Properties of Matter — Page 4 | JEE Physics

Q31

Spherical balls of radius

R

are falling in a viscous fluid of viscosity

\eta

with a velocity

v.

The retarding viscous force acting on the spherical ball is

A inversely proportional to both radius

R

and velocity

v

B directly proportional to both radius

R

and velocity

v

C directly proportional to

R

but inversely proportional to

v

D inversely proportional to

R,

but directly proportional to velocity

v

Correct Answer

Option B

Solution

From Stoke's law, viscous force acting on the ball falling into a viscous fluid

F = 6\pi \eta Rv

$\therefore$

F \propto R

and

F \propto v

hence

F

is directly proportional to radius & velocity.

Q32

If two soap bubbles of different radii are connected by a tube

A air flows from the smaller bubble to the bigger

B air flows from bigger bubble to the smaller bubble till the sizes are interchanged

C air flows from the bigger bubble to the smaller bubble till the sizes become equal

D there is no flow of air.

Correct Answer

Option A

Solution

Pressure inside the bubble, P

= {p_0} + {{4T} \over R}

So

P \propto {1 \over R}

where R is the radius of the bubble.

It means pressure inside a smaller bubble is greater than the inside of a bigger bubble.

So when two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the size of bigger bubble will increase.

Q33

If

S

is stress and

Y

is young's modulus of material of a wire, the energy stored in the wire per unit volume is

A

{{{S^2}} \over {2Y}}

B

2{S^2}Y

C

{S \over {2Y}}

D

{{2Y} \over {{S^2}}}

Correct Answer

Option A

Solution

Energy stored per unit volume of wire,

E = {1 \over 2} \times \,stress\, \times \,strain

$\therefore$

E = {1 \over 2} \times \,stress\, \times \,{{stress} \over Y} = {1 \over 2}{{{S^2}} \over Y}

[ As Young's modulus(Y) =

{{Stress} \over {Strain}}

$\therefore$ Strain =

{{Stress} \over Y}

]

Q34

A

20

cm

long capillary tube is dipped in water. The water rises up to

8

cm.

If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be

A

10

cm

B

8

cm

C

20

cm

D

4

cm

Correct Answer

Option C

Solution

In freely falling elevator

g

= 0 Water fills the tube entirely in gravity less condition.

Hence, length of water column in the capillary tube is 20 cm.

Q35

A spherical solid ball of volume

V

is made of a material of density

{\rho _1}

. It is falling through a liquid of density

{\rho _2}\left( {{\rho _2} < {\rho _1}} \right)

. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed

v,

i.e.,

{F_{viscous}} = - k{v^2}\left( {k > 0} \right).

The terminal speed of the ball is

A

\sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}

B

{{{Vg{\rho _1}} \over k}}

C

\sqrt {{{Vg{\rho _1}} \over k}}

D

{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}

Correct Answer

Option A

Solution

The forces acting on the ball - (1) mg =

V{\rho _1}g

downward direction (2) Thrust upward direction ( By Archimedes principle ) (3) Force of friction ( Buoynat force) upward direction The ball reaches to its terminal speed

\left( {{v_t}} \right)

when acceleration = 0. So, weight $=$ Buoyant force $+$ Viscous force $\therefore$

V\rho {}_1g = V{\rho _2}g + kv_t^2

$\therefore$

{v_t} = \sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}

Q36

Two wires are made of the same material and have the same volume. However wire

1

has cross-sectional area

A

and wire

2

has cross-sectional area

3A.

If the length of wire

1

increases by

\Delta x

on applying force

F,

how much force is needed to stretch wire

2

by the same amount?

A

4F

B

6F

C

9F

D

F

Correct Answer

Option C

Solution

As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section

3A

will be

\ell /3

(same volume as wire

1

). For wire

1,

y = {{F/A} \over {\Delta x/\ell }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)

For wire

2.

Y = {{F'/3A} \over {\Delta x/\left( {\ell /3} \right)}}........(ii)

From

(i)

and

(ii),

{F \over A} \times {\ell \over {\Delta x}} = {{F'} \over {3A}} \times {\ell \over {3\Delta x}} \Rightarrow F' = 9F

Q37

Water is flowing continuously from a tap having an internal diameter

8 \times {10^{ - 3}}\,\,m.

The water velocity as it leaves the tap is

0.4\,\,m{s^{ - 1}}

. The diameter of the water stream at a distance

2 \times {10^{ - 1}}\,\,m

below the tap is close to :

A

7.5 \times {10^{ - 3}}m

B

9.6 \times {10^{ - 3}}m

C

3.6 \times {10^{ - 3}}m

D

5.0 \times {10^{ - 3}}m

Correct Answer

Option C

Solution

From Bernoulli's theorem,

{P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0

{v_2} = \sqrt {v_1^2 + 2gh}

= \sqrt {0.16 + 2 \times 10 \times 0.2}

= 2.03\,m/s

From equation of continuity

{A_2}{v_2} = {A_1}{v_1}

\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}

\Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m

Q38

A uniform cylinder of length

L

and mass

M

having cross-sectional area

A

is suspended, with its length vertical, from a fixed point by a mass-less spring such that it is half submerged in a liquid of density

\sigma

at equilibrium position. The extension

{x_0}

of the spring when it is in equilibrium is:

A

{{Mg} \over k}

B

{{Mg} \over k}\left( {1 - {{LA\sigma } \over M}} \right)

C

{{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)

D

{{Mg} \over k}\left( {1 + {{LA\sigma } \over M}} \right)

Correct Answer

Option C

Solution

From figure,

k{x_0} + {F_B} = Mg

k{x_0} + \sigma {L \over 2}Ag = Mg

[ as mass $=$ density $\times$ volume ]

\Rightarrow k{x_0} = Mg - \sigma {L \over 2}Ag

\Rightarrow {x_0} = {{Mg - {{\sigma LAg} \over 2}} \over k}

= {{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)

Q39

The pressure that has to be applied to the ends of a steel wire of length

10

cm

to keep its length constant when its temperature is raised by

{100^ \circ }C

is: (For steel Young's modulus is

2 \times {10^{11}}\,\,N{m^{ - 2}}

and coefficient of thermal expansion is

1.1 \times {10^{ - 5}}\,{K^{ - 1}}

)

A

2.2 \times {10^8}\,\,Pa

B

2.2 \times {10^9}\,\,Pa

C

2.2 \times {10^7}\,\,Pa

D

2.2 \times {10^6}\,\,Pa

Correct Answer

Option A

Solution

Young's modulus

Y = {{stress} \over {strain}}

stress = Y \times strain

Stress

in steel wire $=$ Applied

pressure

Pressure

$=$

stress

$=$

Y \times \,strain

Strain = {{\Delta L} \over L} = \alpha \Delta T

(As length is constant)

= 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100

= 2.2 \times {10^8}Pa

Q40

An open glass tube is immersed in mercury in such a way that a length of

8

cm

extends above the mercury level. The open end of the tube is then closed and scaled and the tube is raised vertically up by additional

46

cm

. What will be length of the air column above mercury in the tube now? (Atmospheric pressure

=76

cm

of

Hg

)

A

16

cm

B

22

cm

C

38

cm

D

6

cm

Correct Answer

Option A

Solution

Length of the air column above mercury in the tube is,

P + x = {P_0}

\Rightarrow P = \left( {76 - x} \right)

\Rightarrow 8 \times A \times 76 = \left( {76 - x} \right) \times A \times \left( {54 - x} \right)

$\therefore$

x=38

Thus, length of air column

=54-38=16cm.