Properties of Matter Questions — NEET Physics

Q1

The height of liquid column raised in a capillary tube of certain radius when dipped in liquid A vertically is,

5 \mathrm{~cm}

. If the tube is dipped in a similar manner in another liquid

\mathrm{B}

of surface tension and density double the values of liquid

\mathrm{A}

, the height of liquid column raised in liquid

\mathrm{B}

would be __________ m.

A 0.05

B 0.20

C 0.5

D 0.10

Correct Answer

Option A

Solution

height of capillary rise

= {{2s\cos \theta } \over {\rho gR}}

When in A 5 cm

= {{2{s_A}\cos \theta } \over {{\rho _A}gR}}

When in B

h = {{2{s_B}\cos \theta } \over {{\rho _B}gR}}

{s_B} = 2{s_A}

and

{\rho _B} = 2{\rho _A}

h = {{2 \times 2{s_A} \times \cos \theta } \over {2{\rho _A}gR}} = 5

cm

Q2

A submarine is designed to withstand an absolute pressure of 100 atm . How deep can it go below the water surface? (Consider the density of water

=1000 \mathrm{~kg} \mathrm{~m}^{-3}

,

1 \mathrm{~atm}=1 \times 10^5 \mathrm{~Pa}

and gravitational acceleration

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

)

A 990 m

B 9900 m

C 99 m

D 9000 m

Correct Answer

Option A

Solution

$P=P_0+\rho g h$

\begin{aligned} & \Rightarrow 100 \times 10^5=10^5+10^3 \times 10 \times h \\ & \Rightarrow 10^7=10^5+10^4 h \\ & \Rightarrow 10^3=10+h \\ & \Rightarrow h=1000-10=990 \mathrm{~m} \end{aligned}

Q3

Consider a water tank shown in the figure. It has one wall at

x=L

and can be taken to be very wide in the

z

direction. When filled with a liquid of surface tension

S

and density

\rho

, the liquid surface makes angle

\theta_0\left(\theta_0 \ll 1\right)

with the

x

-axis at

x=L

. If

y(x)

is the height of the surface then the equation for

y(x)

is: (take

\theta(x)=\sin \theta(x)=\tan \theta(x)=\dfrac{d y}{d x}, g

is the acceleration due to gravity)

A

\dfrac{d^2 y}{d x^2}=\sqrt{\dfrac{\rho g}{s}}

B

\dfrac{d y}{d x}=\sqrt{\dfrac{\rho g}{s}} x

C

\dfrac{d^2 y}{d x^2}=\dfrac{\rho g}{s} x

D

\dfrac{d^2 y}{d x^2}=\dfrac{\rho g}{s} y

Correct Answer

Option D

Solution

\begin{aligned} & R O C=\text{ Radius of curvature at point } A \\ & \text{ Curvature }=\frac{1}{R O C}=\frac{\left|\frac{d^2 y}{d x^2}\right|}{\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{3}{2}}}=\frac{\left|\frac{d^2 y}{d x^2}\right|}{(1+0)^{\frac{3}{2}}}=\frac{d^2 y}{d x^2} \quad\left[\because \frac{d y}{d x}=\tan \theta=0\right] \\ & \Delta P=S \times \text{ curvature } \\ & \Rightarrow \rho g y=S \frac{d^2 y}{d x^2} \\ & \therefore \frac{d^2 y}{d x^2}=\frac{\rho g y}{S} \end{aligned}

Alternate Solution : For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(\mathrm{S} \sin (\theta+\mathrm{d} \theta)-\mathrm{S} \sin \theta) \mathrm{d}=\mathrm{mg}$ Angle is small $\therefore \sin \theta \approx \theta$

\begin{aligned} & \Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S} \\ & \tan \theta=\frac{d y}{d x} \Rightarrow \text{ Differentiating wrt } x \\ & \sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2} \end{aligned}

Put $d \theta$ from (2) in (1) $d$ take $\cos \theta \approx 1$ , we get

\frac{d^2 y}{d x^2}=\frac{\rho g y}{S}

Q4

An ideal fluid is flowing in a non-uniform cross-sectional tube

X Y

(as shown in the figure) from end

X

to end

Y

. If

K_1

and

K_2

are the kinetic energy per unit volume of the fluid at

X

and

Y

respectively, then the correct option is :

A

K_1=K_2

B

2 K_1=K_2

C

K_1>K_2

D

K_1 < K_2

Correct Answer

Option C

Solution

According to Bernoulli's principle, Kinetic energy per unit volume + Potential energy per unit volume + Pressure $=$ Constant

\frac{1}{2} \rho V^2+\rho g h+P=\text{ constant }

Apply Bernoulli's principle at point

X

and

Y

,

\begin{aligned} & P+K_1+\rho g(0)=P+K_2+\rho g(\mathrm{~h}) \\ & K_1=K_2+\rho g h \\ & K_1>K_2 \end{aligned}

Q5

The maximum elongation of a steel wire of

1 \mathrm{~m}

length if the elastic limit of steel and its Young's modulus, respectively, are

8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}

and

2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}

, is:

A 4 mm

B 0.4 mm

C 40 mm

D 8 mm

Correct Answer

Option A

Solution

First, we need to find the maximum force that can be applied to the steel wire within its elastic limit.

This force can be calculated using the given area under stress and the stress limit provided by the elastic limit of steel.

Let's assume the area of cross-section of the wire is

A

. The force exerted can be given by:

F = \sigma \times A

where:

\sigma = 8 \times 10^8 \mathrm{~N/m}^2

(Elastic limit of steel) To calculate the elongation (

\Delta L

) under this force, we use Hooke's Law, which relates force, elongation, cross-sectional area, original length, and Young's modulus as follows:

F = \frac{Y \times A \times \Delta L}{L}

where:

Y = 2 \times 10^{11} \mathrm{~N/m}^2

(Young's modulus of steel),

L = 1 \mathrm{~m}

(original length of the wire). Substituting for

F

from the earlier expression and rearranging the formula, we get:

\sigma \times A = \frac{Y \times A \times \Delta L}{L}

\sigma = \frac{Y \times \Delta L}{L}

\Delta L = \frac{\sigma \times L}{Y}

\Delta L = \frac{8 \times 10^8 \times 1}{2 \times 10^{11}} \mathrm{~meters}

\Delta L = 0.004 \mathrm{~meters}

\Delta L = 4 \mathrm{~mm}

Thus, the maximum elongation of the wire within the elastic limit is

4 \mathrm{~mm}

. The answer is: Option A - 4 mm.

Q6

A thin flat circular disc of radius

4.5 \mathrm{~cm}

is placed gently over the surface of water. If surface tension of water is

0.07 \mathrm{~N} \mathrm{~m}^{-1}

, then the excess force required to take it away from the surface is

A 19.8 mN

B 198 N

C 1.98 mN

D 99 N

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Excess force }=T \times 2 \pi R \\ & =\frac{7}{100} \times 2 \times 3.14 \times \frac{4.5}{100} \\ & =197.82 \times 10^{-4} \\ & =19.8 \times 10^{-3} \mathrm{~N} \\ & =19.8 \mathrm{~mN} \end{aligned}

Q7

A metallic bar of Young's modulus,

0.5 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}

and coefficient of linear thermal expansion

10^{-5}{ }^{\circ} \mathrm{C}^{-1}

, length

1 \mathrm{~m}

and area of cross-section

10^{-3} \mathrm{~m}^2

is heated from

0^{\circ} \mathrm{C}

to

100^{\circ} \mathrm{C}

without expansion or bending. The compressive force developed in it is :

A

5 \times 10^3 \mathrm{~N}

B

50 \times 10^3 \mathrm{~N}

C

100 \times 10^3 \mathrm{~N}

D

2 \times 10^3 \mathrm{~N}

Correct Answer

Option B

Solution

Given the properties and conditions of the metallic bar, we are required to calculate the compressive force developed due to heating.

Key inputs include the Young's modulus (E), coefficient of linear thermal expansion (α), change in temperature (

\Delta T

), and the original dimensions of the bar.

First, compute the linear expansion of the bar if it were free to expand.

The change in length (

\Delta L

) due to thermal expansion can be computed through the formula:

\Delta L = \alpha L_0 \Delta T

Given:

\alpha = 10^{-5} { }^{\circ} C^{-1}

L_0 = 1 \text{ m}

\Delta T = 100^{\circ} C

Thus:

\Delta L = 10^{-5} \times 1 \times 100 = 0.001 \text{ m}

This is the change in length that the bar would undergo if not constrained.

However, in this scenario, the bar is constrained and does not actually expand.

This constraint induces a compressive stress (constrained thermal stress) in the bar, which can be calculated using the formula relating stress, Young's modulus, and strain:

\sigma = E \epsilon

Where the strain ( $\epsilon$ ) under constrained conditions due to thermal expansion is:

\epsilon = \frac{\Delta L}{L_0} = \frac{0.001}{1} = 0.001

Therefore:

\sigma = 0.5 \times 10^{11} \times 0.001 = 5 \times 10^7 \mathrm{ N/m}^2

This stress is the force per unit area.

To find the compressive force, we need to multiply this stress by the cross-sectional area of the bar:

F = \sigma A = 5 \times 10^7 \mathrm{ N/m}^2 \times 10^{-3} \mathrm{ m}^2 = 5 \times 10^4 \mathrm{ N}

Thus, the compressive force developed in the bar is

50 \times 10^3 \mathrm{~N}

. Hence, the correct answer is Option B:

50 \times 10^3 \mathrm{~N}

Q8

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length

100 \mathrm{~cm}

to stretch it by

1 \mathrm{~mm}

is (if Young's modulus of the wire

=2.0 \times 10^{11} \mathrm{Nm}^{-2}

) :

A

10^{11}

B

10^{17}

C

10^7

D

10^5

Correct Answer

Option D

Solution

\begin{aligned} & \frac{\text{ E.P.E }}{\text{ Volume }}=\frac{1}{2}(\text{ stress)(strain) } \\ & =\frac{1}{2}(Y)(\text{ strain })^2 \\ & =\frac{1}{2}(\mathrm{Y})\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)^2 \\ & =\frac{1}{2}\left(2 \times 10^{11}\right)\left(\frac{1 \times 10^{-3}}{100 \times 10^{-2}}\right)^2 \\ & =10^5 \end{aligned}

Q9

Which of the following statement is not true?

A Coefficient of viscosity is a scalar quantity

B Surface tension is a scalar quantity

C Pressure is a vector quantity

D Relative density is a scalar quantity

Correct Answer

Option C

Solution

Pressure is a scalar quantity.

Q10

The viscous drag acting on a metal sphere of diameter

1 \mathrm{~mm}

, falling through a fluid of viscosity

0.8 \mathrm{~Pa}

s with a velocity of

2 \mathrm{~m} \mathrm{~s}^{-1}

is equal to :

A

15 \times 10^{-3} \mathrm{~N}

B

30 \times 10^{-3} \mathrm{~N}

C

1.5 \times 10^{-3} \mathrm{~N}

D

20 \times 10^{-3} \mathrm{~N}

Correct Answer

Option A

Solution

\begin{aligned} & F=6 \pi \mathrm{r} \eta v \\ & =(6)(3.14)\left(\frac{1 \times 10^{-3}}{2}\right)\left(0.8 \times 10^{-1}\right)(2) \\ & =1.5 \times 10^{-3} \mathrm{~N} \end{aligned}