Properties of Matter — Page 2 | NEET Physics

Q11

The amount of energy required to form a soap bubble of radius

2 \mathrm{~cm}

from a soap solution is nearly: (surface tension of soap solution

=0.03 \mathrm{~N} \mathrm{~m}^{-1}

)

A

5.06 \times 10^{-4} \mathrm{~J}

B

3.01 \times 10^{-4} \mathrm{~J}

C

50.1 \times 10^{-4} \mathrm{~J}

D

30.16 \times 10^{-4} \mathrm{~J}

Correct Answer

Option B

Solution

E=2 T\left(4 \pi \mathrm{R}^{2}\right)

=2(0.03)(4)(3.14)\left(2 \times 10^{-2}\right)^{2}

=3.01 \times 10^{-4} \mathrm{~J}

Q12

The venturi-meter works on :

A Bernoulli's principle

B The principle of parallel axes

C The principle of perpendicular axes

D Huygen's principle

Correct Answer

Option A

Solution

A venturi meter is a device used to measure the flow rate of a fluid through a pipe.

The venturi meter works on Bernoulli's principle, which states that for an incompressible, non-viscous fluid flowing through a pipe, the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant.

Q13

Let a wire be suspended from the ceiling (rigid support) and stretched by a weight

W

attached at its free end. The longitudinal stress at any point of cross-sectional area

A

of the wire is :

A

W / A

B

W / 2 A

C Zero

D

2 W / A

Correct Answer

Option A

Solution

When a wire is suspended from the ceiling and stretched by a weight W attached at its free end, the force acting on the wire is equal to the weight W.

This force causes longitudinal stress in the wire.

Longitudinal stress is defined as the force acting on an object divided by its cross-sectional area.

In this case, the force acting on the wire is W, and the cross-sectional area of the wire is A.

Thus, the longitudinal stress at any point of the wire with cross-sectional area A is given by :

\text{Longitudinal Stress} = \frac{\text{Force}}{\text{Area}} = \frac{W}{A}

Q14

Two copper vessels A and B have the same base area but of different shapes. A takes twice the volume of water as that B requires to fill upto a particular common height. Then the correct statement among the following is :

A Vessel B weighs twice that of A.

B Pressure on the base area of vessels A and B is same.

C Pressure on the base area of vessels A and B is not same.

D Both vessels A and B weigh the same.

Correct Answer

Option B

Solution

In hydrostatic condition,

P = {P_0} + \rho gh

Here, P : absolute pressure at depth h, P 0 is atmospheric pressure Since, height of liquid in both vessel is same therefore pressure on the base of both vessel will be same.

Q15

The terminal velocity of a copper ball of radius 5 mm falling through a tank of oil at room temperature is 10 cm s

-

1 . If the viscosity of oil at room temperature is 0.9 kg m

-

1 s

-

1 , the viscous drag force is :

A 4.23

\times

10

-

6 N

B 8.48

\times

10

-

3 N

C 8.48

\times

10

-

5 N

D 4.23

\times

10

-

3 N

Correct Answer

Option B

Solution

F viscous = $-$ 6 $\pi$ $\eta$ av F viscous = 6 $\times$ 3.14 $\times$ 0.9 $\times$ 5 $\times$ 10 $-$ 3 $\times$ 10 $\times$ 10 $-$ 2 = 84.78 $\times$ 10 $-$ 4 = 8.478 $\times$ 10 $-$ 3 N

Q16

A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball (v) as a function of time (t) is

A A

B B

C C

D D

Correct Answer

Option B

Solution

Initial speed of ball is zero and it finally attains terminal speed

Q17

If a soap bubble expands, the pressure inside the bubble

A Decreases

B Increases

C Remains the same

D Is equal to the atmospheric pressure

Correct Answer

Option A

Solution

Excess pressure inside the bubble

= \Delta P - {{4T} \over R}

{P_{in}} = {P_{out}} + {{4T} \over R}

as 'R' increases 'P' decreases

Q18

The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerin becomes constant after some time. If the density of glycerin is

{d \over 2}

, then the viscous force acting on the ball will be :

A 2Mg

B

{{Mg} \over 2}

C Mg

D

{{3} \over 2}

Mg

Correct Answer

Option B

Solution

Let F v be the viscous force and F B be the Bouyant force acting on the ball.

Then, when body moves with constant velocity Mg = F B + F v [a = 0] F v = Mg – F B =

dVg - {d \over 2}Vg

(Mg = dVg) V = volume of ball. =

{d \over 2}Vg

$\Rightarrow$ F v =

{{Mg} \over 2}

Q19

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in Water. The mass of water that will rise in this tube is :

A 5.0 g

B 10.0 g

C 20.0 g

D 2.5 g

Correct Answer

Option B

Solution

The weight of water in the capillary is balanced by the surface tension force. mg = 2 $\pi$ r.T.cos $\theta$ Since constant, m is directly proportional to r. m $\propto$ r

{{{m_2}} \over {{m_1}}} = {{{r_2}} \over {{r_1}}}

{{{m_2}} \over 5} = {{2r} \over r}

$\therefore$

{m_2}

= 10 grams.

Q20

A wire of length L, are of cross section A is hanging from a fixed support. The length of the wire changed to L 1 when mass M is suspended from its free end. The expression for Young's modulus is :

A

{{Mg\left( {{L_1} - L} \right)} \over {AL}}

B

{{MgL} \over {A{L_1}}}

C

{{MgL} \over {A\left( {{L_1} - L} \right)}}

D

{{Mg{L_1}} \over {AL}}

Correct Answer

Option C

Solution

Stress = {{Mg} \over A}

Strain = {{\Delta L} \over L} = {{{L_{ - 1}} - L} \over L}

Young's modulus =

{{Stress} \over {Strain}}

=

{{MgL} \over {A\left( {{L_1} - L} \right)}}