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# Mechanical Properties of Fluids

#### Multiple Choice Questions

11.

A raft of wood of mass 120 kg floats in water. The weight that can be put on the raft to make it just sink should be (draft = 600 kg/m3) :

• 80 kg

• 50 kg

• 60 kg

• 30 kg

A.

80 kg

The volume of the raft is given by

If the raft is immersed fully in water, then the weight of water displaced is

= Vdwater

= 0.2 × 103 = 200 kg

Hence, the additional weight which can be put on the raft
= 200 kg - 120 kg = 80 kg

12.

The old age arteries carrying blood in the human body become barrow resulting in an increase in blood pressure. This follows from

• Pascal's law

• Stoke's law

• Bernoulli's principle

• Archimedes principle

C.

Bernoulli's principle

According to continuity theorem as

av = constant

as area increases velocity decrease. So when blood flows from narrow arteries to wider one, velocity decreases.

According to Bernoulli's principle

P +   = constant

As the velocity decrease, pressure increases. Thus when arteries become narrow, blood pressure increases. Thus when arteries become narrow, blood pressure increase.

13.

A tank is filled with water upto height H. When a hole is made at a distance h below the level of water, what will be the horizontal range of water jet?

• ${2}\sqrt{\mathrm{h}\left(\mathrm{H}-\mathrm{h}\right)}$

A.

${2}\sqrt{\mathrm{h}\left(\mathrm{H}-\mathrm{h}\right)}$

Applying Bernoulli's theorem,

the velocity of water at point A

v = $\sqrt{2\mathrm{gh}}$

Time taken to reach point C is t

so,

Now, horizontal range R = vt

14.

An ice-cube of density 900 kg/m3 is floating in the water of density 1000 kg/m3 The percentage of volume of the ice-cube outside the water is:

• 20 %

• 35 %

• 10 %

• 25 %

C.

10 %

The percentage of the volume of ice-cube outside the water is

= 10 %

15.

Eight small drops, each of equal charges and radii are coalesce to form a big drop. The capacity of bigger drop in comparison to each smaller drop will be :

• 4 times

• 2 times

• 32 times

• 8 times

B.

2 times

Volume of a bigger drop = 8 $×$ volume of smaller drop

Capacity of a smaller drop C = $4{\mathrm{\pi \epsilon }}_{0}\mathrm{r}$

Capacity of a bigger drop C$4{\mathrm{\pi \epsilon }}_{0}\mathrm{R}$

Hence,    C= 2C

16.

Mercury boils at 367°C. However, mercury thermometers are made such that they can measure temperature upto 500°C. This is done by:

• maintaining vacuum above the mercury column in the stem of the thermometer

• filling nitrogen gas at high pressure above the mercury column

• filling oxygen gas at high pressure above the mercury column

• filling nitrogen gas at low pressure above the mercury column

B.

filling nitrogen gas at high pressure above the mercury column

If we fill nitrogen gas at high pressure above mercury level, the boiling point of mercury is increased which can extend the range up to 500°C.

17.

The surface tension of the soap solution is T. The work done in forming a soap bubble of radius r, is :

• $8{\mathrm{\pi r}}^{2}\mathrm{T}$

• $4{\mathrm{\pi r}}^{2}\mathrm{T}$

• $\frac{4}{3}{\mathrm{\pi r}}^{2}\mathrm{T}$

• $\frac{8}{3}{\mathrm{\pi r}}^{2}\mathrm{T}$

A.

$8{\mathrm{\pi r}}^{2}\mathrm{T}$

The work done in making a soap bubble is given by

W=T×A

where,T = surface tension

A = 2×$4{\mathrm{\pi r}}^{2}$

Therefore, W= T× $8{\mathrm{\pi r}}^{2}$

= $8{\mathrm{\pi r}}^{2}\mathrm{T}$

18.

Assertion: Smaller drops of liquid resist deforming forces better than the larger drops.

Reason: Excess pressure inside a drop is directly proportional to its surface area

• If both assertion and reason are true and reason is the correct explanation of assertion

• If both assertion and reason are true but reason is not the correct explanation of assertion

• If assertion is true but reason is false

• If assertion is true but reason is false

C.

If assertion is true but reason is false

The excess pressure inside the small drop is large as compared to the large drop because of which smaller drop of liquid resists deforming force better than the large drop.

Excess pressure = $\frac{2\mathrm{T}}{\mathrm{r}}$

where T = surface tension

r = radius of liquid drop

Therefore excess pressure is inversely proportional to its radius and hence the surface area.

19.

The radius R of the soap bubble is doubled under isothermal condition. If T be the surface tension of soap bubble, the required surface energy in doing so is given by:

• 32πR2T

• 24πR2T

• 8πR2T

• 4πR2T

B.

24πR2T

Initial surface energy

U1 = 2 × 4πR2T = 8πR2T

Final surface energy

U2 = 2 × 4π(2R)2T = 32πR2T

Hence, required energy

= 32πR2T - 8πR2T = 24πR2T

20.

A sphere of mass Mand radius R is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to

• R2

• R

• 1/R

• 1/R2

A.

R2

Stoke's law

FD = 6$\mathrm{\pi }$ η r vT

FD is the drag force on the sphere falling through the fluid in newtons (N)

η is the viscosity of the fluid in kilogram-per-meter-per-second

r is the radius of the sphere in meters (m)

vT is the terminal velocity in meter-per-second (m/s)

6$\mathrm{\pi }$ηr vT$\frac{4}{3}\mathrm{\pi }$R2 (ρ $-$ σ ) g

∴             v ∝ R2

As M is given,

ρ =

But as  R3 increases, M also increases

ρ is a constant

σ also is constant.

∴   v, the terminal velocity ∝ R2