# Fluid Statics

 Question 1
A pressure measurement device fitted on the surface of a submarine, located at a depth H below the surface of an ocean, reads an absolute pressure of 4.2 MPa. The density of sea water is 1050 $kg/m^3$, the atmospheric pressure is 101 kPa, and the acceleration due to gravity is 9.8 $m/s^2$. The depth H is _______ m (round off to the nearest integer).
 A 128 B 398 C 478 D 256
GATE ME 2021 SET-1   Fluid Mechanics
Question 1 Explanation:

\begin{aligned} &P_{A}=4.2 \mathrm{MPa}\;\;\;\;\text { [Absolute pressure] }\\ P_{\text {atm }} &=101 \mathrm{kPa} \\ \rho &=1050 \mathrm{~kg} / \mathrm{m}^{3} \\ g &=9.8 \mathrm{~m} / \mathrm{s}^{2} \\ P_{A} &=P_{\text {atm }}+\rho \mathrm{gH} \\ 4.2 \times 10^{6} &=\left(101 \times 10^{3}\right)+[1050 \times 9.81 \times \mathrm{H}] \\ H &=397.94 \text { or } 398 \mathrm{~m} \end{aligned}
 Question 2
In the space above the mercury column in a barometer tube, the gauge pressure of the vapour is
 A positive, but more than one atmosphere B negative C zero D positive, but less than one atmosphere
GATE ME 2020 SET-2   Fluid Mechanics
Question 2 Explanation:

In space above mercury column in barometer, ideally perfect vaccum is expected. But due to evaporation of mercury , the pressure in that space is equal to vapour pressure of mercury [nearly 0.2 Pa (abs)] . As absolute pressure in that space is less than atmospheric pressure, gauge pressure is negative.
 Question 3
Which of the following conditions is used lo determine the stable equilibrium of all partially submerged floating bodies?
 A Centre of buoyancy must be above the centre of gravity B Centre of buoyancy must be below the centre of gravity C Metacentre must be at a higher level than the centre of gravity D Metacentre must be at a lower level than the centre of gravity
GATE ME 2020 SET-2   Fluid Mechanics
Question 3 Explanation:
Metacentre must be higher level than the centre of gravity
 Question 4
For the stability of a floating body the
 A centre of buoyancy must coincide with the centre of gravity B centre of buoyancy must be above the centre of gravity C centre of gravity must be above the centre of buoyancy D metacenter must be above the centre of gravity
GATE ME 2017 SET-2   Fluid Mechanics
Question 4 Explanation:

$GM=BM-BG$

(i) $GM \gt 0$, stable
(ii) $GM = 0$, Neutral
(iii) $GM \gt 0$, unstable

NOTE:: Metacentre (M) must always be above the centre of gravity.
 Question 5
Consider a frictionless, massless and leak-proof plug blocking a rectangular hole of dimensions 2R x L at the bottom of an open tank as shown in the figure. The head of the plug has the shape of a semi-cylinder of radius R. The tank is filled with a liquid of density $\rho$ 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
 A $2\rho R^{2}gL\left( 1-\frac{\pi }{4}\right )$ B $2\rho R^{2}gL\left( 1+\frac{\pi }{4}\right )$ C $\pi R^{2}\rho gL$ D $\frac{\pi }{2}\rho R^{2}gL$
GATE ME 2016 SET-2   Fluid Mechanics
Question 5 Explanation:

\begin{aligned} \text{Volume, } V_{total}&=2R \times L \times R\\ &=2R^2L\;\;...(i)\\ \text{Volume: }V_1&=\frac{1}{2}\times \pi R^2 \times L\;\;...(ii)\\ F&= \text{Weight of fluid}\\ &=[V_{total}-V_{1}]\rho g\\ &=\left [ 2R^2L-\frac{\pi R^2L}{2} \right ]\rho g\\ &=\left [ 1-\frac{\pi}{4} \right ]2R^2Lg\rho \\ &=2\rho R^2gL\left [ 1-\frac{\pi}{4} \right ] \end{aligned}
 Question 6
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 $m/s^{2}$
 A 540 B 258 C 741 D 865
GATE ME 2016 SET-2   Fluid Mechanics
Question 6 Explanation:
Let V= Volume of metalic body
45% of volume is in oil
i.e. $V_{oil}=$ 45% of V=0.45V
and $V_{ater}=$ 0.55V
For equilibrium condition,
Net buoyant force $=(F_B)_{water}+(F_B)_{oil}$
\begin{aligned} M_bg&=\rho _w V_{water}\;g+\rho _{oil}V_{oil}\;g \\ M_b&= \rho _w V_{water}+\rho _{oil}V_{oil}\\ \rho _b V&= \rho _w \times 0.55V+\rho _{oil} \times 0.45V\\ \rho _b&= \rho _w \times 0.55+\rho _{oil} \times 0.45\\ &=1000 \times 0.55+700 \times 0.45 \\ &= 865\;kg/m^3 \end{aligned}
 Question 7
An inverted U-tube manometer is used to measure the pressure difference between two pipes A and B, as shown in the figure. Pipe A is carrying oil (specific gravity = 0.8) and pipe B is carrying water. The densities of air and water are 1.16kg/$m^{3}$ and 1000kg/$m^{3}$, respectively. The pressure difference between pipes A and B is __________kPa.
Acceleration due to gravity g = 10 m/$s^{2}$ .
 A -2.199 B -1.895 C -5.264 D -2.978
GATE ME 2016 SET-1   Fluid Mechanics
Question 7 Explanation:

Given data:
Specific gravity of oil,
$s^{oil} = 0.8$
$\therefore \quad \rho_{\mathrm{oil}}=0.8 \times 1000=800 \mathrm{kg} / \mathrm{m}^{3}$
Density of air: $\rho_{\text {air }}=1.16 \mathrm{kg} / \mathrm{m}^{3}$
Density of water,
$\rho_{\text {water }}=1000 \mathrm{kg} / \mathrm{m}^{3}$
Acceleration due to gravity,
\begin{aligned} g &=10 \mathrm{m} / \mathrm{s}^{2} \\ h_{1} &=80 \mathrm{mm}=0.08 \mathrm{m} \\ h_{2} &=200 \mathrm{mm}=0.2 \mathrm{m} \\ h_{3} &=100 \mathrm{mm}=0.1 \mathrm{m} \end{aligned}
Pressure at section 1 on left limb = Pressure at
section 1 on right limb
\begin{aligned} P_{A}-\rho_{\text {oil }} g h_{2}-\rho_{\text {air }} g h_{1}&=p_{B}-\rho_{\text {water }} g\left(h_{1}+h_{2}+h_{3})\right. \\ p_{A}-800 \times 10 \times 0.2&-1.16 \times 10 \times 0.08 =p_{B}-\\ &1000 \times 10(0.08+0.2+0.1) \\ p_{A}-1600-0.928&=p_{B}-3800\\ p_{A}-p_{B} &=-2199.07 \mathrm{Pa} \\ &=-2.199 \mathrm{kPa} \end{aligned}
 Question 8
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
GATE ME 2016 SET-1   Fluid Mechanics
Question 8 Explanation:
For floating body Buoyancy force acts through the centre of buoyancy which is C.G. for displaced volume.
 Question 9
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$^{\circ}$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 _______
 A 5303.688N B 1548.586N C 4859.256N D 4859.654N
GATE ME 2014 SET-2   Fluid Mechanics
Question 9 Explanation:
Step-I: Mass of balloon
\begin{aligned} \text { Volume: } \quad V &=\frac{4}{3} \pi r^{3}=\frac{4 \times 3.14 \times 5^{3}}{3} \\ &=523.33 \mathrm{m}^{3} \\ \rho_{He} &=\frac{P}{R T}=\frac{100 \times 10^{3}}{2080 \times 288} \\ &=0.16693 \mathrm{kg} / \mathrm{m}^{3} \\ \text { Total mass } &=0.16693 \times 523.33 \\ &=87.359 \mathrm{kg} \end{aligned}
Step-II : Mass of air
\begin{aligned} \rho &=\frac{P}{R T}=\frac{100 \times 10^{3}}{289 \times 288} \\ &=1.2 \mathrm{kg} / \mathrm{m}^{3} \end{aligned}
Mass of displaced volume of air
$=1.2 \times 523.33=628 \mathrm{kg}$
Step-III : Net buoyant mass
$=628-87.359=540.64 \mathrm{kg}$
$\therefore$Maximum allowable weight
\begin{aligned} =540.64 \times 9.81 \\ =5303.688 \mathrm{N} \end{aligned}
 Question 10
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 _______
 A 445 B 248 C 985 D 288
GATE ME 2014 SET-2   Fluid Mechanics
Question 10 Explanation:
Forbuble, $\quad \Delta P=\frac{2 \sigma}{r}$
Where
$\Delta P$=pressure difference
$Surface tension : [latex]\sigma=0.072 \mathrm{N} / \mathrm{m}$
Radiu Ills of bubble: $r=\frac{0.001}{2} \mathrm{m}=0.0005 \mathrm{m}$
$\therefore \quad \Delta P=\frac{2 \times 0.072}{0.0005}=288 \mathrm{N} / \mathrm{m}^{2}$
There are 10 questions to complete.

### 1 thought on “Fluid Statics”

1. ANSWER FOR QN:20 IS B
WHY BECAUSE IN MOHR’S CIRCLE DIAGRAM NORMAL STRESS IS POINTED ON NEGATIVE X-AXIS.