Properties of Matter Questions — JEE 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 solid steel ball of diameter 3.6 mm acquired terminal velocity

2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s}

while falling under gravity through an oil of density

925 \mathrm{~kg} \mathrm{~m}^{-3}

. Take density of steel as

7825 \mathrm{~kg} \mathrm{~m}^{-3}

and g as

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

. The viscosity of the oil in SI unit is

A 2.18

B 1.68

C 2.38

D 1.99

Correct Answer

Option D

Solution

To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid: $v_T = \dfrac{2}{9} \dfrac{(\rho_s - \rho_f) r^2 g}{\eta}$ 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 $\eta$ : $\eta = \dfrac{2}{9} \cdot \dfrac{(\rho_s - \rho_f) \cdot r^2 \cdot g}{v_T}$ Substituting the known values into the equation: $\eta = \dfrac{2}{9} \cdot \dfrac{(7825 - 925) \cdot (1.8 \times 10^{-3})^2 \cdot 9.8}{2.45 \times 10^{-2}}$ 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.

Q3

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Clothes containing oil or grease stains cannot be cleaned by water wash. Reason (R): Because the angle of contact between the oil/grease and water is obtuse. In the light of the above statements, choose the correct answer from the option given below.

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

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

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

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

Correct Answer

Option A

Solution

Due to obtuse angle of contact the water doesn't wet the oiled surface properly and cannot wash it also.

$\Rightarrow$ Assertion is correct and Reason given is a correct explanation.

Q4

Under the same load, wire A having length

5.0 \mathrm{~m}

and cross section

2.5 \times 10^{-5} \mathrm{~m}^{2}

stretches uniformly by the same amount as another wire B of length

6.0 \mathrm{~m}

and a cross section of

3.0 \times 10^{-5}

\mathrm{m}^{2}

stretches. The ratio of the Young's modulus of wire A to that of wire

B

will be :

A

1: 2

B

1: 4

C

1: 1

D

1: 10

Correct Answer

Option C

Solution

$\Delta \ell=\dfrac{F \ell}{S Y}$ $F$ is same for both wire and $\Delta \ell$ is also same

\begin{aligned} & \frac{\Delta \ell}{F}=\frac{\ell}{S Y} \\\\ & \Rightarrow \frac{\ell_{A}}{S_{A} Y_{A}}=\frac{\ell_{B}}{S_{B} Y_{B}} \\\\ & \Rightarrow \frac{5}{2.5 \times Y_{A}}=\frac{6}{3 \times Y_{B}} \\\\ & \Rightarrow \frac{Y_{A}}{Y_{B}}=1 \end{aligned}

Q5

Water flows into a large tank with flat bottom at the rate of 10–4m3s–1. Water is also leaking out of a hole ofarea 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is -

A 2.9 cm

B 5.1 cm

C 4 cm

D 1.7 cm

Correct Answer

Option B

Solution

Since height of water column is constant therefore, water inflow rate (Qin) = water outflow rate Qin = 10 $-$ 4 m3s $-$ 1 Qout = Au = 10 $-$ 4 $\times$

\sqrt {2gh}

10 $-$ 4 = 10 $-$ 4

\sqrt {20 \times h}

h =

{1 \over {20}}m

h = 5cm

Q6

A air bubble of radius 1 cm in water has an upward acceleration 9.8 cm s–2. The density of water is 1 gm cm–3 and water offers negligible drag force on the bubble. The mass of the bubble is (g = 980 cm/s2).

A 1.52 gm

B 4.51 gm

C 3.15 gm

D 4.15 gm

Correct Answer

Option D

Solution

B - mg = ma $\Rightarrow$ m =

{B \over {g + a}}

=

{{V{\rho _w}g} \over {g + a}}

=

{{V{\rho _w}} \over {1 + {a \over g}}}

=

{{{4 \over 3}\pi {{\left( 1 \right)}^3} \times 1} \over {1 + {{9.8} \over {980}}}}

= 4.15 gm

Q7

The normal density of a material is

\rho

and its bulk modulus of elasticity is K. The magnitude of increase in density of material, when a pressure P is applied uniformly on all sides, will be :

A

{{\rho K} \over P}

B

{{PK} \over \rho }

C

{{\rho P} \over K}

D

{K \over {\rho P}}

Correct Answer

Option C

Solution

Bulk modulus

K = {{ - \Delta P} \over {{{\Delta v} \over v}}} = {{ - \Delta Pv} \over {\Delta v}}

We know,

\rho = {M \over V}

So,

{{ - \Delta \rho } \over \rho } = {{\Delta v} \over v}

K = {{ - \Delta P} \over {\left( { - {{\Delta \rho } \over \rho }} \right)}} = {{\rho \Delta P} \over {\Delta \rho }}

\Delta \rho = {{\rho \Delta P} \over K}

\Delta \rho = {{\rho P} \over K}

Q8

The terminal velocity (vt) of the spherical rain drop depends on the radius (r) of the spherical rain drop as :

A r1/2

B r

C r2

D r3

Correct Answer

Option C

Solution

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

\Rightarrow {v_t} = C{r^2}

where C is a constant or

{v_t} \propto {r^2}

Q9

A balloon has mass of

10 \mathrm{~g}

in air. The air escapes from the balloon at a uniform rate with velocity

4.5 \mathrm{~cm} / \mathrm{s}

. If the balloon shrinks in

5 \mathrm{~s}

completely. Then, the average force acting on that balloon will be (in dyne).

A 3

B 9

C 12

D 18

Correct Answer

Option B

Solution

{F_{avg}} = \mu \times {v_{rel}}

= {{10} \over 5} \times 4.5 = 9

Q10

For a solid rod, the Young's modulus of elasticity is

3.2 \times 10^{11} \mathrm{Nm}^{-2}

and density is

8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}

. The velocity of longitudinal wave in the rod will be.

A

3.65 \times 10^3 \mathrm{~ms}^{-1}

B

6.32 \times 10^3 \mathrm{~ms}^{-1}

C

18.96 \times 10^3 \mathrm{~ms}^{-1}

D

145.75 \times 10^3 \mathrm{~ms}^{-1}

Correct Answer

Option B

Solution

$v=\sqrt{\dfrac{Y}{\rho}}=\sqrt{\dfrac{3.2 \times 10^{11}}{8 \times 10^{3}}}$

\begin{aligned} & =\sqrt {0.4 \times {{10}^8}} \\\\ & = \sqrt {40 \times {{10}^6}} \\\\ & =6.32 \times 10^{3} \mathrm{~m} / \mathrm{s} \end{aligned}