Test: Pressure & Fluid Statics Level - 2 - Mechanical Engineering MCQ

Equating pressure values at 1 and 1′

P1 = P1′

P1 = ρg(h + R sin θ) ⋯ ①

By mass conservation Ah = aR

π 4 / D2 h = π 4 / d2R

h / R = d2 / D2

h / R = (5 / 50)²

^{h / R= 0.01}

^{If h term is neglected in ① as h ≪ R}

^{P1 ≃ ρgR sin θ}

^{which will be an approximate value of P1}

^{∴ Error = Exact − Approximate}

^{% Error = Exact − Approximate / Exact × 100}

^{% Error = ρg(h + R sin θ) − ρgR sin θ / ρg(h + R sin θ) × 100}

^{% Error = h / h + R sin θ × 100}

^{% Error = 1 / 1 +R / h sin θ × 100}

^{% Error = 1 / 1 + 100 ×1 / 2 × 100}

^{% Error = 1.96%}

pair = ρmg(h6 + h5) − ρw g(h5 + h4) +ρmg(h4 − h3) − ρwg(h2 − h3) +ρmgh2 − ρwgh1

pair = ρmg(h6 + h5 + h4 − h3 + h2) −ρwg(h5 + h4 + h2 − h3 − h1)

pair = [13600 × (0.5 + 0.1 + 0.7 − 0.3 + 0.5) −1000 × (0.1 + 0.7 + 0.5 − 0.3 − 0.4)] × 9.81

pair = 194238 Pa Or 194.238 kPa

FB / A1 = F2 / A2 ⟹ FB × A2 / A1 = F2 ⋯ ①

rom FBD of handle

Equating Moments of F and FB about C

F × 33 = FB × 3

Or FB = F × 11 ⋯ ②

Using equation ① & ② and using

F = 100 N, d1 = 1.5 cm and d2 = 5 cm

We get, F2 = 12,222.22 N or 12.2 kN

mg = FBw + FBHg

ρb × 10³ × g = ρw × 10² × xg + ρHg × 10² × (10 − x)g

6.8 × 10 = x + 13.6(10 − x)

68 = x + 136 − 13.6 x

68 = 12.6 × x

x = 5.39 cm ≈ 5.4 cm

h_p = h_c +I_g / Ahc

But this formula can be a little cumbersome in terms of finding IG, i.e. moment of inertia

about the centroidal axis. Therefore we can also use

h_p = I1−1 / Ahc

1-1 is the line of intersection of planar surface with top surface of liquid.

In this case 1-1 is the diameter of the semicircle.

∴ I1−1 = πd4 / 128 = πR4 / 8

i.e. moment of inertia of the semicircle about the diameter.

h_p = πR4 / 8 / πR2 / 2 × 4R / 3π

[For semi-circle, centroid lies at a distance of 4R/3π from diameter]

h_p = 3πR / 16

So p1 = ρw × g × 0.2 (Gauge Pressure)

(p1)Gauge = 1962 Pa

For calculating force on container

F = pA − mg

F = 1962 × π / 4 × (0.3)² − 79

F = 59.68 N

⇒ pA − pB = ρ1ghB − ρ3ghB + ρBghB − ρAghA

Expressed in terms of head of water

h = (S1 − S3 + SB)hB − SAhA

For ideal gas, PV = mRT or p = ρRT

∴ ρ = p / RT

Substituting value of ρ

dp / dy = −p / RT × g

⇒ dp / p = −g / RTdy

⟹ ln p = −g / RTy + C

or p = C e −gy / RT

Also at zero scale level, i.e.

!t y = 0, p = p0

⇒ p0 = C

Substituting the same

p = p0 e −gy / RT

where FB = Buoyant force T = Tension Clearly T = FB

As the balloon is flexible, and as the pressure changes, the volume of the balloon will also change, so will the buoyant force experienced by it. So

FB1 = ρ × 4 / 3 πr13 × g and FB2 = ρ × 4 / 3 πr23 × g

Given, p = C / D2 ⟹ D = √C / p

⇒ r1 = 1 / 2√ C / p1 and r2 = 1 / 2 √C / p2

% change in Buoyant force

=FB2 − FB1 / FB1× 100

= (FB2 / FB1− 1) × 100

= [(p1 / p2)3/2 − 1] × 100

Putting p1 = 100 kPa and p2 = 1.6 kPa we get the % change in Buoyant Force = −98.4%.

The negative sign indicates that Buoyant Force decreases.

we know dp dh = ρg

⇒ dp = (WO + C(h) 1/2) dh

Integrating on both sides p = WOh + Ch3/2 /3× 2 + C2

p = WOh + 2/3C(h)3/2 + C2

Taking gauge pressure at top surface equal to zero

C2 = 0

hence, p = WOh + 2/3C(h)3/

The total force on gate = ρg hc × !rea = 103 × g × (2 + 1.5) × (3 × 6) = 618.03 kN The center of Pressure will be at the center of the pressure diagram

1 / 3× (2 × 5ρg + 2ρg) / 7ρg × 3 = 12 / 7 = 1.714 m

So the center of pressure is (2 + 1.714)

= 3.714 m from the free surface

Starting from point A the governing manometric equation is

pA + w h1 − wm (h3 + h4 sin 45°)

= pB ∴ Pressure difference, pA − pB

= 9810 × 13.6 (0.45 + 0.2 sin 45°) −9810 × 0.3

= 78905 − 2943

= 75962 N/m2 = 75.96 kN/m

Consider FBD of cylinder

W = FB

W = FB,Oil + FB,Water

W = ρ0Vd0g + ρwVdwg

ρbVbg = ρ0Vd0g + ρwVdwg

900 × A × 10 = 600 × A × x + 1000 × A(10 − x)

9000 = 600 x + 10000 − 1000 x

∴ −1000 = −400 x

x = 1000 / 400

x = 2.5 cm

ρ0g × 0.5 + ρwg × 0.5 = 8.892 kPa

The volume of pressure prism shall give total pressure force

∴ Fp = 1 / 2 × 3.924 × 0.5 × 1 + (3.924 + 8.842 / 2) × 0.5 × 1

Fp = 0.981 + 3.204 Fp

= 4.185 kN

Assuming width of the tank is w perpendicular to the plane of paper

FBottom = ρg(2w × w × w) = 2ρgw³

Fany side = ρg(2w × w) × w / 2= ρgw³

so, FBottom / Fside = ρg2w³ / ρgw³ = 2

pc = ρghc

⇒ hc = pc / ρg

hc = 1.962 × 105 / 870 × 9.81

hc = 22.98 m

For a submerged surface

hp = hc + IG / Ahc

⇒ hp − hc = IG / Ahc

hp − hc = πd⁴/64 / πd²/4 × 22.98 = 4² × 4 / 64 × 22.98

hp − hc = 0.043 m = 4.3 cm

Therefore using hydro-static law

dp / dh = ρg

dp / dh = (400 + 10h)g dp

= (400 + 10h)g dh

p = (400h + 5h2)g

Therefore pressure at a point 8 m below is

p(h = 8) = (400 × 8 + 5 × 82) × 9.81

p(h = 8) = 34531 Pa

≃ 34.531 kPa

In single column manometer as area of one limb (!2 ) is much greater than the other limb (!1 ). The liquid column (h1) in !1 is neglected creating some inherent error in area equal to

A1 / A2 − A1 × 100.

Compound manometer is a combination of multiple manometers arranged in series and is used for measuring large pressure values.

Hence all 3 statements are correct.

tan θ = a / g

⇒20 / 20 = a / g

⇒ a = g

Consider FBD of water body

∴ where p1 and p2 are pressure forces exerted by rear end and bottom of container.

Therefore p1 = ma = ρVa

= 1000 × (0.1 × 0.2 × 1) × 9.81

= 196.2 N