Pressure & Fluid Statics Level - 1 - Free MCQ Practice Test with solutions,

By equating pressure at A-A’ in the two arms of manometer p = patm + ρgh − ρgas gx Note: as the density of gas is not given. The term ρgas × g x (where x: height of gas column and ρgas is density of gas) can be neglected.

p = 96 × 10³ + 850 × 9.81 × 0.55

p = 100.58 kPa

∴ p ≃ 100.6 kPa

Considering Equilibrium T1 = mg

When the block is completely immersed in water

Where FB = Buoyancy Force because of water

T2 = mg − FB

% reduction in tension = T1 − T2 / T1 × 100

= (mg − mg + FB / mg) × 100

= FB / mg x 1000

= ρseawater × Volume of block / g ρconcrete × Volume of block × g × 100

=

ρseawater / ρconcrete × 100 = 1025 / 2300 × 100 = 44.56%

≃ 45%

ρg3h + 2ρg 3h / 2 + 3ρgh = 3ρgH

3h + 3h + 3h = 3H

H / h = 9 / 3 = 3

Equating Pressures at ① and ②

pA + ρwg (h + 10) / 100 = pB + ρwg (h) / 100 + ρmg ( 10 / 100)

pA − pB = (ρm − ρw)g ( 10 / 100)

pA − pB = (2000 − 1000) × 9.81 × 0.1

pA − pB = 981 Pa

The gauge will read gauge pressure of air in the container

∴ Equating gauge pressure at ① and ②

p1 = p2

pair + 1.2 m of H2O = 0 + 0.2 m of Mercury

pair = 0.2 m of Mercury − 1.2 m of H2O

pair = 0.2 × 9.81 × 13600 − 1.2 × 9.81 × 1000

pair = 14911.2 Pa

Expressing this pressure in terms of head (hw) of water

pair = ρwghw

∴ hw = pair / ρwg

hw = 14911.2 / 1000 × 9.81

hw = 1.52 m

hp = hc + IG /

Therefore

1. hp > hc, i. e. P lies below C

2. hp is independent of fluid density or Bulk Modulus. [This is not true if fluid is compressible

3. Also hp − hc = IG / Ahc

i. e. hc ↑ (hp − hc) ↓

Hence all these statements are correct and option (D) is the correct answer.

i. e. T = Fp + W

Where T is the tension in the spring balance (i.e. the measurement) W is the weight of the bucket Fp is the pressure force on the bucket by water.

Note: Fp depends on the pressure at the base and thus height of water in the bucket. When the iron block is immersed in water the level of water in the bucket will rise thereby increasing the pressure at the base and thus Fp and T. i.e. the reading will increase.

As the pressure at a point does not depend upon the volume of fluid, but depends only upon the depth of the point from the top surface of the fluid. Therefore pressure at the base and thus the reading in the spring balance will remain constant.

For equilibrium

FB1 = (mb + ms)g

ρwV1g = (mb + ms)g

Where V1 is the volume of displaced liquid When the cargo is dumped into water

FB2 = mboat × g

while, FB′2 < msteel="" g="" (as="" steel="" sinks="" in="">

FB2 = ρw V2 × g, FB′2 = ρw V2′ g

Now FB2 + FB′2 = ρw(V2 + V2′ )g < (mb="" +="">

while, FB1 = ρw(V1)g = (mb + ms)g

So V1 > (V2 + V2′ )

i.e in second case, the total volume of water displaced is lower than the first case, hence the level of water in tub goes down.

1-1 is the initial line of action of buoyant force.

And Metacenter (M) is the intersection of 1-1 with the new line of action [After displacement].

As is seen the displacement of B to B′ generates a couple opposite to displacement and thus restores the body to initial equilibrium position.

In this case M lies above G.

But in other cases when M lies below G, the couple is over turning and the body is unstable.

For equilibrium Buoyant force = Weight of iceberg

ρseawater × Vsubmerged × g = ρice × VTotal × g

Vsub merged / VTotal =ρice / ρsea water = 917 / 1042 = 0.88 = 88%

Which means the part of the iceberg visible above the sea surface is only 12% of the total volume. Hence the saying “Tip of the iceberg” means a lot is hidden below than what is visible above the surface.

Even though oil is poured first, but oil being lighter will go up and the water layer will settle at the bottom. Finally in equilibrium, the situation shall be as shown

pA = ρoil × g × 0.2 + ρw × g × 0.25

pA = 4022.1 Pa

∗ patm = 74 cm of mercury

Δp = ρair × g × H

Δp = ρHg × g(0.74 − 0.6)

(Pressure difference between Manali & Delhi) Equating,

H = ρHg (0.74 − 0.6) / ρair =13.6 × 10³ × 0.14 / 1.2

H = 1586.667 m

i.e. Manali is 1586.667 m above Delhi.

Taking the free surface of the blood (in bottle) at 0 gauge pressure, (atmospheric pressure)

BPGauge = ρf × g × 1.2

BPGauge = 1020 × g × 1.2 = 12 kPa

The Blood pressure is 12 kPa (Gauge)

(i) The maximum Buoyant Force in Lake

= ρ × Volume × g = 1000 × 150 × g

The weight of the empty boat

= 8560 × g

Max load it can carry

= (150 × 10³ − 8560)g

= 141.4 × 10³ kg

or 141.4 Tones

(ii) For Sea, maximum Buoyant force

= ρsea × Volume × g = 154.5 × 10³g

Max load it can carry in sea

= (154.5 × 10³ − 8560)g = 146 Tones

For Equilibrium

FB = T + mg

or T = FB − mg

FB = ρ!hg ; mg = ρsHAg

T = ρ!hg − ρsHAg

T = (ρh − ρsH)Ag

Based on free body diagram, the reading (T) in spring balance is

T = mg − FB

Therefore reading in air

Ta = mg − FBa

Reading in water

Tw = mg − FBw

Loss in reading

Ta − Tw = FBw − FBa

Ta − Tw = ρwVbg − ρaVbg

Ta − Tw = (ρw − ρa)Vbg

Given that loss of reading for 2 objects is same

therefore (ρw − ρa)V_b1g = (ρw − ρa)V_b2g

∴ V_b1 = V_b2 i.e. volume of the two objects is same.

Given ρg = 1000 kg⁄m³

Total Force on plate

= ρghc(Area) = ρg (1 + 2 sin 45 / 2) (2 × 1) = 3414.2 kg force

hp = hc + IG / Ahc sin2 θ

IG = bd3 / 12

= 1 × 2³ / 12 = 2 / 3

A= 1 × 2 = 2

θ = 45°

hc = 1 +2 sin 45° / 2= 1 + 1 / √2

hc = 1.707 m

hp = 1.707 + 2/3 / 2 × 1.707×1 / 2

hp ≃ 1.9 m

∴ pA = ρgh

In case of vacuum, the atmospheric air from outside will seep inside disturbing the system. Hence vacuum pressure cannot be measured.

In case of gases, a free surface is not formed and reading is not obtained.

In case of high pressure, the required length of the piezometer tube is very long making it unsuitable for use.