Properties of Matter — Page 15 | JEE Physics

Q141

Young's modulus is determined by the equation given by

\mathrm{Y}=49000 \dfrac{\mathrm{m}}{\mathrm{l}} \dfrac{\mathrm{dyne}}{\mathrm{cm}^2}

where

M

is the mass and

l

is the extension of wire used in the experiment. Now error in Young modules

(Y)

is estimated by taking data from

M-l

plot in graph paper. The smallest scale divisions are

5 \mathrm{~g}

and

0.02 \mathrm{~cm}

along load axis and extension axis respectively. If the value of

M

and

l

are

500 \mathrm{~g}

and

2 \mathrm{~cm}

respectively then percentage error of

Y

is :

A 2%

B 0.02%

C 0.5%

D 0.2%

Correct Answer

Option A

Solution

To determine the percentage error in Young's modulus, we need to first understand the propagation of errors in the given formula.

Given the equation:

\mathrm{Y}=49000 \frac{\mathrm{M}}{\mathrm{l}} \frac{\mathrm{dyne}}{\mathrm{cm}^2}

where:

M

is the mass (with its value given as

500 \mathrm{~g}

)

l

is the extension (with its value given as

2 \mathrm{~cm}

) The errors in the measurements are determined by the smallest scale divisions on the graph paper, which are:

5 \mathrm{~g}

for the load axis

0.02 \mathrm{~cm}

for the extension axis To find the percentage error in Young's modulus (

Y

), we need to compute the relative errors in the measurements

M

and

l

, and then propagate these errors through the given formula. The relative error in

M

is:

\frac{\Delta M}{M} = \frac{5 \mathrm{~g}}{500 \mathrm{~g}} = 0.01

The relative error in

l

is:

\frac{\Delta l}{l} = \frac{0.02 \mathrm{~cm}}{2 \mathrm{~cm}} = 0.01

Since

Y

is proportional to

M

and inversely proportional to

l

, the overall percentage error in

Y

is the sum of the percentage errors in

M

and

l

:

\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l} = 0.01 + 0.01 = 0.02

To express this as a percentage, we multiply by 100:

\text{Percentage error in } Y = 0.02 \times 100 = 2\%

Thus, the percentage error in Young's modulus

Y

is: Option A: 2%

Q142

Given below are two statements : Statement I : When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary. The contact angle may be

0^{\circ}

. Statement II : The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well. In the light of the above statement, choose the correct answer from the options given below.

A Statement I is true and Statement II is false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are true

D Both Statement I and Statement II are false

Correct Answer

Option B

Solution

Both statements given above have implications relating to the phenomena of capillarity, which involves the interaction between a liquid and a solid (in this case, a capillary tube).

Let's break down each statement for clarity.

Statement I: When a capillary tube is dipped into a liquid and the liquid neither rises nor falls in the capillary, it suggests a scenario where the adhesive forces (between the liquid and the solid) and the cohesive forces (among the liquid molecules) are in perfect balance.

The contact angle, which is the angle formed by the tangent to the liquid surface at the point of contact with the wall of the tube, plays a crucial role here.

A contact angle of

0^{\circ}

implies complete wetting, meaning the liquid spreads out to maximize contact with the solid.

However, the statement that the liquid neither rises nor falls specifically with a contact angle of

0^{\circ}

seems inaccurate. In reality, when the contact angle is

0^{\circ}

, it denotes perfect wetting, and the liquid tends to rise in the capillary tube.

Thus, the precision of Statement I could be disputed based on the typical behavior of liquids in the context of capillarity and contact angles.

Statement II: The contact angle really is a property of both the solid and the liquid.

It depends on the nature of the solid surface (whether it's hydrophilic or hydrophobic) and the type of liquid.

This is because the contact angle reflects the degree of interaction between the liquid and solid surfaces, which is influenced by characteristics such as surface tension of the liquid and the surface energies of both the liquid and solid.

Thus, Statement II accurately describes the nature of the contact angle as being dependent on both the material of the solid and the liquid.

Considering the analysis above: Statement I suggests a specific scenario that does not align accurately with the principles of capillarity and wettability, particularly at a contact angle of

0^{\circ}

, which typically would result in the liquid rising in the capillary, not staying neutral.

Statement II accurately reflects the factors influencing the contact angle between a liquid and a solid surface.

Therefore, the correct option based on the given statements and their analysis would be: Option B: Statement I is false but Statement II is true.

Q143

Pressure inside a soap bubble is greater than the pressure outside by an amount : (given :

\mathrm{R}=

Radius of bubble,

\mathrm{S}=

Surface tension of bubble)

A

\dfrac{S}{R}

B

\dfrac{4 \mathrm{~S}}{\mathrm{R}}

C

\dfrac{4 \mathrm{R}}{\mathrm{S}}

D

\dfrac{2 S}{R}

Correct Answer

Option B

Solution

The difference in pressure inside a soap bubble as compared to the outside is due to the surface tension created by the soap film on the bubble.

This difference in pressure can be calculated using the formula that relates the surface tension of the soap bubble to the radius of the bubble.

The correct formula for the pressure difference (

\Delta P

) across a soap bubble is given by:

\Delta P = \frac{4 S}{R}

Here,

S

is the surface tension of the bubble and

R

is the radius of the bubble.

The factor of 4 comes from the fact that a soap bubble has two surfaces (an inner and an outer surface), and for each surface, the Laplace pressure (which contributes to the pressure difference due to surface tension) is given by

\frac{2S}{R}

. Thus, for two surfaces, you double this amount, resulting in the

\frac{4 S}{R}

term. Therefore, the correct answer is: Option B

\frac{4 \mathrm{~S}}{\mathrm{R}}

Q144

The fractional compression

\left( \dfrac{\Delta V}{V} \right)

of water at the depth of 2.5 km below the sea level is __________ %. Given, the Bulk modulus of water =

2 \times 10^9

N m

^{-2}

, density of water =

10^3

kg m

^{-3}

, acceleration due to gravity

g = 10

m s

^{-2}

.

A 1.0

B 1.25

C 1.75

D 1.5

Correct Answer

Option B

Solution

The fractional compression $\left( \dfrac{\Delta V}{V} \right)$ 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: $B = \dfrac{\rho gh}{\left(\dfrac{\Delta V}{V}\right)}$ Rearranging the formula to solve for the fractional compression: $\dfrac{\Delta V}{V} \times 100 = \dfrac{\rho gh}{B} \times 100$ Substituting in the given values: $\dfrac{1000 \times 10 \times 2.5 \times 10^3}{2 \times 10^9} \times 100\%$ Calculating the result gives: $= 1.25\%$ Thus, the fractional compression of water at this depth is 1.25%.

Q145

A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity)

A 2 Mg

B

\dfrac{3}{2} \mathrm{Mg}

C

\dfrac{\mathrm{Mg}}{2}

D Mg

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{f}=0 \\ & \Rightarrow \mathrm{mg}-\frac{\mathrm{mg}}{2}-\mathrm{f}=0 \\ & \therefore \mathrm{f}=\frac{\mathrm{mg}}{2} \end{aligned}

Q146

A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water?(Given: density of water = 1000 kg m-3)

A 400 cm3

B 600 cm3

C 1400 cm3

D 4000 cm3

Correct Answer

Option B

Solution

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

Q147

Given below are two statements: Statement I: The hot water flows faster than cold water Statement II: Soap water has higher surface tension as compared to fresh water. In the light 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

Let's analyze the statements one by one:

\textbf{Statement I: "The hot water flows faster than cold water."}

When water is heated, its viscosity decreases.

In simple terms, hot water is "thinner" than cold water.

Lower viscosity means there is less internal friction in the fluid, which allows it to flow more easily and, consequently, faster under the same conditions.

Therefore, under identical pressure conditions, hot water indeed flows faster than cold water.

This makes Statement I true.

\textbf{Statement II: "Soap water has higher surface tension as compared to fresh water."}

Soap is a surfactant.

Surfactants reduce the surface tension of the liquid by interfering with the cohesive forces between the water molecules.

Fresh (pure) water has a relatively high surface tension (about 72.8 mN/m at room temperature), but when soap is added, the surface tension is decreased.

Therefore, soap water has lower, not higher, surface tension compared to pure water.

This makes Statement II false.

Based on this analysis, the correct answer is: Option D: Statement I is true but Statement II is false.

Q148

A massless spring gets elongated by amount

x_1

under a tension of 5 N . Its elongation is

x_2

under the tension of 7 N . For the elongation of

\left(5 x_1-2 x_2\right)

, the tension in the spring will be,

A 20 N

B 39 N

C 11 N

D 15 N

Correct Answer

Option C

Solution

We can solve this using Hooke's Law, which states:

F = kx,

where: •

F

is the force (tension), •

k

is the spring constant, and •

x

is the extension of the spring. Follow these steps: For the first scenario (extension

x_1

under 5 N):

5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}.

For the second scenario (extension

x_2

under 7 N):

7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}.

Equate the two expressions for

k

:

\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad \frac{x_1}{x_2} = \frac{5}{7} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1.

To find the tension for the extension of

\left(5x_1 - 2x_2\right)

, use Hooke's Law again:

\text{Tension, } F = k \left(5x_1 - 2x_2\right).

Substitute

k = \frac{5}{x_1}

:

F = \frac{5}{x_1} \left(5x_1 - 2x_2\right).

Substitute the expression for

x_2

:

F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1\right)\right) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1\right).

Simplify the expression inside the parentheses:

5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}\right)x_1 = \frac{11}{5}x_1.

Now substitute back:

F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 \text{ N}.

Thus, the tension in the spring for the extension

\left(5x_1-2x_2\right)

is

11 \text{ N}.

The correct option is Option C.

Q149

Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is

P_1

. The reading of the pressure gauge falls to

P_2

when the valve is opened. The speed of water flowing in the pipe is proportional to

A

\left(P_1-P_2\right)^2

B

\sqrt{\mathrm{P}_1-\mathrm{P}_2}

C

P_1-P_2

D

\left(P_1-P_2\right)^4

Correct Answer

Option B

Solution

When water flows from a region of higher pressure to a region of lower pressure, the difference in pressure is converted into kinetic energy of the flowing water.

According to Bernoulli's principle for horizontal flow (where gravitational effects can be ignored), this conversion can be expressed as:

\frac{1}{2}\rho v^2 = P_1 - P_2

Here, $\rho$ is the density of water,

v

is the speed of the water,

P_1

is the initial pressure when the valve is closed,

P_2

is the pressure after the valve is opened. By solving for

v

, we have:

v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}

This equation shows that the speed of the water,

v

, is proportional to the square root of the pressure difference:

v \propto \sqrt{P_1 - P_2}

Thus, the correct relationship is given by the square root of the pressure difference.

Q150

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density

1000 \mathrm{~kg} / \mathrm{m}^3

. If the pressure inside the bubble is

2100 \mathrm{~N} / \mathrm{m}^2

greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use

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

)

A 0.02

B 0.05

C 0.25

D 0.1

Correct Answer

Option B

Solution

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}}.