Properties of Matter — Page 16 | JEE Physics

Q151

The amount of work done to break a big water drop of radius '

R

' into 27 small drops of equal radius is 10 J . The work done required to break the same big drop into 64 small drops of equal radius will be

A 20 J

B 10 J

C 5 J

D 15 J

Correct Answer

Option D

Solution

To solve the problem, we need to compare the increase in surface area when the big drop is split into small drops.

Step 1: Determine the radius of the small drops For 27 small drops: Total volume is conserved: $27 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi R^3 \implies r^3 = \dfrac{R^3}{27} \implies r = \dfrac{R}{3}.$ For 64 small drops: Similarly, $64 \times \dfrac{4}{3}\pi r'^3 = \dfrac{4}{3}\pi R^3 \implies r'^3 = \dfrac{R^3}{64} \implies r' = \dfrac{R}{4}.$ Step 2: Calculate the surface area before and after the break-up Initial surface area of the big drop: $A_{\text{initial}} = 4\pi R^2.$ For 27 drops: Surface area of one small drop: $4\pi\left(\dfrac{R}{3}\right)^2 = \dfrac{4\pi R^2}{9}.$ Total surface area: $A_{27} = 27 \times \dfrac{4\pi R^2}{9} = 3 \times 4\pi R^2 = 12\pi R^2.$ Increase in surface area: $\Delta A_{27} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2.$ For 64 drops: Surface area of one small drop: $4\pi\left(\dfrac{R}{4}\right)^2 = 4\pi \dfrac{R^2}{16} = \dfrac{\pi R^2}{4}.$ Total surface area: $A_{64} = 64 \times \dfrac{\pi R^2}{4} = 16\pi R^2.$ Increase in surface area: $\Delta A_{64} = 16\pi R^2 - 4\pi R^2 = 12\pi R^2.$ Step 3: Relate work done to the change in surface area Since the work done is proportional to the increase in surface area: $\text{Work} \propto \Delta A.$ We know that breaking into 27 drops requires 10 J corresponding to an increase of $8\pi R^2$ .

Thus, if $W$ is the work done, $W_{27} = k \cdot 8\pi R^2 = 10\, \text{J},$ where $k$ is the proportionality constant.

For 64 drops: $W_{64} = k \cdot 12\pi R^2.$ Dividing the two equations: $\dfrac{W_{64}}{10} = \dfrac{12\pi R^2}{8\pi R^2} = \dfrac{12}{8} = \dfrac{3}{2}.$ Thus, $W_{64} = 10 \times \dfrac{3}{2} = 15\, \text{J}.$ Final Answer: 15 J (Option D)

Q152

In the experiment for measurement of viscosity '

\eta

' of given liquid with a ball having radius

R

, consider following statements. A. Graph between terminal velocity V and R will be a parabola. B. The terminal velocities of different diameter balls are constant for a given liquid. C. Measurement of terminal velocity is dependent on the temperature. D. This experiment can be utilized to assess the density of a given liquid. E. If balls are dropped with some initial speed, the value of

\eta

will change. Choose the correct answer from the options given below:

A C, D and E Only

B A, B and E Only

C A, C and D Only

D B, D and E Only

Correct Answer

Option C

Solution

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.

Q153

Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid rises, the coefficient of viscosity increases. C. As the temperature of gas increases, the coefficient of viscosity increases D. The onset of turbulence is determined by Reynold's number. E. In a steady flow two stream lines never intersect. Choose the correct answer from the options given below:

A B, C, D Only

B C, D, E Only

C A, D, E Only

D A, B, C Only

Correct Answer

Option B

Solution

(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 $R_e > 3000 \Rightarrow$ 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.

Q154

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is

P \times 10^{11} \, \text{Nm}^{-2}

, where the value of

P

is: (Take

g = 3\pi \, \text{m/s}^2

)

A 2.5

B 25

C 10

D 5

Correct Answer

Option D

Solution

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).

Q155

Two cylindrical vessels of equal cross sectional area of

2 \mathrm{~m}^2

contain water upto heights 10 m and 6 m , respectively. If the vessels are connected at their bottom then the work done by the force of gravity is (Density of water is

10^3 \mathrm{~kg} / \mathrm{m}^3

and

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2

)

A

{ }1 \times 10^5 \mathrm{~J}

B

4 \times 10^4 \mathrm{~J}

C

8 \times 10^4 \mathrm{~J}

D

6 \times 10^4 \mathrm{~J}

Correct Answer

Option C

Solution

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).

Q156

Two wires A and B are made of same material having ratio of lengths

\dfrac{L_A}{L_B}=\dfrac{1}{3}

and their diameters ratio

\dfrac{d_A}{d_B}=2

. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

A

3: 4

B

1: 12

C

1: 3

D

1: 6

Correct Answer

Option B

Solution

Given: The ratio of lengths: $\dfrac{L_A}{L_B} = \dfrac{1}{3}$ The ratio of diameters: $\dfrac{d_A}{d_B} = 2$ 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: $\Delta L = \dfrac{F \cdot L}{A \cdot Y}$ 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 $B$ : $\Delta L_A = \dfrac{F_A \cdot L_A}{A_A \cdot Y_A}$ $\Delta L_B = \dfrac{F_B \cdot L_B}{A_B \cdot Y_B}$ Since $F_A = F_B$ and $Y_A = Y_B$ , the ratio of their elongations becomes: $\dfrac{\Delta L_A}{\Delta L_B} = \dfrac{L_A \cdot A_B}{L_B \cdot A_A}$ The cross-sectional area $A$ is a function of diameter, $A = \dfrac{\pi}{4} d^2$ .

Therefore, $A_A = \dfrac{\pi}{4} d_A^2 \quad \text{and} \quad A_B = \dfrac{\pi}{4} d_B^2$ Substituting these into the elongation ratio: $\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{L_A}{L_B}\right) \left(\dfrac{\dfrac{\pi}{4} d_B^2}{\dfrac{\pi}{4} d_A^2}\right) = \left(\dfrac{L_A}{L_B}\right) \left(\dfrac{d_B}{d_A}\right)^2$ Substitute the given ratios: $\dfrac{\Delta L_A}{\Delta L_B} = \left(\dfrac{1}{3}\right)\left(\dfrac{1}{2}\right)^2 = \left(\dfrac{1}{3}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{12}$ Thus, the ratio of the elongations of wires $A$ and $B$ is $\dfrac{1}{12}$ .

Q157

Two water drops each of radius '

r

' coalesce to form a bigger drop. If '

T

' is the surface tension, the surface energy released in this process is :

A

4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{1 / 3}\right]

B

4 \pi \mathrm{r}^2 \mathrm{~T}[1+\sqrt{2}]

C

4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{2 / 3}\right]

D

4 \pi \mathrm{r}^2 \mathrm{~T}[\sqrt{2}-1]

Correct Answer

Option C

Solution

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: $2 \times \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \pi r^3$ 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.

Q158

Consider a completely full cylindrical water tank of height 1.6 m and of cross-sectional area

0.5 \mathrm{~m}^2

. It has a small hole in its side at a height 90 cm from the bottom. Assume, the crosssectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is:

\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)

A

2 \mathrm{~m} / \mathrm{s}

B

5 \mathrm{~m} / \mathrm{s}

C

3 \mathrm{~m} / \mathrm{s}

D

4 \mathrm{~m} / \mathrm{s}

Correct Answer

Option D

Solution

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

Q159

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of

10^5 \mathrm{~N}

at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement

\theta

of the rod axis from its original position would be : (shear moduli,

G=10^{10} \mathrm{~N} / \mathrm{m}^2

)

A

1 / 160 \pi

B

1 / 2 \pi

C

1 / 4 \pi

D

1 / 40 \pi

Correct Answer

Option A

Solution

\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}

Q160

Two liquids

A

and

B

have

\theta_A

and

\theta_B

as contact angles in a capillary tube. If

K=\cos \theta_A / \cos \theta_B

, then identify the correct statement:

A K is negative, then liquid A and liquid B have convex meniscus.

B

K

is negative, then liquid

A

and liquid

B

have concave meniscus.

C K is zero, then liquid A has convex meniscus and liquid B has concave meniscus.

D K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.

Correct Answer

Option D

Solution

The contact angle,

\theta,

of a liquid in a capillary tube determines its meniscus shape: If

\theta 0

and the liquid wets the tube, giving a concave meniscus. If

\theta > 90^\circ,

then

\cos \theta 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 Consider Option D: It states that if

K

is negative, then liquid A has a concave meniscus (so

\theta_A 0

) and liquid B has a convex meniscus (so

\theta_B > 90^\circ

,

\cos \theta_B Hence, the ratio becomes negative:

K = \frac{(+)}{(-)} Therefore, the correct statement is: Option D