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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energy = 4VT − R 1 r 1 is released
energy = 3VT + R 1 r 1 is absorbed
energy = 3VT − R 1 r 1 is released
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.
VR2 n2r2
VR2 nr2
VR2 n3r2
V2R nr
Solution
Volume inflow rate = volume anflow rate 𝜋𝑅2𝑉= 𝑛𝜋𝑟2 ⇒ 𝑣= 𝜋𝑅2𝑉 𝑛𝜋𝑟2 = 𝑉𝑅2 𝑛𝑟2
1.70
(2 ) 2.35
3.0
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.
1,−1,−1
−1,−1,1
−1,−1,−1
1,1,1
Solution
Vc= ηx ρyrz Critical velocity is given by 𝑉𝑐= 𝑅𝜂 2𝜌𝑟 So, 𝑥 = 1 𝑦 = −1 𝑧 = −1
Water rises upto the tip of capillary tube and then starts overflowing like a fountain
Water rises upto the top of capillary tube and stays there without overflowing
Water rises upto a point a little below the top and stays there
Water does not rise at all
Solution
Water will not overflow but will change its radius of curvature.
{1 +(n + 1)p}ρ
{2 +(n + 1)p}ρ
{2 +(n − 1)p}ρ
{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
650 kg m –3
425 kg m –3
800 kg m –3
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
r 5
r 2
r 3
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.
W 𝑚—1 𝐾—1
J m 𝐾—1
J 𝑚—1 𝐾—1
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
1 : 1
rA : rB
vA : vB
rB : rA
Solution
ω = 2π 𝑇 ⇒ 𝑇1= 𝑇2 or ω1= ω2
2.5 g
5.0 g
10.0 g
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.
Mg 2
Mg
3 Mg 2
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
B
C
D
Solution
Initial speed of ball is zero and it finally attains terminal speed
Decreases
Increases
Remains the same
Is equal to the atmospheric pressure
Solution
Excess pressure inside the bubble = TP R 4∆= in out 4TPP R= + as ‘R’ increases ‘P’ decreases
50.1×10⁻⁴ J
30.16×10⁻⁴ J
5.06×10⁻⁴ J
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.
The principle of perpendicular axes
Huygen’s principle
Bernoulli’s principle
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.
Solution
ROC = Radius of curvature at point A
Curvature =
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