Properties of Matter

Young's Modulus is given by :

$ \mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{~L} / \mathrm{L}}=\frac{\mathrm{FL}}{\mathrm{~A} \Delta \mathrm{~L}} $

Where:

• F is the applied force (load)

• L is the original length.

• A is the area of cross-section

• $\quad \Delta \mathrm{L}$ is the change in length (stretching)

According to question, both wires are stretched by the same magnitude ( $\Delta \mathrm{L}_{\mathrm{A}}=\Delta \mathrm{L}_{\mathrm{B}}$ ) under the same given load $\left(\mathrm{F}_{\mathrm{A}}=\mathrm{F}_{\mathrm{B}}\right)$.

Therefore, for both wires:

$ Y \propto \frac{L}{A} $

Let $Y_A$ and $Y_B$ be the Young's moduli for wires $A$ and $B$ :

$ \frac{Y_A}{Y_B}=\frac{\left(\frac{L_A}{A_A}\right)}{\left(\frac{L_B}{A_B}\right)}=\frac{L_A}{L_B} \times \frac{A_B}{A_A} $

The given values :

• $\mathrm{L}_{\mathrm{A}}=6.0 \mathrm{~cm}, \mathrm{~L}_{\mathrm{B}}=5.4 \mathrm{~cm}$

• $\mathrm{A}_{\mathrm{A}}=3.0 \times 10^{-5} \mathrm{~m}^2, \mathrm{~A}_{\mathrm{B}}=4.5 \times 10^{-5} \mathrm{~m}^2$

$ \begin{gathered} \frac{Y_A}{Y_B}=\frac{6.0}{5.4} \times \frac{4.5 \times 10^{-5}}{3.0 \times 10^{-5}} \\ \frac{Y_A}{Y_B}=\frac{5}{3}=\frac{x}{3} \Rightarrow x=5 \end{gathered} $

So, the value of x is 5 . Hence, the correct option is (D).

When an object floats in a liquid, the weight of the object is exactly balanced by the buoyant force (up thrust) exerted by the liquid.

Weight of block is $=\mathrm{W}=\mathrm{m} \cdot \mathrm{g}=\left(\rho_{\mathrm{b}} \times \mathrm{V}_{\text {total }}\right) \times \mathrm{g}$

Buoyant force ( $\mathrm{F}_{\mathrm{B}}$ ) acting on a submerged body is equal to the weight of displaced liquid,

$ F_B=\left(\rho_1 V_{\text {submerged }}\right) \times g $

For flotation :

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

$ \rho_{\mathrm{b}} \mathrm{~V}_{\text {total }} \mathrm{g}=\rho_{\mathrm{l}} \mathrm{~V}_{\text {submerged }} \mathrm{g} $

For a cubical block with a uniform cross-sectional area (A):

Total Volume $=\mathrm{V}_{\text {total }}=\mathrm{AH}$

Submerged Volume $\mathrm{V}_{\text {submerged }}=\mathrm{Ah}$ (where h is the submerged height)

$\Rightarrow $ $ \rho_{\mathrm{b}}(\mathrm{AH})=\rho_{\mathrm{l}}(\mathrm{Ah}) \Rightarrow \mathrm{h}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{l}}}\right) \mathrm{H} $

Putting the given values,

$ \rho_{\mathrm{b}}=600 \mathrm{~kg} / \mathrm{m}^3, \rho_{\mathrm{l}}=900 \mathrm{~kg} / \mathrm{m}^3 \text { and } \mathrm{H}=8.0 \mathrm{~cm} $

$\Rightarrow $ $ \mathrm{~h}=\left(\frac{600}{900}\right) \times 8.0 \mathrm{~cm} \approx 5.33 \mathrm{~cm} $

Therefore, the height of the submerged part of the block is approximately 5.3 cm.

Hence, the correct option is (D).

A metal rod of original length $L$ that is free to move. If we heat this rod by a temperature difference $\Delta T$, it will naturally expand.

The change in length ( $\Delta \mathrm{L}$ ) is given by the law of linear thermal expansion :

$ \Delta \mathrm{L}=\mathrm{L} \cdot \alpha \cdot \Delta \mathrm{~T} $

Where $\alpha=$ Coefficient of Linear Expansion

When a wire fixed at both ends is cooled, it contracts. Since the supports are rigid, they prevent this contraction, creating a tension in the wire.

$ \text { Strain } \varepsilon=\frac{\Delta \mathrm{L}}{\mathrm{~L}}=\alpha \Delta \mathrm{T} $

According to Hooke's Law, Stress is proportional to Strain within the elastic limit. The constant of proportionality is Young's Modulus (Y).

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

$\Rightarrow $ Stress $(\sigma)=\mathrm{Y} \times$ Strain

$ \sigma=\mathrm{Y} \cdot(\alpha \Delta \mathrm{~T}) $

Stress is defined as Force (F) per unit Area (A).

$ \text { Stress }=\frac{\mathrm{F}}{\mathrm{~A}}=\mathrm{Y} \cdot(\alpha \Delta \mathrm{~T}) $

$\Rightarrow $ $\mathrm{F}=(\mathrm{Y} \alpha \Delta \mathrm{T}) \times \mathrm{A}$

$ \Rightarrow \mathrm{F} \propto \Delta \mathrm{~T} \Rightarrow \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{\Delta \mathrm{T}_1}{\Delta \mathrm{~T}_2} $

Initial Temperature $=\mathrm{t}_0=27^{\circ} \mathrm{C}$

Final Temperature $=\mathrm{t}_1=-43^{\circ} \mathrm{C}$

$ \Delta \mathrm{T}_1=27-(-43)=70^{\circ} \mathrm{C} $

Initial tension in the wire is, $\mathrm{F}_1=\mathrm{T}$

Final tension in the wire is $\mathrm{F}_2=1.4 \mathrm{~T}$

Let the new final temperature be $t_2$.

$ \Delta \mathrm{T}_2=27-\mathrm{t}_2 $

$\Rightarrow $ $\frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{1.4 \mathrm{~T}}{\mathrm{~T}}=\frac{27-\mathrm{t}_2}{70}$

$\Rightarrow $ $1.4=\frac{27-t_2}{70}$

$\Rightarrow $ $27-t_2=1.4 \times 70$

$\Rightarrow $ $27-t_2=98$

$\Rightarrow $$ t_2=27-98 $

$\Rightarrow $$ t_2=-71^{\circ} \mathrm{C} $

Using Stokes' Law, the formula for terminal velocity $\left(v_t\right)$ of a spherical body falling through a viscous fluid is :

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

Where :

• $r=$ Radius of the sphere $r=\frac{\text { Diameter }}{2}=\frac{2 \mathrm{~mm}}{2}=1 \mathrm{~mm}=0.1 \mathrm{~cm}$.

• $\rho=$ Density of the sphere $=\rho=10.5 \mathrm{~g} / \mathrm{cm}^3$

• $\sigma=$ Density of the fluid, Density of glycerine $=\sigma=1.5 \mathrm{~g} / \mathrm{cm}^3$

• $\mathrm{g}=$ Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~cm} / \mathrm{s}^2$

• $\eta=$ Coefficient of viscosity $=10$ Poise

1 Poise $=1 \frac{\mathrm{~g}}{\mathrm{~cm} \cdot \mathrm{~s}^3}$, which is the CGS unit of viscosity.

Substituting the values into the terminal velocity equation :

$ v_t=\frac{2 \times(0.1)^2 \times(10.5-1.5) \times 1000}{9 \times 10} $

$\Rightarrow $ $\mathrm{v}_{\mathrm{t}}=\frac{2 \times 0.01 \times 9 \times 1000}{9 \times 10}$

$\Rightarrow $ $ \mathrm{v}_{\mathrm{t}}=2 \mathrm{~cm} / \mathrm{s} $

The sphere accelerates until the viscous drag reaches a point where the upward forces equal the weight of the sphere. At this point, it continues to fall at a constant terminal velocity of $2.0 \mathrm{~cm} / \mathrm{s}$.

Young's Modulus (Y) is defined as the ratio of Stress $(\sigma)$ to Strain $(\varepsilon)$ within the elastic limit :

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

For the given graph,

• 1. Y -axis: $\operatorname{Strain}(\varepsilon)$

• 2. X -axis: Stress $(\sigma)$

Since Strain is on the Y -axis and Stress is on the X -axis, the slope of the line is given by :

$ \text { Slope }=\frac{\Delta \mathrm{Y}}{\Delta \mathrm{X}}=\frac{\text { Strain }}{\text { Stress }} $

$ \text { Slope }=\frac{1}{\mathrm{Y}} $

It shows that the slope is inversely proportional to Young's Modulus, i.e., higher slope corresponds to smaller Young's Modulus (material is more elastic) whereas lower slope shows larger Young's Modulus (material is less elastic).

So, the material with the largest Young's modulus will be the line with the smallest slope.

• 1. Material A has the highest slope → smallest Y.

• 2. Material C has the lowest slope → largest Y.

Therefore, material C has the largest Young's modulus. Hence, the correct option is (D).

The height (h) to which a liquid rises in a capillary tube is given by:

$ \mathrm{h}=\frac{2 \mathrm{~S} \cos \theta}{\mathrm{r} \rho \mathrm{~g}} $

Where :

• S is the surface tension of the liquid.

• $\theta$ is the angle of contact.

• $\mathrm{r}$ is the inner radius of the capillary tube.

• $\rho$ is the density of the liquid.

• $g$ is the acceleration due to gravity.

Assuming the angle of contact ( θ) and gravity ( $g$ ) remain constant, the height $h$ depends on variables $S, r$, and $\rho$ as,

$ h \propto \frac{S}{r \rho} $

Taking the natural logarithm of both sides :

$ \ln (h)=\ln (S)-\ln (r)-\ln (\rho)+\text { constant } $

Differentiating this equation,

$ \frac{\Delta \mathrm{h}}{\mathrm{~h}}=\frac{\Delta \mathrm{S}}{\mathrm{~S}}-\frac{\Delta \mathrm{r}}{\mathrm{r}}-\frac{\Delta \rho}{\rho} $

So, the percentage change in height is,

$ \% \Delta \mathrm{~h}=\% \Delta \mathrm{~S}-\% \Delta \mathrm{r}-\% \Delta \rho $

It is given that all three quantities (radius, density, and surface tension) decrease by $1 \%$. Therefore, their percentage changes are negative :

• Change in surface tension, $\% \Delta \mathrm{~S}=-1 \%$

• Change in inner radius, $\% \Delta \mathrm{r}=-1 \%$

• Change in density, $\% \Delta \rho=-1 \%$

Substituting these values into our percentage change equation :

$ \% \Delta \mathrm{~h}=(-1 \%)-(-1 \%)-(-1 \%) $

$\Rightarrow $ $ \% \Delta h=-1 \%+1 \%+1 \%=1 \% $

The positive sign indicates that the height of the liquid in the tube will increase by $1 \%$. Hence, the correct option is (B).

Statement I

Consider an imaginary cube of water out of still water in a container,

This tiny cube has mass, so gravity is pulling it down. If nothing were pushing back up against it, it would fall to the bottom. However, the fluid is still (in static equilibrium). This means the net force on our imaginary cube must be zero.

The water below our imaginary cube must be pushing up on it to counteract gravity. Similarly, the water on the sides is pushing inward to keep it from spreading.

This pushing force per unit area is exactly what pressure is. Therefore, the surrounding fluid exerts pressure on our imaginary surface inside the bulk of the liquid. Pressure is a scalar property that exists at every single point within a fluid, not just where it touches a solid container.

Therefore, the statement I is false.

Statement II

A molecule deep in the interior is completely surrounded by other identical molecules on all sides. The attractive forces pull it equally in every direction. The net force on an interior molecule is zero.

A molecule on the surface has molecules next to it and below it, but practically none above it. Because it is missing half of its neighbours, there is a net inward force pulling the surface molecule down into the bulk of the liquid.

Now, if it is to bring an interior molecule up to the surface to create more surface area. It has to be pulled against that net inward force. And whenever an object moves against an opposing force by external force, then work must be done.

By the Work-Energy Theorem, the work is stored as Potential Energy. Therefore, molecules on the surface possess higher potential energy than those in the bulk interior.

Any system always prefers the lowest possible energy state. Because having surface molecules costs extra energy, a liquid will naturally try to minimize the number of molecules on its surface. It does this by minimizing its overall surface area. This macroscopic tendency of a liquid surface to shrink and act like a stretched elastic membrane is exactly what we call surface tension.

Therefore, the statement II is true.

So, statement I is false, and Statement II is true.

Hence, the correct option is (A).

When a capillary tube is dipped in a liquid, the liquid rises due to surface tension until the upward force of surface tension is balanced by the weight of the liquid column.

The surface tension T acts along the circumference of the meniscus. If $\theta$ is the contact angle, the vertical component of the force is :

$ \mathrm{F}_{\mathrm{up}}=\mathrm{T} \cos \theta \times(2 \pi \mathrm{r}) $

The weight of the liquid column of height h is :

$ W=\left(\pi r^2 h\right) \rho g $

At equilibrium :

$ \mathrm{T} \cos \theta(2 \pi \mathrm{r})=\pi \mathrm{r}^2 \mathrm{~h} \rho \mathrm{~g} $

$\Rightarrow $ $ \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{~g}} $

If we assume the two liquids refers to the same substance, then:

• $\mathrm{T}_1=\mathrm{T}_2 \ldots$ (i) (Surface tension is a property of the material and temperature).

• $\rho$ and $g$ are constant.

• $\theta$ is constant.

This results the relationship: $\mathrm{h} \propto \frac{1}{\mathrm{r}}$.

We are given that $r_1>r_2 \Rightarrow \frac{r_1}{r_2}>1$.

So, the ratio of heights of capillary rise,

$ \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_2}<1 \Rightarrow \mathrm{~h}_1<\mathrm{h}_2 \ldots (ii)$

From both the relations, $\mathrm{h}_1<\mathrm{h}_2$ and $\mathrm{T}_1=\mathrm{T}_2$.

Hence, the correct option is (B).

To find the final length of the composite rod, we need to calculate the expansion of both the aluminium and steel sections individually and then add them to the original length.

Total Initial Length $\left(\mathrm{L}_1\right)$ is 120 cm at $30^{\circ} \mathrm{C}$.

Since both rods have the same length, the initial length of each rod is, $\mathrm{L}_{\mathrm{Al}}=\frac{120}{2}=60 \mathrm{~cm}$ and $\mathrm{L}_{\mathrm{St}}=\frac{120}{2}=$ 60 cm .

The change in temperature is,

$ \Delta \mathrm{T}=\mathrm{T}_{\text {final }}-\mathrm{T}_{\text {initial }}=100^{\circ} \mathrm{C}-30^{\circ} \mathrm{C}=70^{\circ} \mathrm{C} $

Coefficients of linear expansion ( $\alpha$ ) for aluminium is $\alpha_{\mathrm{Al}}=24 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and that for steel is $\alpha_{\mathrm{St}}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}=12 \times 10^{-6} /{ }^{\circ} \mathrm{C}$.

The linear expansion is $\Delta \mathrm{L}=\mathrm{L}_{\text {initial }} \cdot \alpha \cdot \Delta \mathrm{T}$

For Aluminium :

$ \Delta \mathrm{L}_{\mathrm{Al}}=60 \times\left(24 \times 10^{-6}\right) \times 70 $

$\Rightarrow $ $\Delta \mathrm{L}_{\mathrm{Al}}=60 \times 70 \times 24 \times 10^{-6}$

$\Rightarrow $ $ \Delta \mathrm{L}_{\mathrm{Al}}=4200 \times 24 \times 10^{-6}=100800 \times 10^{-6}=0.1008 \mathrm{~cm} $

For Steel :

$ \Delta \mathrm{L}_{\mathrm{St}}=60 \times\left(12 \times 10^{-6}\right) \times 70 $

$\Rightarrow $ $ \Delta \mathrm{L}_{\mathrm{St}}=4200 \times 12 \times 10^{-6}=50400 \times 10^{-6}=0.0504 \mathrm{~cm} $

The total change in length ( $\Delta \mathrm{L}_{\text {total }}$ ) is :

$ \Delta \mathrm{L}_{\text {total }}=\Delta \mathrm{L}_{\mathrm{Al}}+\Delta \mathrm{L}_{\mathrm{St}} $

$\Rightarrow $ $ \Delta \mathrm{L}_{\text {total }}=0.1008+0.0504=0.1512 \mathrm{~cm} $

So, the final length of the rod is :

$ \mathrm{L}_{\mathrm{final}}=120+0.1512=120.1512 \mathrm{~cm} $

$\Rightarrow $ $ \mathrm{L}_{\mathrm{final}} \approx 120.15 \mathrm{~cm} $

Therefore, the final length of the combination is 120.15 cm .

Hence, the correct option is (C).

Applying the equation of continuity

The rate of flow ( Q ) is constant throughout the tube.

$ \mathrm{Q}=\mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 $

Where :

• $\mathrm{A}_1=$ cross-section area at $\mathrm{A}, \mathrm{A}_1=6 \mathrm{~cm}^2$

• $\mathrm{A}_2=$ cross-section area at $\mathrm{B}, \mathrm{A}_2=3 \mathrm{~cm}^2$

• $\mathrm{V}_1, \mathrm{~V}_2$ are velocities at A and B .

$ 6 v_1=3 v_2 \Rightarrow v_2=2 v_1 $

Applying Bernoulli's principle between two points,

$ P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_2^2 $

For a horizontal tube, the height is the same, $\mathrm{h}_1=\mathrm{h}_2$

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

The pressure difference ( $P_1-P_2$ ) is measured by the difference in the heights of the water columns ( $h$ ) :

$ \mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{gh} $

From relations (ii) and (iii),

$ \rho g h=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) $

$\Rightarrow $ $ 2 g h=v_2^2-v_1^2 $

Substituting $\mathrm{v}_2=2 \mathrm{v}_1$ from relation (i) :

$ 2 g h=\left(2 v_1\right)^2-v_1^2 $

$\Rightarrow $ $2 g h=4 v_1^2-v_1^2=3 v_1^2$

$ \Rightarrow v_1=\sqrt{2 g h / 3} $

The given values are,

• $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~cm} / \mathrm{s}^2$

• $\mathrm{h}=5 \mathrm{~cm}$

Putting the values,

$ v_1=\sqrt{\frac{2 \times 1000 \times 5}{3}}=\sqrt{\frac{10000}{3}}=\frac{100}{\sqrt{3}} \mathrm{~cm} / \mathrm{s} $

So, the rate of flow (Q),

$ Q=A_1 \cdot v_1 $

$\Rightarrow $ $ Q=6 \cdot \frac{100}{\sqrt{3}}=2 \times \sqrt{3} \times \sqrt{3} \times \frac{100}{\sqrt{3}}=200 \sqrt{3} \mathrm{~cm}^3 / \mathrm{s} $

Therefore, the rate of flow of water through the horizontal tube is $200 \sqrt{3} \mathrm{~cm}^3 / \mathrm{s}$.

Hence, the correct option is (D).

The density of liquid is $\rho=600 \mathrm{~kg} / \mathrm{m}^3$.

Area of cross-section at point A is $\mathrm{A}_{\mathrm{A}}=1.0 \mathrm{~cm}^2=1.0 \times 10^{-4} \mathrm{~m}^2$.

Area of cross-section at point $B$ is $A_B=20.0 \mathrm{~mm}^2=20.0 \times 10^{-6} \mathrm{~m}^2=0.2 \times 10^{-4} \mathrm{~m}^2$.

The velocity at A is $\mathrm{v}_{\mathrm{A}}=10 \mathrm{~cm} / \mathrm{s}=0.1 \mathrm{~m} / \mathrm{s}$.

For an incompressible liquid in steady flow, the mass flow rate must remain constant.

Using equation of continuity,

$ A_A v_A=A_B v_B $

Substituting the values :

$ \left(1.0 \times 10^{-4}\right) \times(0.1)=\left(0.2 \times 10^{-4}\right) \times v_B $

$\Rightarrow $ $ \mathrm{v}_{\mathrm{B}}=\frac{1.0 \times 0.1}{0.2}=\frac{0.1}{0.2}=0.5 \mathrm{~m} / \mathrm{s} $

According to Bernoulli's equation for a horizontal flow (where height $h$ is constant), the sum of pressure and kinetic energy per unit volume is constant :

$P_A+\frac{1}{2} \rho v_A^2=P_B+\frac{1}{2} \rho v_B^2$

$\Rightarrow $ $ P_A-P_B=\Delta P_{A B}=\frac{1}{2} \rho\left(v_B^2-v_A^2\right) $

Substituting the given values $\rho=600 \mathrm{~kg} / \mathrm{m}^3, \mathrm{v}_{\mathrm{B}}=0.5 \mathrm{~m} / \mathrm{s}$ and $\mathrm{v}_{\mathrm{A}}=0.1 \mathrm{~m} / \mathrm{s}$

$ P_A-P_B=\frac{1}{2} \times 600 \times\left(0.5^2-0.1^2\right) $

$\Rightarrow $ $P_A-P_B=300 \times(0.25-0.01)$

$\Rightarrow $ $\mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}=300 \times 0.24$

$\Rightarrow $ $ \mathrm{P}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}}=72 \mathrm{~Pa} $

Therefore, the difference in pressures at points A and B is 72 Pa .

Hence, the correct option is (D).

When a spherical object falls through a viscous fluid, it eventually reaches a constant speed called terminal velocity $\left(\mathrm{v}_{\mathrm{t}}\right)$. This occurs when the downward force of gravity is balanced by the upward buoyant force and the viscous drag force.

The force of gravity acting on the sphere is $F_g=m g=\frac{4}{3} \pi r^3 \rho g$, where $\rho$ is the density of the liquid.

Using Archimedean's principle of buoyant force $\mathrm{F}_{\mathrm{b}}=\frac{4}{3} \pi \mathrm{r}^3 \sigma \mathrm{~g}$, where $\sigma$ is the density of the gas.

Using Stokes' Law the viscous drag is $\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{r} \mathrm{v}_{\mathrm{t}}$

At terminal velocity,

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

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

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

For the same liquid and gas, terminal velocity is proportional to the square of the radius :

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

The large drop of radius $R$ is broken into 64 identical droplets of radius $r$. Since the total volume remains constant :

$ V_R=64 \times V_r $

$\Rightarrow $ $\frac{4}{3} \pi R^3=64 \times \frac{4}{3} \pi r^3$

$\Rightarrow $ $\mathrm{R}^3=64 \mathrm{r}^3$

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

Let $\mathrm{v}_1$ be the terminal velocity of the large drop and $\mathrm{v}_2$ be the terminal velocity of a small droplet. Using the relationship $\mathrm{v} \propto \mathrm{r}^2$ :

$ \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{R}^2}{\mathrm{r}^2} $

Substituting $\mathrm{r}=\frac{\mathrm{R}}{4}$ :

$ \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{R}^2}{\left(\frac{\mathrm{R}}{4}\right)^2} $

$\Rightarrow $ $\frac{v_1}{v_2}=\frac{R^2}{R^2 / 16}$

$\Rightarrow $ $ \frac{v_1}{v_2}=16 $

Therefore, the ratio of terminal velocities $\frac{v_1}{v_2}$ is 16 .

Hence, the correct option is (D).

Young's Modulus (Y) is defined as the ratio of tensile stress to tensile strain within the elastic limit.

Stress ( $\sigma$ ) is force (W) per unit cross-sectional area (A).

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

Strain $(\epsilon)$ is extension $(\Delta \mathrm{l})$ per unit original length $(\mathrm{L})$.

$ \epsilon=\frac{\Delta \mathrm{l}}{\mathrm{~L}} $

Thus, Young's Modulus is :

$ Y=\frac{\sigma}{\epsilon}=\frac{W / A}{\Delta l / L} $

$ \mathrm{Y}=\frac{\mathrm{WL}}{\mathrm{~A} \Delta \mathrm{l}} $

The original length is $\mathrm{L}=1 \mathrm{~m}$

The cross-sectional area is $\mathrm{A}=10^{-5} \mathrm{~m}^2$

From the graph, we need a pair of values for W and $\Delta \mathrm{l}$.

Let's take $\mathrm{W}=20 \mathrm{~N}$ and the corresponding extension $\Delta \mathrm{l}=2 \times 10^{-4} \mathrm{~m}$.

Substituting the values in the expression for Young's modulus;

$ Y=\frac{20 \times 1}{10^{-5} \times\left(2 \times 10^{-4}\right)} $

$\Rightarrow $ $Y=\frac{20}{2 \times 10^{-9}}$

$\Rightarrow $ $Y=10 \times 10^9$

$ \mathrm{Y}=1.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^2 $

Therefore, the Young's modulus of the wire is $1.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$. Thus, the correct option is (C).

The capacity of vessel is $528 \mathrm{dm}^3$.

Since $1 \mathrm{dm}=10 \mathrm{~cm} \Rightarrow 1 \mathrm{dm}^3=1000 \mathrm{~cm}^3$

$ \mathrm{V}=528,000 \mathrm{~cm}^3 $

Volume of a cylinder is $\mathrm{V}=\pi \mathrm{r}^2 \mathrm{H}$.

Given $\mathrm{r}=40 \mathrm{~cm}$

$ 528,000=\times H $

$\Rightarrow $ $\frac{528,000}{\pi \times(40)^2}=\mathrm{H}$

$\Rightarrow $ $ \mathrm{H}=105 \mathrm{~cm} $

Depth of hole from top is $\mathrm{h}=70 \mathrm{~cm}$.

From Torricelli's Law velocity of efflux is given as,

$ \mathrm{v}=\sqrt{2 \mathrm{gh}} $

The vessel is on a block of height H and the hole is at distance $(\mathrm{H}-\mathrm{h})$ from the bottom of the vessel.

Total height from ground $\mathrm{Y}=(\mathrm{H}-\mathrm{h})+\mathrm{H}=2 \mathrm{H}-\mathrm{h}$

The time to fall is,

$ t=\sqrt{\frac{2 Y}{g}} $

The horizontal range of projectile is given as

$ \mathrm{R}=\mathrm{v} \times \mathrm{t}=\sqrt{2 \mathrm{gh}} \times \sqrt{\frac{2(2 \mathrm{H}-\mathrm{h})}{\mathrm{g}}}=2 \sqrt{\mathrm{~h}(2 \mathrm{H}-\mathrm{h})} $

The height of water level from the hole is $\mathrm{h}=70 \mathrm{~cm}$

$ 2 \mathrm{H}-\mathrm{h}=2(105)-70=210-70=140 \mathrm{~cm} . $

$\Rightarrow $ $\mathrm{R}=2 \sqrt{70 \times 140}$

$\Rightarrow $ $R=2 \sqrt{70 \times 70 \times 2}$

$\Rightarrow $ $ \mathrm{R}=2 \times 70 \times \sqrt{2}=140 \sqrt{2} \mathrm{~cm} $

Therefore, the horizontal of water falling on the ground is $140 \sqrt{2} \mathrm{~cm}$.

Thus. the correct option is (B).

When a lift of mass M accelerates upward with acceleration a, the tension (T) in the supporting cable must overcome both gravity and provide the upward acceleration.

Using Newton's Second Law ( $\mathrm{F}_{\text {net }}=\mathrm{Ma}$ ):

$ \mathrm{T}-\mathrm{Mg}=\mathrm{Ma} \Rightarrow \mathrm{~T}=\mathrm{M}(\mathrm{~g}+\mathrm{a}) $

The stress $(\sigma)$ is defined as the restoring force (tension) per unit cross-sectional area (A):

$ \sigma=\frac{T}{A}=\frac{M(g+a)}{A} $

For a wire with a circular cross-section of radius r , the area of cross-section is $\mathrm{A}=\pi \mathrm{r}^2$.

$ \sigma_{\max }=\frac{\mathrm{M}\left(\mathrm{~g}+\mathrm{a}_{\max }\right)}{\pi \mathrm{r}^2} $

The mass of lift is $\mathrm{M}=1600 \mathrm{~kg}$

The radius of cable is $\mathrm{r}=4 \mathrm{~mm}=4 \times 10^{-3} \mathrm{~m}$

So, the area is

$ \mathrm{A}=\pi \mathrm{r}^2=3.14 \times\left(4 \times 10^{-3}\right)^2=3.14 \times 16 \times 10^{-6} \mathrm{~m}^2 $

$\Rightarrow $ $ \mathrm{A}=50.24 \times 10^{-6} \mathrm{~m}^2 $

So, the maximum stress is

$ \sigma_{\max }=4 \times 10^8 \mathrm{~N} / \mathrm{m}^2 $

From the maximum stress formula,

$ 4 \times 10^8=\frac{1600\left(10+\mathrm{a}_{\max }\right)}{50.24 \times 10^{-6}} $

$\Rightarrow $ $4 \times 10^8 \times 50.24 \times 10^{-6}=1600\left(10+a_{\max }\right)$

$\Rightarrow $ $200.96 \times 10^2=1600\left(10+a_{\max }\right)$

$\Rightarrow $ $20096=1600\left(10+a_{\max }\right)$

$\Rightarrow $ $10+a_{\max }=\frac{20096}{1600}=12.56$

$\Rightarrow $ $ \mathrm{a}_{\max }=12.56-10=2.56 \mathrm{~m} / \mathrm{s}^2 $

Therefore, the maximum acceleration of the lift is $2.56 \mathrm{~m} / \mathrm{s}^2$.

Thus, the correct option is (A).

Young's Modulus is defined as the ratio of Stress $(\sigma)$ to Strain $(\epsilon)$.

$ Y=\frac{\text { Stress }}{\text { Strain }}=\frac{\sigma}{\epsilon} $

Stress is force per unit area, $\sigma=\frac{\text { Force (F) }}{\text { Area (A) }}$

Strain is change in length per unit original length,

$ \epsilon=\frac{\text { Change in length }(\Delta \mathrm{L})}{\text { Original length }(\mathrm{L})} $

So, $Y=\frac{F / A}{\Delta L / L}$

$ \Rightarrow \Delta \mathrm{L}=\frac{\mathrm{FL}}{\mathrm{AY}} $

Since the wires are joined in series and kept under tension, the Force (F) acting on both wires is the same.

The area (A) is the same for both.

It is given that the elongation in both the wires is equal, $\Delta \mathrm{L}_{\mathrm{A}}=\Delta \mathrm{L}_{\mathrm{B}}$.

Since $\Delta \mathrm{L}_{\mathrm{A}}=\Delta \mathrm{L}_{\mathrm{B}}:$

$\frac{F L_A}{A Y_A}=\frac{F L_B}{A Y_B}$

$\Rightarrow \frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{A}}}=\frac{\mathrm{L}_{\mathrm{B}}}{\mathrm{Y}_{\mathrm{B}}}$

$\Rightarrow $ $ \frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{~L}_{\mathrm{B}}}=\frac{\mathrm{Y}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{B}}} $

The ratio of Young's modulus is $\frac{Y_A}{Y_B}=\frac{20}{11}$ and the length of wire $A$ is $L_A=2.2 \mathrm{~m}$

$\frac{2.2}{\mathrm{~L}_{\mathrm{B}}}=\frac{20}{11}$

$\Rightarrow $ $\mathrm{L}_{\mathrm{B}}=2.2 \times \frac{11}{20}$

$\Rightarrow $ $\mathrm{L}_{\mathrm{B}}=\frac{24.2}{20}$

$\Rightarrow $ $ \mathrm{L}_{\mathrm{B}}=1.21 \mathrm{~m} $

Therefore, the length of wire B is 1.21 m . Hence, the correct option is (C).

For a liquid drop, surface energy is

$E = S \times \text{surface area}$

where $S$ is the surface tension.

Step 1: Initial total surface area of 8 small drops

Each small drop has radius $r$.

So, surface area of one drop is

$4\pi r^2$

For $8$ drops, total initial surface area is

$A_i = 8 \times 4\pi r^2 = 32\pi r^2$

Step 2: Radius of the bigger drop

When $8$ drops coalesce, volume is conserved.

Volume of one small drop is

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

So total initial volume is

$8 \times \frac{4}{3}\pi r^3$

Let radius of bigger drop be $R$. Then,

$\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$

$R^3 = 8r^3$

$R = 2r$

Step 3: Final surface area of bigger drop

$A_f = 4\pi R^2 = 4\pi (2r)^2 = 16\pi r^2$

Step 4: Decrease in surface area

$\Delta A = A_i - A_f = 32\pi r^2 - 16\pi r^2 = 16\pi r^2$

Step 5: Surface energy released

Surface energy released is

$\Delta E = S \Delta A = S \times 16\pi r^2$

$\Delta E = 16\pi r^2 S$

So, the correct answer is:

$\boxed{16\pi r^2 S}$

Hence, Option B is correct.

Young’s modulus is a property of the material itself. It does not depend on the length or radius of the wire.

We know,

$ Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}} $

or,

$ Y = \frac{F/A}{\Delta L/L} = \frac{FL}{A \Delta L} $

Here, $A = \pi r^2$ is the area of cross-section.

Even if the radius $r$ and length $L$ are doubled, Young’s modulus $Y$ of the steel wire does not change, because it depends only on the material, that is, steel.

So, the correct answer is:

$ \boxed{\text{Option C: remains unchanged}} $

For a string, elongation is given by

$ \Delta L = \frac{FL}{AY} $

where

• $F$ = tension in the string

• $L$ = length of the string

• $A$ = area of cross-section

• $Y$ = Young's modulus

We need the ratio

$ \frac{\Delta L_A}{\Delta L_B} $

So,

$ \frac{\Delta L_A}{\Delta L_B} = \frac{\frac{F_A L_A}{A Y_A}}{\frac{F_B L_B}{A Y_B}} = \frac{F_A}{F_B}\cdot \frac{L_A}{L_B}\cdot \frac{Y_B}{Y_A} $

Since area is same for both strings, it cancels.

Now let us find each ratio.

1. Tension in string $A$

String $A$ supports both masses $M$ and $2M$, so

$ F_A = (M+2M)g = 3Mg $

2. Tension in string $B$

String $B$ supports only the lower mass $2M$, so

$ F_B = 2Mg $

Thus,

$ \frac{F_A}{F_B} = \frac{3Mg}{2Mg} = \frac{3}{2} $

3. Ratio of lengths

Given

$ \frac{L_A}{L_B} = 2 $

4. Ratio of Young's moduli

Given

$ \frac{Y_A}{Y_B} = 0.5 = \frac{1}{2} $

So,

$ \frac{Y_B}{Y_A} = 2 $

Now substitute:

$ \frac{\Delta L_A}{\Delta L_B} = \frac{3}{2}\times 2 \times 2 = 6 $

Hence, the ratio of elongations is

$ \boxed{6} $

So, the correct option is D.

Use the equation of continuity:

$ A_1 v_1 = A_2 v_2 $

Here,

• area of hose, $A_1 = 30 \,\text{cm}^2$

• speed of water in hose, $v_1 = 50 \,\text{cm/s}$

Now the spray gun has 10 perforations, each of area

$ 15 \,\text{mm}^2 $

So total outlet area is

$ A_2 = 10 \times 15 = 150 \,\text{mm}^2 $

Convert this into $\text{cm}^2$:

Since

$ 1 \,\text{cm}^2 = 100 \,\text{mm}^2 $

therefore,

$ 150 \,\text{mm}^2 = \frac{150}{100} = 1.5 \,\text{cm}^2 $

Now apply continuity equation:

$ 30 \times 50 = 1.5 \times v_2 $

$ v_2 = \frac{30 \times 50}{1.5} $

$ v_2 = 1000 \,\text{cm/s} $

Convert into $\text{m/s}$:

$ 1000 \,\text{cm/s} = 10 \,\text{m/s} $

So, the speed of water from each perforation is

$ \boxed{10 \,\text{m/s}} $

Correct option: B

Use the formula for bulk modulus:

$ B = -\frac{\Delta P}{\Delta V / V} $

For decrease in volume, we take only the magnitude:

$ B = \frac{\Delta P}{\Delta V / V} $

So,

$ \frac{\Delta V}{V} = \frac{\Delta P}{B} $

Given:

$ \Delta P = 6.3 \times 10^7 \ \text{N/m}^2 $

$ B = 2.1 \times 10^9 \ \text{N/m}^2 $

Substitute:

$ \frac{\Delta V}{V} = \frac{6.3 \times 10^7}{2.1 \times 10^9} $

$ \frac{\Delta V}{V} = 3 \times 10^{-2} = 0.03 $

Percentage decrease in volume:

$ 0.03 \times 100 = 3\% $

So, the correct answer is:

$ \boxed{3\%} $

Option B

For a string under tension, extension is given by

$ \Delta L=\frac{FL}{AY} $

where

$F$ = tension,

$L$ = length,

$A$ = area of cross-section,

$Y$ = Young's modulus.

Since the two strings are connected in series and mass of strings is neglected, the same force acts in both strings.

The load is $0.8\ \text{kg}$, so tension is

$ F=mg=0.8 \times 10=8\ \text{N} $

Radius of each string:

$ r=0.2\ \text{mm}=0.2\times 10^{-3}\ \text{m}=2\times 10^{-4}\ \text{m} $

So cross-sectional area is

$ A=\pi r^2=\pi (2\times 10^{-4})^2=4\pi \times 10^{-8}\ \text{m}^2 $

Now for string $A$:

$ L_A=0.314\ \text{m}, \qquad Y_A=2\times 10^{10}\ \text{N/m}^2 $

Its extension:

$ \Delta L_A=\frac{8\times 0.314}{(4\pi \times 10^{-8})(2\times 10^{10})} $

$ \Delta L_A=\frac{2.512}{8\pi \times 10^2} =\frac{2.512}{800\pi} $

Using $\pi \approx 3.14$,

$ \Delta L_A=\frac{2.512}{2512}=10^{-3}\ \text{m}=1\ \text{mm} $

Now for string $B$:

Its length and Young's modulus are both twice those of $A$.

So,

$ L_B=2L_A,\qquad Y_B=2Y_A $

Hence

$ \Delta L_B=\frac{F L_B}{A Y_B} =\frac{F(2L_A)}{A(2Y_A)} =\frac{F L_A}{A Y_A} =\Delta L_A $

Therefore,

$ \Delta L_B=1\ \text{mm} $

Net extension of the combination:

$ \Delta L=\Delta L_A+\Delta L_B=1+1=2\ \text{mm} $

Therefore, the correct answer is

$ \boxed{2\ \text{mm}} $

So, Option B is correct.

Let’s break down the solution step by step:

First, note the given values:

Length of the wire, $L = 3 \, \text{m}$

Radius of the wire, $r = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$

Extension of the wire, $\Delta L = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m}$

Mass attached, $m = 50 \, \text{kg}$

Acceleration due to gravity, $g = 3\pi \, \text{m/s}^2$

The force (weight) acting on the wire is:

$F = mg = 50 \times 3\pi = 150\pi \, \text{N}$

Next, calculate the cross-sectional area, $A,$ of the wire:

$A = \pi r^2 = \pi (3 \times 10^{-3})^2 = \pi \times 9 \times 10^{-6} = 9\pi \times 10^{-6} \, \text{m}^2$

Young’s modulus, $E,$ is given by:

$E = \frac{FL}{A \Delta L}$

Substitute the values into the formula:

$ E = \frac{(150\pi) \times 3}{(9\pi \times 10^{-6}) \times (1 \times 10^{-4})} $

Simplify step by step:

Multiply in the numerator:

$150\pi \times 3 = 450\pi$

Multiply in the denominator:

$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$

Cancel the common factor $\pi$ in the numerator and denominator:

$ E = \frac{450}{9 \times 10^{-10}} $

Divide 450 by 9:

$ E = \frac{450}{9} \times \frac{1}{10^{-10}} = 50 \times 10^{10} = 5 \times 10^{11} \, \text{Nm}^{-2} $

The problem states that Young's modulus can be expressed as $P \times 10^{11} \, \text{Nm}^{-2}.$ Here we have:

$ P = 5 $

Thus, the correct option is:

Option D (5).

$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} \\ & \mathrm{~h}=\frac{100}{7} \mathrm{~cm} \\ & \sin 60^{\circ}=\frac{\mathrm{h}}{\ell} \\ & \ell=\frac{\mathrm{h} \times 2}{\sqrt{3}} \\ & \ell=\frac{100}{7} \times \frac{2}{\sqrt{3}} \\ & =\frac{200}{7 \times \sqrt{3}} \\ & =16.49 \mathrm{~cm} \end{aligned}$

Given:

The ratio of lengths: $\frac{L_A}{L_B} = \frac{1}{3}$

The ratio of diameters: $\frac{d_A}{d_B} = 2$

Both wires are subject to the same stretching force, and since they are made of the same material, their Young's modulus ($Y$) is the same.

The elongation ($\Delta L$) of a wire subject to a force is given by:

$ \Delta L = \frac{F \cdot L}{A \cdot Y} $

where $F$ is the force applied, $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus.

For wires $A$ and $B$:

$\Delta L_A = \frac{F_A \cdot L_A}{A_A \cdot Y_A}$

$\Delta L_B = \frac{F_B \cdot L_B}{A_B \cdot Y_B}$

Since $F_A = F_B$ and $Y_A = Y_B$, the ratio of their elongations becomes:

$ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} $

The cross-sectional area $A$ is a function of diameter, $A = \frac{\pi}{4} d^2$. Therefore,

$ A_A = \frac{\pi}{4} d_A^2 \quad \text{and} \quad A_B = \frac{\pi}{4} d_B^2 $

Substituting these into the elongation ratio:

$ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{L_A}{L_B}\right) \left(\frac{\frac{\pi}{4} d_B^2}{\frac{\pi}{4} d_A^2}\right) = \left(\frac{L_A}{L_B}\right) \left(\frac{d_B}{d_A}\right)^2 $

Substitute the given ratios:

$ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2 = \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = \frac{1}{12} $

Thus, the ratio of the elongations of wires $A$ and $B$ is $\frac{1}{12}$.

$\begin{aligned} & \text { Shear moduli }=\frac{\sigma_{\text {shear }}}{\theta} \\ & 10^{10}=\frac{10^5}{\pi \times 16 \times 10^{-4}} \times \frac{1}{\theta} \\ & \theta=\frac{1}{160 \pi} \text { Radian } \end{aligned}$

The contact angle, $\theta,$ of a liquid in a capillary tube determines its meniscus shape:

If $\theta < 90^\circ,$ then $\cos \theta > 0$ and the liquid wets the tube, giving a concave meniscus.

If $\theta > 90^\circ,$ then $\cos \theta < 0$ and the liquid is non-wetting, resulting in a convex meniscus.

If $\theta = 90^\circ,$ then $\cos \theta = 0$ and the meniscus is essentially flat.

We are given the ratio:

$K = \frac{\cos \theta_A}{\cos \theta_B}.$

For $K$ to be negative:

The numerator and denominator must have opposite signs.

This means one of the liquids has $\cos \theta > 0$ (concave meniscus) and the other has $\cos \theta < 0$ (convex meniscus).

Consider Option D:

It states that if $K$ is negative, then liquid A has a concave meniscus (so $\theta_A < 90^\circ$, $\cos \theta_A > 0$) and liquid B has a convex meniscus (so $\theta_B > 90^\circ$, $\cos \theta_B < 0$).

Hence, the ratio becomes negative:

$K = \frac{(+)}{(-)} < 0.$

Therefore, the correct statement is:

Option D

First, when the two vessels are connected, the water levels equalize at

$H = \frac{H_1 + H_2}{2} = \frac{10 + 6}{2} = 8\text{ m}.$

We can get the work done by gravity from the loss of gravitational potential energy:

Initial potential energy (taking the bottom as zero)

For a column of height $H$ and cross-section $A$,

$U = \rho g A\int_0^H z\,dz = \frac{\rho g A H^2}{2}.$

So

$U_i = \frac{\rho g A}{2}\bigl(H_1^2 + H_2^2\bigr) = \frac{10^3\cdot 10\cdot 2}{2}(10^2 + 6^2) = 10^4\,(100 + 36) = 1.36\times10^6\text{ J}.$

Final potential energy (both at height 8 m)

$U_f = 2\;\frac{\rho g A H^2}{2}\;\Big|_{H=8} = \rho g A\,(8^2) = 10^3\cdot10\cdot2\cdot64 = 1.28\times10^6\text{ J}.$

Work done by gravity = drop in potential energy

$W = U_i - U_f = 1.36\times10^6 - 1.28\times10^6 = 0.08\times10^6 = 8\times10^4\text{ J}.$

Answer: 8×10^4 J (Option C).

To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid:

$ v_T = \frac{2}{9} \frac{(\rho_s - \rho_f) r^2 g}{\eta} $

Where:

$ v_T $ is the terminal velocity,

$ \rho_s $ is the density of the steel ball,

$ \rho_f $ is the density of the fluid,

$ r $ is the radius of the ball,

$ g $ is the acceleration due to gravity,

$ \eta $ is the viscosity of the fluid.

Given:

Diameter of the steel ball = 3.6 mm, therefore radius $ r = 1.8 $ mm = $ 1.8 \times 10^{-3} $ m,

Terminal velocity $ v_T = 2.45 \times 10^{-2} $ m/s,

Density of oil $ \rho_f = 925 $ kg/m$^3$,

Density of steel $ \rho_s = 7825 $ kg/m$^3$,

Acceleration due to gravity $ g = 9.8 $ m/s$^2$.

Rearranging the formula to solve for $\eta$:

$ \eta = \frac{2}{9} \cdot \frac{(\rho_s - \rho_f) \cdot r^2 \cdot g}{v_T} $

Substituting the known values into the equation:

$ \eta = \frac{2}{9} \cdot \frac{(7825 - 925) \cdot (1.8 \times 10^{-3})^2 \cdot 9.8}{2.45 \times 10^{-2}} $

Calculating this, we approximate the viscosity $\eta$ to be:

$ \eta \approx 1.99 \, \text{Pa}\cdot\text{s} $

Thus, the viscosity of the oil is approximately 1.99 Pa·s.

Apply Bernouli equation between points 1 & 2

$ \begin{aligned} & P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_2^2+0 \\ & P_0+\frac{m g}{A}+\rho g \frac{70}{100}=P_0+\frac{1}{2} \rho v_2^2 \\ & \frac{5000}{0.5}+10^3 \times 10 \frac{70}{100}=\frac{1}{2} \times 10^3 v_2^2 \\ & 10^3+10^3 \times 7=\frac{10^3}{2} v_2^2 \\ & v_2^2=16 \\ & v_2=4 \mathrm{~m} / \mathrm{s} \end{aligned}$

As the tank area is large $v_1$ is negligible compared $\text { to } v_2$

To determine the surface energy released when two identical water drops coalesce to form a larger drop, let's go through the process step by step:

Volume Conservation:

When two droplets, each with a radius $ R $, merge to form a larger droplet, the total volume is conserved. Therefore:

$ 2 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3 $

Solving for $ r $, the radius of the larger drop, we find:

$ r = 2^{1/3} R $

Initial Surface Energy ($ U_i $):

The initial surface energy of the two smaller droplets is given by the surface area of each multiplied by the surface tension $ T $:

$ U_i = 2 \times 4 \pi R^2 T $

Final Surface Energy ($ U_f $):

The surface energy of the larger droplet is given by:

$ U_f = 4 \pi r^2 T = 4 \pi R^2 T (2^{2/3}) $

Energy Released:

The energy released in the process is the difference between the initial and final surface energy:

$ \text{Heat lost} = U_i - U_f = 4 \pi R^2 T \left[2 - 2^{2/3}\right] $

This expression gives us the amount of surface energy that is released when two identical water drops combine to form a single larger drop.

The fractional compression $\left( \frac{\Delta V}{V} \right)$ of water at a depth of 2.5 km below sea level is calculated as follows:

Given:

Bulk modulus of water, $B = 2 \times 10^9 \, \text{N/m}^2$

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

Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

The relationship between pressure change and volume change using the bulk modulus is given by:

$ B = \frac{\rho gh}{\left(\frac{\Delta V}{V}\right)} $

Rearranging the formula to solve for the fractional compression:

$ \frac{\Delta V}{V} \times 100 = \frac{\rho gh}{B} \times 100 $

Substituting in the given values:

$ \frac{1000 \times 10 \times 2.5 \times 10^3}{2 \times 10^9} \times 100\% $

Calculating the result gives:

$ = 1.25\% $

Thus, the fractional compression of water at this depth is 1.25%.

First, let’s find the weight of the cube and set it equal to the upward buoyant force that keeps the cube floating.

Step 1: Set up the equation for floating
The weight of the cube (mass × gravity) is balanced by the buoyant force (density of water × volume of cube under water × gravity):
$\mathrm{Mg} = \mathrm{F}_{\mathrm{B}} \Rightarrow (400 \times 10^{-3}) = 10^3 \times \mathrm{V}_{\mathrm{d}}$

Step 2: Solve for the volume under water
We find the volume of the cube under water ($\mathrm{V}_\mathrm{d}$):
$\mathrm{V}_{\mathrm{d}} = 400 \times 10^{-6}~\mathrm{m}^3$

Step 3: Find the total volume of the cube
Each edge of the cube is 10 cm, so total volume is:
$ (10 \times 10^{-2})^3 $

Step 4: Find the volume outside water
Subtract the volume under water from the total volume to get the volume outside water:
$(\text {Vol.})_{\text{outside}} = (10 \times 10^{-2})^3 - 400 \times 10^{-6}$

Calculate the answer:
$= 600 \times 10^{-6}~\mathrm{m}^3 = 600~\mathrm{cm}^3$

We know, terminal velocity of a sphere of radius R in a liquid of viscosity $\eta$,

$v = {2 \over 9}{{{R^2}} \over \eta }(\sigma - \rho )$ .... (1)

where, $\sigma$ = mass of density of sphere

$\rho$ = density of liquid

we can see, $v \propto {R^2}$ (for constant $\eta,\sigma$ & $\rho$)

Hence, graph between v and R is parabola.

As v depends on R so the terminal velocities of different diameter balls will be different.

We know, the viscosity of a liquid usually decreases as the temperature increases and $v \propto {1 \over \eta }$

So terminal velocity depends on the temperature. $T \uparrow \Rightarrow \eta \downarrow \Rightarrow v \uparrow $

As the equation $v = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$ involves density of liquid $\rho$. So the experiment can be utilized to asses it.

From (1), $\eta = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$

Here, $\eta$ does not depend on initial speed of the sphere. Hence, option 3 is correct.

(A) Surface tension arises due to the extra potential energy of the molecules at the surface of a liquid compared to the molecules in the interior.

(B) When a liquid is heated, the kinetic energy of it molecules increases, causing them to move more freely and overcome the intermolecular forces that contribute to viscosity, thus lowering it.

(C) Higher temperature in gas means faster moving molecules, resulting in more collisions and therefore an increase in viscosity.

(D) Reynold's number $ = {{Inertial\,force} \over {Visous\,force}}$

${R_e} = {{\rho VD} \over \mu }$,

v = velocity of flow

D = Diameter of pipe

If, $R_e < 2000$, flow is laminar flow

$2000 < R_e < 3000 \Rightarrow$ transition region

$R_e > 3000 \Rightarrow$ turbulant flow.

(E) Tangent at a point to the streamline gives the direction of net velocity of the flow. If the two streamlines intersect each other. It signifies two directions of velocity which cannot be possible. Thus, two streamlines cannot intersect each other. Hence, option 2 is correct.

$ P_{\text{bubble}} = P_{\text{atm}} + \rho g h + \frac{2S}{r} $

Given that the pressure inside the bubble is $P_{\text{atm}} + 2100\,\text{N/m}^2$ and the liquid pressure at a depth of 20 cm is

$ P_{\text{liquid}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.20 = P_{\text{atm}} + 2000\,\text{N/m}^2, $

the pressure difference due solely to surface tension is

$ \Delta P = \Bigl(P_{\text{bombubble}} - P_{\text{liquid}}\Bigr) = \bigl(P_{\text{atm}}+2100\bigr) - \bigl(P_{\text{atm}}+2000\bigr) = 100\,\text{N/m}^2. $

The Laplace pressure difference for a bubble (which has one interface) is given by

$ \Delta P = \frac{2S}{r}. $

The given bubble radius is 0.1 cm, which in SI units is

$ r = 0.1\,\text{cm} = 0.001\,\text{m}. $

Substituting the known values into the Laplace equation gives

$ \frac{2S}{0.001} = 100. $

Solving for $S$:

$ 2S = 100 \times 0.001 = 0.1, $

$ S = \frac{0.1}{2} = 0.05\,\text{N/m}. $

Thus, the surface tension of the liquid is

$ \boxed{0.05\,\text{N/m}}. $