Properties of Matter — Page 2 | JEE Physics

Q11

A small spherical ball of radius

r

, falling through a viscous medium of negligible density has terminal velocity '

v

'. Another ball of the same mass but of radius

2 r

, falling through the same viscous medium will have terminal velocity:

A

4 \mathrm{v}

B

2 \mathrm{~V}

C

\dfrac{v}{4}

D

\dfrac{\mathrm{v}}{2}

Correct Answer

Option D

Solution

Since density is negligible hence Buoyancy force will be negligible At terminal velocity.

\mathrm{Mg} =6 \pi \eta \mathrm{rv}

\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad

(as mass is constant) Now,

\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}

r^{\prime}=2 \mathrm{r}

So,

v^{\prime}=\frac{v}{2}

Q12

Given below are two statements: Statement I : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II : If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate 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 true but Statement II is false

D Statement I is false but Statement II is true

Correct Answer

Option C

Solution

Surface tension will be less as temperature increases

\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}

Height of capillary rise will be smaller in hot water and larger in cold water.

Q13

A spherical ball of radius

1 \times 10^{-4} \mathrm{~m}

and density

10^5 \mathrm{~kg} / \mathrm{m}^3

falls freely under gravity through a distance

h

before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of

h

is approximately: (The coefficient of viscosity of water is

9.8 \times 10^{-6} \mathrm{~N} \mathrm{~s} / \mathrm{m}^2

)

A 2518 m

B 2396 m

C 2249 m

D 2296 m

Correct Answer

Option A

Solution

To solve this problem, we can use the concepts of terminal velocity and the forces acting on the spherical ball.

First, let's analyze the situation step-by-step.

When the ball falls freely under gravity, it achieves a terminal velocity

v_t

in water.

This terminal velocity is reached when the gravitational force is balanced by the drag force and the buoyant force in the water.

The forces acting on the ball are: 1.

Gravitational Force:

F_g = mg

2. Buoyant Force:

F_b = \rho_{\text{water}} V g

3. Drag Force:

F_d = 6 \pi \eta r v_t

Where,

m

is the mass of the ball.

g

is the acceleration due to gravity (

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

).

\rho_{\text{water}}

is the density of water (

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

).

V

is the volume of the ball (

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

). $\eta$ is the coefficient of viscosity of water (

9.8 \times 10^{-6} \, \mathrm{Ns/m^2}

).

r

is the radius of the ball (

1 \times 10^{-4} \, \mathrm{m}

).

v_t

is the terminal velocity. Using the equilibrium condition at terminal velocity:

F_g = F_b + F_d

mg = \rho_{\text{water}} V g + 6 \pi \eta r v_t

First, compute the mass of the ball:

m = \rho_{\text{ball}} \times V = \rho_{\text{ball}} \times \frac{4}{3} \pi r^3

m = 10^5 \, \mathrm{kg/m^3} \times \frac{4}{3} \pi (1 \times 10^{-4} \, \mathrm{m})^3

m = 10^5 \,\mathrm{kg/m^3} \times \frac{4}{3} \pi \times 10^{-12} \,\mathrm{m^3}

m = \frac{4}{3} \pi \times 10^{-7} \, \mathrm{kg}

Next, solve for the terminal velocity

v_t

using the equilibrium equation:

mg = \rho_{\text{water}} \frac{4}{3} \pi r^3 g + 6 \pi \eta r v_t

v_t = \frac{mg - \rho_{\text{water}} \frac{4}{3} \pi r^3 g}{6 \pi \eta r}

v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi (1 \times 10^{-4})^3 \times 9.8}{6 \pi \times 9.8 \times 10^{-6} \times 10^{-4}}

v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi \times 10^{-12} \times 9.8}{6 \pi \times 9.8 \times 10^{-10}}

v_t = \frac{\frac{4}{3} \pi \times 9.8 \times 10^{-7} (1 - 10^{-5})}{6 \pi \times 9.8 \times 10^{-10}}

v_t = \frac{\frac{4}{3} \times 10^{-7}}{6 \times 10^{-10}}

v_t = \frac{4}{18} \times 10^3 \, \mathrm{m/s}

v_t \approx 222.22 \, \mathrm{m/s}

The height

h

required to reach this terminal velocity while the ball falls freely under gravity can be found using the kinematic equation:

v_t = \sqrt{2gh}

h = \frac{v_t^2}{2g}

h = \frac{(222.22)^2}{2 \times 9.8}

h = \frac{49328.88}{19.6}

h \approx 2517.8 \, \mathrm{m}

So, the closest value of

h

is approximately: Option A: 2518 m.

Q14

A small ball of mass

m

and density

\rho

is dropped in a viscous liquid of density

\rho_0

. After sometime, the ball falls with constant velocity. The viscous force on the ball is :

A

m g\left(1-\dfrac{\rho_0}{\rho}\right)

B

m g\left(\dfrac{\rho_0}{\rho}-1\right)

C

m g\left(1-\rho \rho_0\right)

D

m g\left(1+\dfrac{\rho}{\rho_0}\right)

Correct Answer

Option A

Solution

\begin{aligned} & F_V=\left(m g-F_B\right) \frac{m g}{m g}=\left(\frac{\rho V_g-\rho_0 V_g}{\rho V_g}\right) m g \\ & F_v=m g\left(1-\frac{\rho_0}{\rho}\right) \end{aligned}

Q15

A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle ≈ 0°) with 30° inclined with the vertical. The length of water risen in the capillary is _______ cm. (Take

g = 9.8 \text{ m/s}^2

)

A

\dfrac{82}{5}

B

\dfrac{68}{5}

C

\dfrac{57}{2}

D

\dfrac{71}{5}

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} \\ & \mathrm{~h}=\frac{100}{7} \mathrm{~cm} \\ & \sin 60^{\circ}=\frac{\mathrm{h}}{\ell} \\ & \ell=\frac{\mathrm{h} \times 2}{\sqrt{3}} \\ & \ell=\frac{100}{7} \times \frac{2}{\sqrt{3}} \\ & =\frac{200}{7 \times \sqrt{3}} \\ & =16.49 \mathrm{~cm} \end{aligned}

Q16

The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

A 1.16 mm

B 1.36 mm

C 1.00 mm

D 0.90 mm

Correct Answer

Option A

Solution

{{400} \over {{\pi \over 4}{d^2}}} = 379 \times {10^6}

{d^2} = {{4 \times 400 \times {{10}^{ - 6}}} \over {\pi \times 379}} = 0.336 \times {10^{ - 6}} \times 4

d = 2\sqrt {0.336} \times {10^{ - 3}}M \simeq 1.16\,mm

Q17

If the length of a wire is made double and radius is halved of its respective values. Then, the Young's modulus of the material of the wire will :

A remain same

B become 8 times its initial value

C become

\dfrac{1}{4}

of its initial value

D become 4 times its initial value

Correct Answer

Option A

Solution

Young's modulus of matter depends on material of wire and is independent of the dimensions of the wire.

As the material remains same so Young's modulus also remain same.

Q18

Young's modules of material of a wire of length '

L

' and cross-sectional area

A

is

Y

. If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be :

A 4 Y

B 2 Y

C

\mathrm{\dfrac{Y}{4}}

D Y

Correct Answer

Option D

Solution

Young's modulus depends on the material not length and cross sectional area. So young's modulus remains same.

Q19

When an air bubble of radius r rises from the bottom to the surface of a lake its radius becomes

{{5r} \over 4}.

Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :

A 11.2 m

B 8.7 m

C 9.5 m

D 10.5 m

Correct Answer

Option C

Solution

Given, bubble of radius r rises from bottom of lake.

The radius of bubble at top becomes 5r/4.

Therefore, pressure in bubble at bottom is

{P_1} = {P_0} + \rho gh + {{4T} \over r}

Pressure in bubble at top is

{P_2} = {P_0} + {{4T} \over {5r/4}}

Now, we know

{P_1}{V_1} = {P_2}{V_2}

, therefore,

\left( {{P_0} + \rho gh + {{4T} \over 4}} \right){{4\pi } \over 3}{r^3} = \left( {{P_0} + {{4T} \over {5r/4}}} \right){{4\pi } \over 3}{\left( {{{5r} \over 4}} \right)^3}

{P_0} + \rho hg + {{4T} \over r} = \left( {{P_0} + {{4T \times 4} \over {5r}}} \right){{125} \over {64}}

Now given P0 = 10 $\rho$ g, therefore,

10\rho g + \rho gh + {{4T} \over r} = {{125} \over {64}} \times 10\rho g + {{16T} \over {5r}} \times {{125} \over {64}}

Neglecting effect of temperature, we get

10\rho g + \rho gh = {{125} \over {64}} \times 10\rho g

\Rightarrow 10 + h = {{125} \over {64}} \times 10 \Rightarrow h = {{1250} \over {64}} - 10 = 9.53125\,m

$\Rightarrow$ h $\sim$ 9.5 m

Q20

A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be :

A 2.0

B 1.2

C 0.1

D 0.4

Correct Answer

Option A

Solution

y =

{{{\omega ^2}{x^2}} \over {2g}}

=

{{{{\left( {2 \times 2\pi } \right)}^2} \times {{\left( {0.05} \right)}^2}} \over {20}}

\simeq

2 cm