Properties of Matter — Page 14 | JEE Physics

Q131

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A: When you squeeze one end of a tube to get toothpaste out from the other end, Pascal's principle is observed. Reason R: A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container. In the light of the above statements, choose the most appropriate answer from the options given below

A Both A and R are correct but R is NOT the correct explanation of A

B A is not correct but R is correct

C A is correct but R is not correct

D Both A and B are correct and R is the correct explanation of A

Correct Answer

Option D

Solution

Assertion A states that when you squeeze one end of a tube to get toothpaste out from the other end, Pascal's principle is observed.

This is true because when you apply pressure on one end of the tube, the pressure is transmitted uniformly throughout the enclosed incompressible fluid (the toothpaste in this case) and eventually pushes the toothpaste out of the other end.

Reason R provides the definition of Pascal's principle: "A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container." This principle is directly applicable to the situation described in Assertion A.

When you squeeze the tube, you apply pressure on the toothpaste, and this pressure is transmitted uniformly to all parts of the toothpaste, causing it to be pushed out of the other end.

Therefore, Option D is the correct choice as both Assertion A and Reason R are correct statements, and R is the correct explanation of A.

Q132

A big drop is formed by coalescing 1000 small droplets of water. The surface energy will become :

A

\dfrac{1}{100}

th

B

\dfrac{1}{10}

th

C 100 times

D 10 times

Correct Answer

Option B

Solution

To answer this question, we need to understand the relationship between the surface area of the droplets and the surface energy involved.

Surface energy is directly proportional to the surface area of the liquid.

The surface energy,

E

, for a droplet is given by:

E = \gamma \times A

Where:

E

is the surface energy, $\gamma$ is the surface tension of the liquid, and

A

is the surface area of the droplet.

When multiple droplets coalesce, they form a larger droplet with a certain volume.

Since the volume is conserved, the volume of the large droplet will be equal to the sum of the volumes of the small droplets.

Let's denote:

r

as the radius of a small droplet,

R

as the radius of the large droplet,

V_{\text{small}}

as the volume of a small droplet, and

V_{\text{large}}

as the volume of the large droplet. The volume of one small droplet is:

V_{\text{small}} = \frac{4}{3}\pi r^3

The total volume of 1000 small droplets is:

1000 \times V_{\text{small}} = 1000 \times \frac{4}{3}\pi r^3

Since the volume is conserved, the volume of the large droplet formed by the coalescence of 1000 small droplets is:

V_{\text{large}} = 1000 \times \frac{4}{3}\pi r^3

Now, if

R

is the radius of the large droplet, then:

V_{\text{large}} = \frac{4}{3}\pi R^3

Equating the volumes, we have:

\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3

R^3 = 1000 \times r^3

R = 10r

Now, let's look at the surface area. The surface area for a small droplet is

A_{\text{small}} = 4\pi r^2

and for a large droplet is

A_{\text{large}} = 4\pi R^2

. Substitute

R = 10r

:

A_{\text{large}} = 4\pi (10r)^2

A_{\text{large}} = 4\pi \times 100r^2

A_{\text{large}} = 100 \times 4\pi r^2

A_{\text{large}} = 100 \times A_{\text{small}}

So, the large droplet has 100 times the surface area of one small droplet.

The surface energy of 1000 small droplets would be

1000 \times E_{\text{small}}

because each small droplet has an energy

E_{\text{small}} = \gamma \times A_{\text{small}}

. The surface energy of the big droplet is

E_{\text{large}} = \gamma \times A_{\text{large}}

. But we have just shown that

A_{\text{large}} = 100 \times A_{\text{small}}

, so:

E_{\text{large}} = \gamma \times 100 \times A_{\text{small}}

This means the surface energy of the big droplet is 100 times the surface energy of one small droplet.

Since there were 1000 small droplets originally, the surface energy of the big droplet is:

\frac{E_{\text{large}}}{1000 \times E_{\text{small}}} = \frac{\gamma \times 100 \times A_{\text{small}}}{1000 \times \gamma \times A_{\text{small}}} = \frac{1}{10}

Therefore, the correct answer is: Option B: The surface energy will become

\frac{1}{10}

th of the original.

Q133

With rise in temperature, the Young's modulus of elasticity :

A changes erratically

B increases

C decreases

D remains unchanged

Correct Answer

Option C

Solution

The Young's modulus of elasticity, denoted as $E$ , is a measure of the stiffness of a material.

It defines the relationship between stress (force per unit area) and strain (deformation) in a material in the linear (elastic) portion of the stress-strain curve.

As temperature changes, the interatomic distances and bonding energies within a material also change, affecting its mechanical properties, including its elasticity.

For most materials, as the temperature increases, the atoms within the material gain kinetic energy and vibrate more vigorously.

This increased vibration results in a reduction of the forces between atoms, making it easier for the material to deform under a given load.

Hence, generally, the Young's modulus $E$ decreases with increasing temperature because the material becomes softer and less stiff.

Thus, the correct option here would be: Option C: decreases However, it should be noted that the exact relationship between temperature and Young's modulus can vary depending on the material type and its microstructure.

While the general trend for metals and polymers is a decrease in Young's modulus with rising temperature, the rate of decrease and the temperature range over which this occurs can differ significantly between materials.

Some advanced materials and composites may exhibit more complex behavior due to their unique properties.

Q134

Given below are two statements : Statement (I) :Viscosity of gases is greater than that of liquids. Statement (II) : Surface tension of a liquid decreases due to the presence of insoluble impurities. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is correct but statement II is incorrect

B Statement I is incorrect but Statement II is correct

C Both Statement I and Statement II are incorrect

D Both Statement I and Statement II are correct

Correct Answer

Option B

Solution

Gases have less viscosity. Due to insoluble impurities like detergent surface tension decreases

Q135

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The property of body, by virtue of which it tends to regain its original shape when the external force is removed, is Elasticity. Reason (R) : The restoring force depends upon the bonded inter atomic and inter molecular force of solid. In the light of the above statements, choose the correct answer from the options given below :

A (A) is false but (R) is true

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C Both (A) and (R) are true and (R) is the correct explanation of (A)

D (A) is true but (R) is false

Correct Answer

Option C

Solution

The statement given as Assertion (A) is indeed true.

Elasticity is a physical property of a material whereby it can restore it to its original shape or size after deformation (stretching, compressing, or twisting) provided the deformation is within the material's elastic limit.

Once the external force is removed, an elastic material will return to its initial state due to its intrinsic property.

So, we can affirm the validity of Assertion (A).

The statement labelled as Reason (R) is also true.

When a material is deformed due to an applied external force, the molecules within the material are displaced from their equilibrium positions.

The intermolecular and interatomic forces create a restoring force that attempts to bring the molecules back to their original, or equilibrium, positions.

This restoring force is what allows the material to regain its shape when the external force is removed, assuming the material has not been deformed beyond its elastic limit.

Hence, the restoring force indeed depends on the bonded interatomic and intermolecular forces within a solid.

Furthermore, Reason (R) is not just a true statement in isolation, but it is the correct explanation for Assertion (A).

The reason that a material exhibits elasticity (Assertion A) is directly tied to the nature of the restoring force (Reason R), which is a result of the bonded interatomic and intermolecular forces within the material.

Without these restoring forces, the material would not be able to return to its original shape, and thus would not exhibit elasticity.

Therefore, the correct answer to the given question is: Option C Both (A) and (R) are true and (R) is the correct explanation of (A)

Q136

A small liquid drop of radius

R

is divided into 27 identical liquid drops. If the surface tension is

T

, then the work done in the process will be:

A

4 \pi \mathrm{R}^2 \mathrm{~T}

B

8 \pi R^2 \mathrm{~T}

C

\dfrac{1}{8} \pi R^2 T

D

3 \pi R^2 \mathrm{~T}

Correct Answer

Option B

Solution

Volume constant

\begin{aligned} & \frac{4}{3} \pi R^3=27 \times \frac{4}{3} \times \pi r^3 \\ & R^3=27 r^3 \\ & R=3 r \\ & r=\frac{R}{3} \\ & r^2=\frac{R^2}{9} \end{aligned}

\begin{aligned} & \text{ Work done }=T . \Delta A \\ & =27 T\left(4 \pi r^2\right)-T 4 \pi R^2 \\ & =27 T 4 \pi \frac{R^2}{9}-4 \pi R^2 T \\ & =8 \pi R^2 T \end{aligned}

Q137

A wire of length

L

and radius

r

is clamped at one end. If its other end is pulled by a force

F

, its length increases by

l

. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

A 2 times

B 4 times

C 3 times

D

\dfrac{3}{2}

times

Correct Answer

Option A

Solution

\begin{aligned} & Y=\frac{\text{ stress }}{\text{ strain }} \\ & Y=\frac{\frac{\mathrm{F}}{\frac{\pi \mathrm{r}^2}{\ell}}}{\frac{\ell}{\mathrm{L}}} \end{aligned}

\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{\ell}{\mathrm{L}} \quad \text{.... (i)}

\begin{aligned} & \mathrm{Y}=\frac{\frac{\mathrm{F} / 2}{\pi \mathrm{r}^2 / 4}}{\frac{\Delta \ell}{\mathrm{L}}} \\ & \mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4} \end{aligned}

From (i)

\begin{aligned} & \mathrm{Y} \pi \mathrm{r}^2 \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2} \\ & \Delta \ell=2 \ell \end{aligned}

Q138

The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:

A

1: 9

B

1: 27

C

1: 81

D

1: 3

Correct Answer

Option B

Solution

To find the ratio between the volumes of the first and the second soap bubble, we need to understand the relation between the excess pressure inside a soap bubble and its volume.

The excess pressure ( $P$ ) inside a soap bubble is given by the formula:

P = \frac{4T}{r}

where $T$ is the surface tension of the soap solution, and $r$ is the radius of the soap bubble.

For a soap bubble, the factor of 4 comes from having two surfaces (inner and outer), each contributing $2T/r$ to the pressure.

Given that the excess pressure inside the first soap bubble ( $P_1$ ) is thrice the excess pressure inside the second soap bubble ( $P_2$ ), we can write:

P_1 = 3P_2

Substituting the formula for excess pressure, we get:

\frac{4T}{r_1} = 3 \times \frac{4T}{r_2} \Rightarrow \frac{1}{r_1} = 3 \times \frac{1}{r_2} \Rightarrow \frac{r_2}{r_1} = 3

Next, we calculate the ratio between their volumes. The volume ( $V$ ) of a sphere (or a bubble) is given by:

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

Thus, the volume ratio of the first bubble ( $V_1$ ) to the second bubble ( $V_2$ ) is:

\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3

Since we found that $\dfrac{r_2}{r_1} = 3$ , it then follows that:

\frac{V_1}{V_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27}

Therefore, the correct option is: Option B:

1: 27

Q139

A sphere of relative density

\sigma

and diameter

D

has concentric cavity of diameter

d

. The ratio of

\dfrac{D}{d}

, if it just floats on water in a tank is :

A

\left(\dfrac{\sigma-2}{\sigma+2}\right)^{1 / 3}

B

\left(\dfrac{\sigma+1}{\sigma-1}\right)^{1 / 3}

C

\left(\dfrac{\sigma-1}{\sigma}\right)^{1 / 3}

D

\left(\dfrac{\sigma}{\sigma-1}\right)^{1 / 3}

Correct Answer

Option D

Solution

To solve this problem, we consider the buoyancy and weight force acting on the sphere.

For the sphere to just float on water, the weight of the water displaced by the sphere must be equal to the weight of the sphere.

The volume of water displaced by the sphere is equivalent to the outer volume of the sphere minus the volume of the cavity inside it.

The volume of a sphere is given by

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

, where $r$ is the radius of the sphere. For the given sphere, its outer radius is

R = \frac{D}{2}

, and the radius of the cavity is

r = \frac{d}{2}

. Therefore, the volume of the sphere excluding the cavity is:

V_{\text{solid part}} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3

V_{\text{solid part}} = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 - \frac{4}{3}\pi \left(\frac{d}{2}\right)^3

Relative density ( $\sigma$ ) is defined as the ratio of the density of an object to the density of water.

This means the actual density of the sphere is $\sigma \times \text{density of water}$ .

Since the object just floats, the weight of the displaced water is equal to the weight of the solid part of the sphere (ignoring the cavity), which can be mathematically represented as:

\text{Weight of solid part} = \text{Weight of displaced water}

\sigma \cdot \rho_{\text{water}} \cdot V_{\text{solid part}} \cdot g = \rho_{\text{water}} \cdot V_{\text{displaced water}} \cdot g

Since the sphere is floating, $V_{\text{displaced water}} = \dfrac{4}{3}\pi R^3$ , the equation simplifies to:

\sigma \left(\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3

Cancelling out common terms gives:

\sigma (R^3 - r^3) = R^3

Given that $\sigma$ is the relative density, we can rearrange the equation to solve for the ratio of $D/d$ or equivalently $R/r$ :

\sigma = \frac{R^3}{R^3 - r^3}

Solving for $R/r$ :

R^3 (1 - \sigma) + \sigma r^3 = 0

\frac{R^3}{r^3} = \frac{\sigma}{\sigma - 1}

Since $R = \dfrac{D}{2}$ and $r = \dfrac{d}{2}$ , the ratio of $D/d$ is equal to the ratio of $R/r$ , thus:

\frac{D}{d} = \left(\frac{\sigma}{\sigma - 1}\right)^{1/3}

This matches Option D:

\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}

Q140

Given below are two statements : Statement I : The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well. Statement II : The rise of a liquid in a capillary tube does not depend on the inner radius of the tube. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false.

B Both Statement I and Statement II are true.

C Statement I is false but Statement II is true.

D Statement I is true but Statement II is false.

Correct Answer

Option D

Solution

Option D, "Statement I is true but Statement II is false," is the correct choice.

Here's an explanation for both statements: Statement I: True The contact angle between a solid and a liquid is indeed a measure of the wettability of the solid surface by the liquid.

The contact angle is determined by the nature of both the solid and the liquid.

It is a function of the interfacial tensions between solid-liquid ( $\gamma_{\text{SL}}$ ), solid-vapor ( $\gamma_{\text{SV}}$ ), and liquid-vapor ( $\gamma_{\text{LV}}$ ).

This relationship can be understood through Young's equation: $\cos \theta = \dfrac{\gamma_{\text{SV}} - \gamma_{\text{SL}}}{\gamma_{\text{LV}}}$ Where $\theta$ is the contact angle.

Therefore, since the interfacial tensions vary with the materials in contact, the contact angle is indeed a property of the materials of both the solid and the liquid.

Statement II: False The rise of a liquid in a capillary tube is strongly dependent on the inner radius of the tube.

This relationship is described by the Jurin's Law, which states the height ( $h$ ) to which a liquid will rise (or fall) in a capillary tube is inversely proportional to the radius ( $r$ ) of the tube, among other factors.

The law is given by: $h = \dfrac{2\gamma \cos \theta}{\rho g r}$ Where: $\gamma$ is the liquid-air surface tension, $\theta$ is the contact angle, $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, $r$ is the radius of the capillary tube.

As seen from the equation, $h$ is inversely proportional to $r$ .

Therefore, the rise of the liquid indeed depends on the inner radius of the tube, making Statement II false.