Bernoulli's equation, which is applied to describe the behavior of fluid flow. For a fluid in steady flow, the principle states that the sum of the pressure potential energy density, kinetic energy density, and the gravitational potential energy density has the same value at all points along a streamline. Since the plane is in level flight, we can ignore changes in gravitational potential energy. Bernoulli's equation can be written as:

$ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2, $

where,

• $P_1$ and $P_2$ are the static pressures on the lower and upper surfaces of the wing respectively,
• $v_1$ is the airspeed over the lower surface of the wing,
• $v_2$ is the airspeed over the upper surface of the wing,
• $\rho$ is the density of air,
• $P_2 > P_1$ because $v_2 > v_1$.

First, let's convert the airspeeds to $ \mathrm{m/s} $:

$ v_1 = 180 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 50 \mathrm{m/s}, $

$ v_2 = 252 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 70 \mathrm{m/s}. $

The pressure difference between the lower and upper wing surfaces can thus be calculated using Bernoulli's equation:

$ \Delta P = P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2. $

Substitute the values (with $\rho = 1 \mathrm{~kg/m}^3$):

$ \Delta P = \frac{1}{2} (1 \mathrm{~kg/m}^3) (70 \mathrm{m/s})^2 - \frac{1}{2} (1 \mathrm{~kg/m}^3) (50 \mathrm{m/s})^2, $

$ \Delta P = \frac{1}{2} (4900 \mathrm{~kg/m \cdot s}^2) - \frac{1}{2} (2500 \mathrm{~kg/m \cdot s}^2), $

$ \Delta P = \frac{1}{2} (2400 \mathrm{~kg/m \cdot s}^2), $

$ \Delta P = 1200 \mathrm{~N/m}^2. $

The lift force generated by the pressure difference over one wing is $ \Delta P \times \text{wing area} $, and since there are two wings, we must double the lift force generated by one wing to find the total lift force sustaining the plane. The weight of the plane is effectively the lift force when in level flight at constant speed. So:

$ \text{Lift} = 2 \times \Delta P \times \text{wing area}, $

$ \text{Lift} = 2 \times 1200 \mathrm{~N/m}^2 \times 40 \mathrm{~m}^2, $

$ \text{Lift} = 2 \times 48000 \mathrm{N}, $

$ \text{Lift} = 96000 \mathrm{N}. $

To find the mass $m$ of the plane, we use Newton's second law, where lift force is equal to the weight ($ mg $, where $g$ is the acceleration due to gravity):

$ m g = \text{Lift}, $

$ m = \frac{\text{Lift}}{g}, $

$ m = \frac{96000 \mathrm{N}}{10 \mathrm{m/s}^2}, $

$ m = 9600 \mathrm{kg}. $

Therefore, the mass of the plane is 9600 kg.

$\begin{aligned} & \mathrm{T}=\left(\frac{2 \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=\frac{80}{3} \mathrm{~N} \\ & \mathrm{~A}=\pi \mathrm{r}^2=16 \pi \times 10^{-10} \mathrm{~m}^2 \\ & \text { Strain }=\frac{\Delta \ell}{\ell}=\frac{\mathrm{F}}{\mathrm{AY}}=\frac{\mathrm{T}}{\mathrm{AY}} \\ & =\frac{80 / 3}{16 \pi \times 10^{-10} \times 2 \times 10^{11}}=\frac{1}{12 \pi} \\ & \alpha=12 \end{aligned}$

$\begin{aligned} & \beta=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \\ & \Delta \mathrm{P}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\ & \rho \mathrm{gh}=-\beta \frac{\Delta \mathrm{V}}{\mathrm{V}} \\ & 10^3 \times 10 \times \mathrm{h}=-9 \times 10^8 \times\left(-\frac{0.02}{100}\right) \\ & \Rightarrow \mathrm{h}=18 \mathrm{~m} \end{aligned}$

$\begin{aligned} & \rho\left({ }_3^4 \pi r^3\right) 1000={ }_3^4 \pi R^3 \rho \\ & R=10 r \\ & E_1=1000 \times 4 \pi r^2 \times S \\ & E_2=4 \pi(10 r)^2 S \\ & \frac{E_1}{E_2}=\frac{10}{1}, x=10 \end{aligned}$

$\begin{aligned} & a=\frac{10}{3} \mathrm{~m} / \mathrm{s}^2 \\\\ & 30-T_1=3 \times a \\\\ & T_1=20 \mathrm{~N} \\\\ & \text { strain }=\frac{\text { stress }}{Y} \\\\ & =2 \times 10^{-4} \end{aligned}$

$\mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell}$

$\begin{aligned} & \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \mathrm{V}=\mathrm{A} \ell \Rightarrow \ell=\frac{\mathrm{V}}{\mathrm{A}} \\ & \Delta \ell=\frac{\mathrm{FV}}{\mathrm{A}^2 \mathrm{Y}} \end{aligned}$

$Y ~\& V$ is same for both the wires

$\begin{aligned} & \Delta \ell \propto \frac{\mathrm{F}}{\mathrm{A}^2} \\ & \frac{\Delta \ell_1}{\Delta \ell_2}=\frac{\mathrm{F}_1}{\mathrm{~A}_1^2} \times \frac{\mathrm{A}_2^2}{\mathrm{~F}_2} \\ & \Delta \ell_1=\Delta \ell_2 \\ & \mathrm{~F}_1 \mathrm{~A}_2^2=\mathrm{F}_2 \mathrm{~A}_1^2 \\ & \frac{\mathrm{F}_1}{\mathrm{~F}_2}=\frac{\mathrm{A}_1^2}{\mathrm{~A}_2^2}=\left(\frac{4}{1}\right)^2=16 \end{aligned}$

$\begin{aligned} & \mathrm{F}=\frac{1}{2} \rho\left(\mathrm{v}_1^2-\mathrm{v}_2^2\right) \mathrm{A} \\ & \mathrm{F}=\frac{1}{2} \times 1.2 \times\left(70^2-65^2\right) \times 2 \\ & =810 \mathrm{~N} \end{aligned}$

$\begin{aligned} & \text { Change in pressure }=\frac{1}{2} \rho \mathrm{v}^2 \\ & 4.5 \times 10^4-2.0 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\ & 2.5 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\ & \mathrm{v}^2=50 \\ & \mathrm{v}=\sqrt{50} \\ & \text { Velocity of water }=\sqrt{\mathrm{V}}=\sqrt{50} \\ & =\mathrm{V}=50 \end{aligned}$

$\begin{aligned} & \mathrm{B}=-\frac{\Delta \mathrm{P}}{\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)} \\ & -\left(\frac{\Delta \mathrm{V}}{\mathrm{V}}\right)=\frac{\rho \mathrm{gh}}{\mathrm{B}}=\frac{1000 \times 10 \times 4000}{2 \times 10^9} \\ & =2 \times 10^{-2}[-\mathrm{ve} \text { sign represent compression }] \end{aligned}$

We can use the Bernoulli equation and continuity equation to solve this problem. The Bernoulli equation is given by:

$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$

The continuity equation is given by:

$A_1 v_1 = A_2 v_2$

From the given data, we have:

$P_1 - P_2 = 3 \mathrm{~Nm}^{-2}$ $A_1 = 10 \mathrm{~cm}^2 = 10 \times 10^{-4} \mathrm{~m}^2$ $A_2 = 5 \mathrm{~cm}^2 = 5 \times 10^{-4} \mathrm{~m}^2$ $\rho = 1.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$

Rearrange the continuity equation to solve for $v_2$:

$v_2 = \frac{A_1}{A_2} v_1 = 2v_1$

Substitute $v_2$ and rearrange the Bernoulli equation:

$P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$

$3 = \frac{1}{2} \times 1.25 \times 10^3 (4v_1^2 - v_1^2)$

Now, solve for $v_1$:

$3 = \frac{1}{2} \times 1.25 \times 10^3 \times 3v_1^2$ $v_1^2 = \frac{3}{1.875 \times 10^3}$

$v_1 = \sqrt{\frac{3}{1.875 \times 10^3}}$

$v_1 \approx 0.0400 \mathrm{~m} \mathrm{~s}^{-1}$

Now, calculate the rate of flow of glycerin through the pipe (volume flow rate) using $v_1$ and $A_1$:

$Q = A_1 v_1$

$Q = 10 \times 10^{-4} \times 0.0400$

$Q = 4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$

So, the rate of flow of glycerin through the pipe is $4 \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}$, and the value of $x$ is 4.

To determine the fundamental frequency of the vibrating wire, we need to first find the tension (T) in the wire and the wave velocity (V) in the wire.

1. Tension in the wire (T): We used Young's modulus (Y) to relate the stress and strain in the wire. The formula for stress is:

$\text{stress} = Y \times \text{strain}$

Here, the strain is the extension ($\Delta L$) divided by the original length (L):

$\text{strain} = \frac{\Delta L}{L}$

Now, the tension (T) in the wire is the product of stress and cross-sectional area (A):

$T = \text{stress} \times A$

Combining the above equations, we get the expression for tension:

$T = \frac{Y \Delta L}{L} \times A$

1. Wave velocity in the wire (V): The linear mass density ($\mu$) of the wire is given by:

$\mu = \frac{m}{L}$

We need to find the ratio $\frac{T}{\mu}$, which represents the square of the wave velocity. Using the expressions for tension and linear mass density, we get:

$\frac{T}{\mu} = \frac{Y \Delta L}{L} \times \frac{A}{m} = \frac{Y \Delta L}{L} \times \frac{1}{\rho}$

Here, $\rho$ is the density of the wire material. Plugging in the given values, we find the value of $\frac{T}{\mu}$, which is:

$\frac{T}{\mu} = 6.4 \times 10^3$

Now, we find the wave velocity (V) by taking the square root of $\frac{T}{\mu}$:

$V = \sqrt{T/\mu} = 80 \mathrm{~m/s}$

1. Fundamental frequency (f): Finally, we find the fundamental frequency of the vibrating wire using the formula:

$f = \frac{V}{2L}$

Plugging in the values, we get the fundamental frequency (f) as:

$f = 80 \mathrm{~Hz}$

So, the fundamental frequency of vibration in the wire is 80 Hz.

Given:

1. When tension $T_1 = 100 \mathrm{~N}$, extension $= l_1 - l_0$.
2. When tension $T_2 = 120 \mathrm{~N}$, extension $= l_2 - l_0$.

Now, let's write the equations using Hooke's Law:

$100 = k(l_1 - l_0)$

$120 = k(l_2 - l_0)$

Divide the first equation by the second equation:

$\frac{5}{6} = \frac{l_1 - l_0}{l_2 - l_0}$

Given the relationship between $l_1$ and $l_2$:

$10l_2 = 11l_1$

Now, let's solve for $l_0$:

$5l_2 - 5l_0 = 6l_1 - 6l_0$

$l_0 = 6l_1 - 5l_2$

Substitute the relationship between $l_1$ and $l_2$:

$l_0 = 6l_1 - 5\left(\frac{11l_1}{10}\right)$

$l_0 = 6l_1 - \frac{11l_1}{2}$

$l_0 = \frac{l_1}{2}$

Therefore, the natural length of the wire is $\frac{1}{x}l_1 = \frac{2}{1}l_1 = 2l_1$.

The value of $x$ is $2$.

The terminal velocity of a small spherical object moving under the action of gravity through a fluid medium is given by Stokes' Law, which is stated as:

$v = \frac{2}{9} \frac{r^2 g (\rho_p - \rho_f)}{\eta}$,

where:

• $v$ is the velocity of the object (in this case, the air bubble),
• $r$ is the radius of the object,
• $g$ is the acceleration due to gravity,
• $\rho_p$ is the density of the object (negligible in this case, as it's an air bubble),
• $\rho_f$ is the density of the fluid (the solution), and
• $\eta$ is the coefficient of viscosity of the fluid.

Since we are neglecting the density of the air bubble, the formula simplifies to:

$v = \frac{2}{9} \frac{r^2 g \rho_f}{\eta}$.

Rearranging for $\eta$, we get:

$\eta = \frac{2}{9} \frac{r^2 g \rho_f}{v}$.

Given that $r = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$, $g = 10 \, \text{ms}^{-2}$, $\rho_f = 1750 \, \text{kg/m}^{3}$, and $v = 0.35 \, \text{cm/s} = 0.35 \times 10^{-2} \, \text{m/s}$, we can substitute these values into the formula to find $\eta$:

$\eta = \frac{2}{9} \frac{(3 \times 10^{-3})^2 \times 10 \times 1750}{0.35 \times 10^{-2}} = 10 \, \text{Pas}$.

Therefore, the coefficient of viscosity of the solution is $10 \, \text{Pas}$.

To find the mass $m$ of the metal block, we need to consider the tensile stress developed in the wire. The formula for tensile stress is:

$\text{Tensile Stress} = \frac{\text{Force}}{\text{Area}}$

The force acting on the wire is the weight of the metal block, which can be represented as $F = mg$.

The cross-sectional area of the wire, given its diameter $d = 14 \, mm$, can be calculated using the formula for the area of a circle:

$A = \pi (\frac{d}{2})^2 = \pi (\frac{14}{2})^2 \, mm^2$

Now, convert the area to $m^2$:

$A = \pi (\frac{14 \times 10^{-3}}{2})^2 \, m^2$

We are given that the tensile stress developed in the wire is $7 \times 10^5 \, Nm^{-2}$. Using the tensile stress formula, we can write:

$7 \times 10^5 \, Nm^{-2} = \frac{mg}{A}$

Now, solve for the mass $m$:

$m = \frac{7 \times 10^5 \, Nm^{-2} \cdot A}{g}$

Substitute the values of A and g into the equation:

$m = \frac{7 \times 10^5 \, Nm^{-2} \cdot \pi (\frac{14 \times 10^{-3}}{2})^2 \, m^2}{9.8 \, ms^{-2}}$

After calculating, we get:

$m \approx 11 \, kg$

Therefore, the mass of the metal block is approximately $11 \, kg$.