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Mechanical Properties of Fluids

Mechanical Properties of FluidsNEET Physics · Class 11 · NCERT Chapter 9

17 NEET previous-year questions on Mechanical Properties of Fluids, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.

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All (17)
Mechanical Properties of Fluids (17)

A

energy = 4VT       − R 1 r 1 is released

B

energy = 3VT       + R 1 r 1 is absorbed

C

energy = 3VT       − R 1 r 1 is released

D

energy is neither released nor absorbed

Solution

The change in surface energy when small drops coalesce into one larger drop is given by . Since , we have . Substituting, . Since is negative, energy is released, so option (c) is correct.

A

VR2 n2r2

B

VR2 nr2

C

VR2 n3r2

D

V2R nr

Solution

Volume inflow rate = volume anflow rate 𝜋𝑅2𝑉= 𝑛𝜋𝑟2 ⇒ 𝑣= 𝜋𝑅2𝑉 𝑛𝜋𝑟2 = 𝑉𝑅2 𝑛𝑟2

A

1.70

B

(2 ) 2.35

C

3.0

D

1.50

Solution

www.vedantu.com 38 Power = F⃗ .V⃗⃗ =PAV⃗⃗ = ρgh AV =13.6 ×103 ×10 ×150 ×10−3 ×0.5 ×10−3 60 watt⁄ =102 60 watt=1.70 watt.

A

1,−1,−1

B

−1,−1,1

C

−1,−1,−1

D

1,1,1

Solution

Vc= ηx ρyrz Critical velocity is given by 𝑉𝑐= 𝑅𝜂 2𝜌𝑟 So, 𝑥 = 1 𝑦 = −1 𝑧 = −1

A

Water rises upto the tip of capillary tube and then starts overflowing like a fountain

B

Water rises upto the top of capillary tube and stays there without overflowing

C

Water rises upto a point a little below the top and stays there

D

Water does not rise at all

Solution

Water will not overflow but will change its radius of curvature.

A

{1 +(n + 1)p}ρ

B

{2 +(n + 1)p}ρ

C

{2 +(n − 1)p}ρ

D

{1 +(n − 1)p}ρ

Solution

www.vedantu.com 22 f0 = mg PA(1 − P)Lg+ nρApLg = dALg ρ(1 − p)+ nρp = d [1 − p +np]ρ = d [1 +(n − 1)p]ρ = d

A

650 kg m –3

B

425 kg m –3

C

800 kg m –3

D

928 kg m –3

Solution

hoil ρoil g = hwater ρwater g 140 × ρoil = 130 × ρwater ρoil = 313 × 1000 kg/m14 ρoil = 928 kg m –3

A

r 5

B

r 2

C

r 3

D

r 4

Solution

The rate of production of heat is given by , where is the viscous force and is the terminal velocity. For a sphere, , and (Stokes' law). Therefore, , so option (d) is correct.

A

W 𝑚—1 𝐾—1

B

J m 𝐾—1

C

J 𝑚—1 𝐾—1

D

Wm 𝐾—1 N E W S 2 vm/s 10 m/s θ 10Ω i1 V1 V1 10V 10Ω i2 A2 V2 10V 10Ω 2 SPACE FOR ROUGH WORK

Solution

𝑄 𝛥𝑡=𝐴𝐾′(𝛥𝜃 𝛥𝑥) (𝐽/𝑠) 𝑚−𝐾=𝐾′ J/s/m/K Or W−m−1−K−1

A

1 : 1

B

rA : rB

C

vA : vB

D

rB : rA

Solution

ω = 2π 𝑇 ⇒ 𝑇1= 𝑇2 or ω1= ω2

A

2.5 g

B

5.0 g

C

10.0 g

D

20.0 g

Solution

The mass of water that rises in a capillary tube is proportional to the cross-sectional area of the tube, which is . For a tube of radius , the cross-sectional area is times larger. Therefore, the mass of water that rises in the tube of radius is . However, the correct answer is , as the height of water rise is inversely proportional to the radius, so the total mass is . Option (c) is correct.

A

Mg 2

B

Mg

C

3 Mg 2

D

2 Mg

Solution

When the ball reaches terminal velocity, the net force is zero. The forces acting are the weight , the buoyant force (since the density of glycerine is ), and the viscous force . At terminal velocity, . However, the correct interpretation of the options suggests that the viscous force is , but the closest matching option is (a) , which should be interpreted as . Thus, the correct option is (a).

A

A

B

B

C

C

D

D

Solution

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

A

Decreases

B

Increases

C

Remains the same

D

Is equal to the atmospheric pressure

Solution

Excess pressure inside the bubble = TP R 4∆= in out 4TPP R= + as ‘R’ increases ‘P’ decreases

A

50.1×10⁻⁴ J

B

30.16×10⁻⁴ J

C

5.06×10⁻⁴ J

D

3.01×10⁻⁴ J

Solution

The energy required to form a soap bubble is given by , where is the radius and is the surface tension. Substituting and , , so option (d) is correct.

A

The principle of perpendicular axes

B

Huygen’s principle

C

Bernoulli’s principle

D

The principle of parallel axes

Solution

The venturi-meter operates on Bernoulli’s principle, which relates the pressure and velocity of a fluid in a pipe. As the fluid passes through a constricted section, its velocity increases and pressure decreases, allowing for flow rate measurement. Thus, option (c) is correct.

A

B

C

D

Solution

ROC = Radius of curvature at point A

Curvature =

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