1. In a hand operated liquid sprayer (figure shown below) the liquid from the container rises to the top of the tube because of
(a) Capillary effect
(b) Suction produced by the air jet at the top end of tube
(c) Suction produced by piston during the backward stroke
(d) Pumping of the air in to the container
(1 Mark, 1990)

Ans: b
Explanation:
Air jet produces vacuum at the top of the tube. This vacuum produces suction, which rises the liquid.
2.The crosssectional area of one limb of a Utube manometer (figure shown below) is made 500 time larger than the other, so that the pressure difference between the two limbs can be determined by measuring ‘h’ on one limb of the manometer. The percentage error involved is
(a) 1.0
(b) 0.5
(c) 0.2
(d) 0.05
(2 Mark, 1990)

Ans: c
Explanation:
In single column manometer:
A\times \Delta h=a\times hWhere, A = Crosssectional area of the reservoir
a = Crosssectional area of the limb
The term containing \Delta h is ignored in the calculation.
So, % error = \frac { \Delta h }{ h } \times 100=\frac { a }{ A } \times 100
=> % error = \frac { 1 }{ 500 } \times 100 = 0.2%
3. A circular plate 1 m in diameter is submerged vertically in water such its upper edge is 8 m below the free surface of water. The total hydrostatic pressure force on one side of plate is
(a) 6.7 kN
(b) 65.4 kN
(c) 45.0 kN
(d) 77.0 kN
(2 Mark, 1988)

Ans: b
Explanation:
The total hydrostatic pressure force (F) = \rho ghA
=> F = 1000 * 9.81 * (8 + 0.5) * \frac { \pi }{ 4 } { 1 }^{ 2 }
=> F = 65.4 kN
4. Shown below are three cylindrical gates which restrain water in a 2D channel. Which gate experiences the maximum vertical component, the minimum vertical component and the maximum horizontal component of the hydrostatic force?
(1 Mark, 1993)

Ans: Max. VC – A, Min. VC – C, Max. HC – C
5. Bodies in flotation to be in stable equilibrium, the necessary and sufficient condition is that the centre of gravity is located below the ____.
(1 Mark, 1994)

Ans: Metacentre
6. A fluid is said to be Newtonian when the shear stress is:
(a) Directly proportional to the velocity gradient
(b) Inversely proportional to the velocity gradient
(c) Independent of the velocity gradient
(d) None of the above
(1 Mark, 1995)

Ans: a
Explanation:
According to Newton’s law of viscosity:
\tau \alpha \frac { du }{ dy }Where, \tau = Shear stress
and \frac { du }{ dy } = Velocity gradient
7. The force F needed to support the liquid of density d and the vessel on top in the figure below is:
(a) gd [ha − (H − h) A]
(b) gdHa
(c) gdHA
(d) gd(H − h)A
(2 Mark, 1995)

Ans: c
Explanation:
Many students may think:
Total pressure force = Total weight of the liquid
=> F = gd[ha – (H – h)A]
(But this is wrong.)
The actual answer is:
Total pressure force (F) = gdHA
This is known as Hydrostatic paradox.
8. A mercury manometer is sued to measure the static pressure at a point in a water pipe as shown in Fig.2.9. The level difference of mercury in the two limbs is 10 mm. The gauge pressure at that point is
(a) 1236 Pa
(b) 1333 Pa
(c) zero
(d) 98 Pa
(1 Mark, 1996)
9. The dimension of surface tension is
(a) ML^{−1}
(b) L^{2} T^{−1}
(c) ML ^{1}T ^{−1 }
(d) MT^{2}
(1 Mark, 1996)

Ans: d
Unit of surface tension : N/m
Dimension of Surface tension: \frac { \left[ { M }{ L }{ T }^{ 2 } \right] }{ { L }} = \left[ { M }{ T }^{ 2 } \right]
10. Refer the following figure, the absolute pressure of gas A in the bulb is:
(a) 771.2 mm Hg
(b) 752.65 mm Hg
(c) 767.35 mm Hg
(d) 748.8 mm Hg
(2 Mark, 1997)

Ans: a
Taking the reference line DE:
{ P }_{ A }+{ \rho }_{ w }g{ h }_{ 1 }={ \rho }_{ m }gh+{ \rho }_{ w }g{ h }_{ 2 }=> { P }_{ A }+1000\times 9.81\times 0.17=13600\times 9.81\times 0.02+1000\times 9.81\times 0.05
=> P_{A }= 1491.12 N/m^{2 }[Gauge pressure]
=> P_{A }= \frac { 1491.12 }{ 13600\times 9.81 } = 0.0112 m of Hg = 11.2 mm of Hg
Absolute pressure (P_{A}) = 11.2 + 760 = 771.2 mm of Hg
11. If ‘P’ is the gauge pressure within a spherical droplet, the gauge pressure within a bubble of the same fluid and of same size will be
(a) P/4
(b) P/2
(c) P
(d) 2P
(1 Mark, 1999)

Ans: d
Explanation:
Gauge pressure in a spherical droplet (P_{s})= \frac { 2\sigma }{ r }
Gauge pressure in bubble (P_{b}) = \frac { 4\sigma }{ r }
So, P_{b }= 2 * P_{s }= 2 * P = 2P
12. Kinematic viscosity of air at 20°C is given to be 1.6 \times 10^{5 }m^{2}/s. Its kinematic viscosity at 70°C will be vary approximately
(a) 2.2 \times 10^{5 }m^{2}/s
(b) 1.6 \times 10^{5 }m^{2}/s
(c) 1.2 \times 10^{5 }m^{2}/s
(d) 3.2 \times 10^{5 }m^{2}/s
(1 Mark, 1999)

Ans: a
Explanation:
Kinematic viscosity increases with increases in temperature.
Most appropriate answer is option a.
14. The SI unit of kinematic viscosity (ν) is
(a) m^{2}/sec
(b) kg/(msec)
(c) m/sec^{2}
(d) m^{3}/sec^{2}
(1 Mark, 2001)

Ans: a
Option ‘a’ is correct answer.
15. A static fluid can have
(a) Nonzero normal and shear stress
(b) Negative normal stress and zero shear stress
(c) Positive normal stress and zero shear stress
(d) Zero normal stress and nonzero shear stress
(1 Mark, 2001)

Ans: c
Static fluid exert only normal force not shear force.
16. The horizontal and vertical hydrostatic forces F_{x} and F_{y} on the semicircular gate, having a width w into the plane of figure, are
(a) F_{x} = ρghrw and F_{y} = 0
(b) F_{x} = 2ρghrw and F_{y} = 0
(c) F_{x} = ρghrw and F_{y} = ρgwr^{2}/h
(d) F_{x} = 2ρghrw and F_{y} = πρgwr^{2}/2
(2 Mark, 2001)

Ans: d
Horizontal force can be calculated by taking the projection of semicircular gate.(xy in the figure)
F_{x} = \rho ghA=\rho gh\times (2r\times w)=2\rho ghrw
F_{y }= \rho g\times (Volume of ABCDEA – Volume of ABDEA)
=> F_{y }= \rho g\times (Volume of BDCB)
=> F_{y }= \rho g\times \frac { \pi { r }^{ 2 } }{ 2 } \times w = \pi \rho gw{ r }^{ 2 }/2
17. A cylindrical body of crosssectional area A, height H and density ρ_{s}, is immersed to a depth h in a liquid of density ρ and tied to the bottom with a string. The tension in the string is
(a) ρghA
(b) (ρ_{s }– ρ)ghA
(c) (ρ – ρ_{s})ghA
(d) (ρh – ρ_{s}H)gA
(1 Mark, 2003)
18. The pressure gauges G_{1} and G_{2} installed on the system show pressures of P_{G1 }= 5.00 bar and P_{G2}=1.00 bar. The value of unknown pressure P is
(Atmospheric Pressure 1.01 bar)
(a) 1.01 bar
(b) 2.01 bar
(c) 5.00 bar
(d) 7.01 bar
(2 Mark, 2004)

Ans: d
For pressure gauge G2, atmospheric pressure is = 1.01 bar
Given, P_{G2}=1.00 bar
Absolute pressure in the right side box = 1 + 1.01 = 2.01 bar
For pressure gauge G1, atmospheric pressure is = 2.01 bar
Given, P_{G1}=5.00 bar
Absolute pressure in the left side box = 5 + 2.01 = 7.01 bar
19. An incompressible fluid (kinematic viscosity, 7.4 × 10^{7} m^{2}/s, specific gravity, 0.88) is held between two parallel plates. If the top plate is moved with a velocity of 0.5 m/s while the bottom one is held stationary, the fluid attains a linear velocity profile in the gap of 0.5 mm between these plates; the shear stress in Pascals on the surface of top plate is
(a) 0.651×10^{3}
(b) 0.651
(c) 6.51
(d) 0.651×10^{3}
(1 Mark, 2004)

Ans: b
Explanation:
Shear stress (\tau )=\mu \frac { du }{ dy }
=> \tau =(\nu \times \rho )\frac { du }{ dy }
=> \tau =(7.4\times { 10 }^{ 7 }\times 880)\frac { 0.5 }{ 0.5\times { 10 }^{ 3 } }
=> \tau = 0.651
20. For a Newtonian fluid
(a) Shear stress is proportional to shear strain
(b) Rate of shear stress is proportional to shear strain
(d) Rate of shear stress is proportional to rate of shear strain
(2 Mark, 2006)

Ans: c
Explanation:
We know: \tau \alpha \frac { d\phi }{ dt }
=> Shear stress is proportional to rate of shear strain.
20. A journal bearing has shaft diameter of 40mm and a length of 40mm. The shaft is rotating at 20 rad/s and the viscosity of the lubricant is 20 mPa.s. The clearance is 0.020mm .The loss of torque due to the viscosity of the lubricant is approximately
(a) 0.040 Nm
(b) 0.252 Nm
(c) 0.400 Nm
(d) 0.652 Nm
(2 Mark, 2008)

Ans: a
Shear force on the shaft (F) = \mu A\frac { du }{ dy }
= \mu \times \pi dl\times \frac { du }{ dy }
= 20\times { 10 }^{ 3 }\times \pi \left( 40\times { 10 }^{ 3 } \right) \left( 40\times { 10 }^{ 3 } \right) \times \frac { 20\times 20\times { 10 }^{ 3 } }{ 0.02\times { 10 }^{ 3 } }
=> F = 2.1 N
Torque = F\times r = 2.01\times 0.02 = 0.04 Nm
21. For the stability of a floating body, under the influence of gravity alone, which of the following is TRUE?
(a) Metacentre should be below centre of gravity
(b) Metacentre should be above centre of gravity
(c) Metacentre and centre of gravity must lie on the same horizontal line
(d) Metacentre and centre of gravity must lie on the same vertical line
(1 Mark, 2010)

Ans: b
For stability in floating body, the metacenter must lie above the centre of gravity.
21. A lightly loaded full journal bearing has a journal of 50 mm, bush bore of 50.05 mm and bush length of 20 mm. if rotational speed of journal is 1200 rpm and average viscosity of liquid lubricant is 0.03 Pas, the power loss (in W) will be
(a) 37
(b) 74
(c) 118
(d) 237
(2 Mark, 2010)

Ans: a
Tangential velocity of the shaft (u) = \frac { \pi dn }{ 60 }
=> u = \frac { \pi \times 50\times { 10 }^{ 3 }\times 1200 }{ 60 } = 3.14 m/s
Radial clearance (dy) = \frac { Dd }{ 2 }
=> dy = \frac { 50.0550 }{ 2 } = 0.025 mm
Shear stress \left( \tau \right) =\mu \frac { du }{ dy }
=> \tau =0.03\times \frac { 3.14 }{ 0.025\times { 10 }^{ 3 } }
=> \tau =3768N/{ m }^{ 2 }
Shear force on the shaft (F) = \tau \times \pi dl
=> F = 3768\times \pi \left( 50\times { 10 }^{ 3 } \right) \left( 20\times { 10 }^{ 3 } \right)
=> F = 11.83 N
Power (P) = F\times v=F\times \frac { \pi dn }{ 60 }
=>P = 11.83\times \frac { \pi \left( 50\times { 10 }^{ 3 } \right) 1200 }{ 60 }
=> P = 37.1 W
22. A hinged gate of length 5 m, inclined at 30° with the horizontal and with water mass on its left, is shown in the figure below. Density of water is 1000 kg/m^{3}. The minimum mass of the gate in kg per unit width (perpendicular to the plane of paper), required to keep it closed ?
(a) 5000
(b) 6600
(c) 7546
(d) 9623
(2 Mark, 2013)

Ans: d
Width = 1 unit
Centre of gravity:
\overline { X } =2.5\times sin{ 30 }^{ 0 } = 1.25 m
Total pressure force on the gate \left( { F }_{ T } \right) =\rho gA\overline { X }
=> { F }_{ T }=1000\times 9.81\times (5\times 1)\times 1.25 = 61.3125 kN
Center of pressure:
\overline { h } =\overline { X } +\frac { { I }_{ G } }{ A\overline { X } } { sin }^{ 2 }\theta=> \overline { h } =1.25+\frac { \frac { 1 }{ 12 } \times 1\times { 5 }^{ 3 } }{ (5\times 1)\times 1.25 } \times { sin }^{ 2 }{ 30 }^{ 0 }
=> \overline { h } = 1.667 m
Taking moment about O,
W\times 2.5cos{ 30 }^{ 0 }={ F }_{ T }\times \left( \frac { \overline { h } }{ sin{ 30 }^{ 0 } } \right)=> W = 94396.7 N
=> mass (m) = W/9.81 = 9623 kg
23. An aluminium alloy (density 2600 kg/m^{3}) casting is to be produced. A cylindrical hole of 100 mm diameter and 100 mm length is made in the casting using sand core (density 1600 kg/m^{3}). The net buoyancy force (in newton) acting on the core is __.
(2 Mark, 2014[1])

Ans: 7.7
Explanation:
Net buoyancy force (F_{b}) = \left( { \rho }_{ al }{ \rho }_{ c } \right) gV
=> F_{b }= \left( 26001600 \right) \times 9.81\times \frac { \pi }{ 4 } { d }^{ 2 }l
=> F_{b }= 1000\times 9.81\times \frac { \pi }{ 4 } \times { 0.1 }^{ 2 }\times 0.1
=> F_{b }= 7.7 N
24. In a simple concentric shaftbearing arrangement, the lubricant flows in the 2 mm gap between the shaft and the bearing. The flow may be assumed to be a plane Couette flow with zero pressure gradient. The diameter of the shaft is 100 mm and its tangential speed is 10 m/s. The dynamic viscosity of the lubricant is 0.1 kg/m.s. The frictional resisting force (in newton) per 100 mm length of the bearing is ____.
(2 Mark, 2014[1])

Ans: 15.7
Explanation:
Surface area where the shear force acts:
A = \pi dL = \pi \times 0.1\times 0.1 = 0.0314 m^{2}
Frictional resisting force F=\mu A\frac { du }{ dy }
=> F=0.1\times 0.0314\times \frac { 10 }{ 0.002 } = 15.7 N
25. For a completely submerged body with centre of gravity ‘G’ and centre of buoyancy ‘B’, the condition of stability will be
(a) G is located below B
(b) G is located above B
(c) G and B are coincident
(d) Independent of the locations of G and B
(1 Mark, 2014[1])

Ans: a
Explanation:
For a completely submerged body, the condition of stability is:
G located below B.
26. A spherical balloon with a diameter of 10 m, shown in the figure below is used for advertisements. The balloon is filled with helium (R_{He} = 2.08 kJ/kg.K) at ambient conditions of 15°C and 100 kPa. Assuming no disturbances due to wind, the maximum allowable weight (in newton) of balloon material and rope required to avoid the fall of the balloon (R_{air} = 0.289 kJ/kg.K) is ___.
(2 Mark, 2014[2])

Ans: 5303.68
Explanation:
Density of air ({ \rho }_{ air })=\frac { { R }_{ air }T }{ P }
=> { \rho }_{ air }=\frac { { 10 }^{ 5 } }{ 289\times 288 } = 1.2 kg/m^{3}
Density of hillium ({ \rho }_{ He })=\frac { { R }_{ He }T }{ P }
=> { \rho }_{ air }=\frac { { 10 }^{ 5 } }{ 289\times 288 } = 0.167 kg/m^{3}
The maximum allowable weight of balloon material and rope = Buoyancy force by the air – total weight of helium inside the balloon
=> W_{max} = F_{b} – W_{He}
=> W_{max} = { \rho }_{ air }\times V\times g{ \rho }_{ He }\times V\times g
=> W_{max }= { (\rho }_{ air }{ \rho }_{ He })\times V\times g
=> W_{max }= (1.20.167)\times \frac { 4 }{ 3 } \pi \times { 5 }^{ 3 }\times 9.81
=> W_{max }= 5306 N
27. The difference in pressure (in N/m^{2}) across an air bubble of diameter 0.001 m immersed in water (surface tension = 0.072 N/m) is _____.
(1 Mark, 2014[2])

Ans: 288
Air bubble immersed in water is like water droplet in the air.
We use same formula for both of the case.
The difference in pressure \left( \Delta P \right) =\frac { 2\sigma }{ R }
=> \Delta P=\frac { 2\times 0.072 }{ 0.0005 }
=> \Delta P= 288
28. Temperature of nitrogen in a vessel of volume 2 m^{3} is 288 K. A Utube manometer connected to the vessel shows a reading of 70 cm of mercury (level higher in the end open to atmosphere). The universal gas constant is 8314 J/kmolK, atmospheric pressure is 1.01325 bar, acceleration due to gravity is 9.81 m/s^{2} and density of mercury is 13600 kg/m^{3}. The mass of nitrogen (in kg) in the vessel is __.
(2 Mark, 2015[1])

Ans: 4.56
Explanation:
Chracteristic gas constant is
{ R }_{ C }=\frac { { R }_{ U } }{ Molar\quad mass }=\frac { 8314 }{ 28 } = 296 J/kgK
Absolute pressure in the vessel = Atmospheric pressure + Gauge pressure
=> { P }_{ abs }={ P }_{ atm }+\rho \times g\times h
=> { P }_{ abs }=101325+13600\times 9.81\times 0.7
=> { P }_{ abs } = 194716.2 N/m^{2}
Using ideal gas equation: PV = mR_{C}T
=> m = \frac { PV }{ { R }_{ C }T } =\frac { 194716.2\times 2 }{ 296\times 288 }
=> m = 4.56 kg
28. An inverted Utube manometer is used to measure the pressure difference between two pipes A and B, as shown in figure. Pipe A is carrying oil (specific gravity = 0.8) and pipe B is carrying water. The densities of air and water are 1.16 kg/m^{3} and 1000 kg/m^{3} The pressure difference pipes A and B are ______ kPa.
Acceleration due to gravity g = 10m/s^{2}.
(2 Mark, 2016[1])

Ans: 2.16
{ P }_{ A }{ \rho }_{ o }g{ h }_{ o }{ \rho }_{ a }g{ h }_{ a }={ P }_{ B }{ \rho }_{ w }g{ h }_{ w }=> { P }_{ A }{ P }_{ B }={ \rho }_{ o }g{ h }_{ o }+{ \rho }_{ a }g{ h }_{ a }{ \rho }_{ w }g{ h }_{ w }
=> { P }_{ A }{ P }_{ B }=800\times 9.81\times 0.2+1.16\times 9.81\times 0.08{ 1000\times 9.81\times 0.38 }
=> { P }_{ A }{ P }_{ B }= 2.16 kPa
29. For a floating body, buoyant force acts at the
(a) Centroid of the floating body
(b) Center of gravity of the body
(c) Centroid of the fluid vertically below the body
(d) Centroid of the displaced fluid
(1 Mark, 2016[1])

Ans: d
Explanation:
For a floating body, buoyant force acts at the centroid of the displaced fluid. This is also known as centre of buoyancy.
30. Assuming constant temperature condition and air to be an ideal gas, the variation in atmospheric pressure with height calculated from fluid statics is
(a) Linear
(b) Exponential
(c) Quadratic
(d) Cubic
(1 Mark, 2016[2])

Ans: b
According to Hydrostatic law:
\frac { dp }{ dz } =\rho g…(1)
Ideal gas Equation: p=\rho RT…(2)
From Equation (1) and (2),
\frac { dp }{ p } =\left( \frac { g }{ RT } \right) dz=> ln\frac { p }{ { p }_{ o } } =\left( \frac { g }{ RT } \right) z
=> p={ p }_{ o }{ e }^{ \left( \frac { gz }{ RT } \right) }
From the above equation,
The pressure varies (decreases) exponentially with altitude, if temperature variation is considered negligible.
31. The large vessel shown in the figure contains oil and water. A body is submerged at the interface of oil and water such that 45 percent of its volume is in oil while the rest is in water. The density of the body is _________ kg/m^{3}. The specific gravity of oil is 0.7 and density of water is 1000 kg/m^{3}.
Acceleration due to gravity g = 10 m/s^{2}.
(2 Mark, 2016[2])

Ans: 865
Explanation:
Body is in equilibrium.
So, W={ F }_{ b1 }+{ F }_{ b2 }
=> { \rho }_{ b }\times g\times V={ \rho }_{ w }\times g\times 0.55V+{ \rho }_{ o }\times g\times 0.45V
=> { \rho }_{ b }={ 0.55\rho }_{ w }+{ 0.45\rho }_{ o }
=> { \rho }_{ b }={ 0.55\times 1000 }+0.45\times 700
=> { \rho }_{ b } = 865 kg/m^{3}
32. Consider a frictionless, massless and leakproof plug blocking a rectangular hole of dimensions 2R × L at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semicylinder of radius R. The tank is filled with a liquid of density ρ up to the tip of the plug. The gravitational acceleration is g. Neglect the effect of the atmospheric pressure.
The force F required to hold the plug in its position is
(a) 2ρR^{2}gL\left( 1\frac { \pi }{ 4 } \right)
(b) 2ρR^{2}gL\left( 1+\frac { \pi }{ 4 } \right)
(c) \piR^{2}ρgL
(d) \frac { \pi }{ 2 }ρR^{2}gL
(2 Mark, 2016[2])
33. Consider fluid flow between two infinite horizontal plates which are parallel (the gap between them being 50 mm). The top plate is sliding parallel to the stationary bottom plate at a speed of 3 m/s. The flow between the plates is solely due to the motion of the top plate. The force per unit area (magnitude) required to maintain the bottom plate stationary is _________ N/m^{2}.
Viscosity of the fluid μ = 0.44 kg/ms and density ρ = 888 kg/m^{3}.
(2 Mark, 2016[2])

Ans: 26.4
Explanation:
We know: \tau =\mu \frac { du }{ dy }
=> \tau =0.44\times \frac { 3 }{ 0.05 }
=> \tau = 26.4 N/m^{2}
34. A solid block of 2.0 kg mass slides steadily at a velocity V along a vertical wall as shown in the figure below. A thin oil film of thickness h = 0.15 mm provides lubrication between the block and the wall. The surface area of the face of the block in contact with the oil film is 0.04 m^{2}. The velocity distribution within the oil film gap is linear as shown in the figure. Take dynamic viscosity of oil as 7×10^{3} Pas and acceleration due to gravity as 10 m/s^{2}. Neglect weight of the oil. The terminal velocity V (in m/s) of the block is ______ (correct to one decimal place).
(2 Mark, 2018[1])

Ans: 10.71
Explanation:
The block slides steadily at a velocity V.
At this condition:
Weight of the block = Frictional force on the body
=> mg = \frac { \mu V }{ h } A
=> 2\times 10=\frac { 7\times { 10 }^{ 3 }\times V }{ 0.15\times { 10 }^{ 3 } } \times 0.04
=> V = 10.71 m/s
35. Two immiscible, incompressible, viscous fluids having same densities but different viscosities are contained between two infinite horizontal parallel plates, 2 m apart as shown below. The bottom plate is fixed and the upper plate moves to the right with a constant velocity of 3 m/s. With the assumptions of Newtonian fluid, steady, and fully developed laminar flow with zero pressure gradient in all directions, the momentum equations simplify to \frac { { d }^{ 2 }u }{ d{ y }^{ 2 } } =0.
If the dynamic viscosity of the lower fluid,{ \mu }_{ 2 }, is twice that of the upper fluid,{ \mu }_{ 1 }, then the velocity at the interface (round off to two decimal places) is ___________ m/s.
(2 Mark, 2019[1])

Ans: 1
Shear stress is same at the both side of middle plate.
{ \tau }_{ 1 }={ \tau }_{ 2 }=> \frac { { \mu }_{ 2 }{ V }_{ 1 } }{ 1 } =\frac { { \mu }_{ 1 }\left( { V }_{ 2 }{ V }_{ 1 } \right) }{ 1 }
=> { 2\mu }_{ 1 }{ V }_{ 1 }={ \mu }_{ 1 }\left( { V }_{ 2 }{ V }_{ 1 } \right)
=> 3{ V }_{ 1 }={ V }_{ 2 } = 3 m/s
=> { V }_{ 1 } = 1 m/s
36. A cube of side 100 mm is placed at the bottom of an empty container on one of its faces. The density of the material of the cube is 800 kg/m^{3}. Liquid of density 1000 kg/m^{3} is now poured into the container. The minimum height to which the liquid needs to be poured into the container for the cube to just lift up is _____ mm.
(2 Mark, 2019[1])