Mechanical Properties of Fluids Test

Result

Mechanical Properties of Fluids Test - 53

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Weekly Quiz Competition

Question 1
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The terminal velocity of solid sphere of radius $$0.1$$ m moving in air in vertically downward direction, is: $$(\eta =1.8\times 10^{-5} Ns/m^2$$, density of sphere $$=1000 kg/m^3$$ and $$g=10 m/s^2$$)

A
$$2\times 10^5$$ m/s

B
$$1.2\times 10^2$$ cm/s

C
$$4\times 10^2$$ cm/s

D
None of these

Solution
Question 2
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How many cylinders of hydrogen at atmospheric pressure are require to fill a balloon whose volume is $$500\ m^3$$, if hydrogen is stored in cylinder of volume $$0.05\ m^3$$ at an absolute pressure of $$15\times 10^5 \ Pa$$?

A
$$700$$

B
$$675$$

C
$$605$$

D
$$710$$

Solution
Question 3
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A solid sphere falls with a terminal velocity of $$20m/s^{-1}$$ in air. If it is allowed to fall in vacuum

A
terminal velocity will be $$20 m/s^{-1}$$.

B
terminal velocity will be less than $$20 m/s^{-1}$$

C
terminal velocity will be more than $$20 m/s^{-1}$$

D
there will be no terminal velocity

Solution

Terminal velocity is the maximum velocity attained with constant value. For constant value of velocity, there must be no net force acting on object.

In Air: Terminal velocity is obtained, when the net downward forse ( due to gravity) becomes equal to net upward force ( buoyancy, drag force).

But in vaccum, only acting force is gravitational force. Hence it's a accelerating motion. Thus Velocity is not constant and so there will be no terminal velocity.
Question 4
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In a steady incompressible flow of a liquid.

A
The speed does not change if the area of cross-section changes

B
The speed increases if the area of cross-section increases

C
The speed decreases if the area of cross-section increases

D
Bubbles are produced when the area of the cross-section increases

Solution

In steady incompressible fluid, Velocity of fluid flow does not change with time about any crossection, but it can increase or decrease along the direction of flow according to the continuity equation.
Hence when the Speed of fluid flow decrease than area of crossection would be increased, and vice-versa.
Option C
Question 5
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At $$20^o$$C, to attain the terminal velocity how fast will an aluminium sphere of radii $$1$$ mm fall through water? [Assume flow to be laminar flow and specific gravity $$(Al)=2.7$$, $$\eta_{water}=8\times 10^{-4}$$Pa]

A
$$5$$ m/s

B
$$4.7$$ m/s

C
$$4$$ m/s

D
$$2$$ m/s

Solution
Question 6
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A hole is made at the bottom of a tank filled with water $$\left(density={10}^{3}kg/m^{3}\right)$$. If the total pressure at the bottom of the the tank is $$3atm\left(1atm=10^{5}{N/m}^{2}\right)$$, then the velocity of efflux is

A
$$\sqrt{400}m/s$$

B
$$\sqrt{200}m/s$$

C
$$\sqrt{600}m/s$$

D
$$\sqrt{500}m/s$$

Solution

We know that velocity of the efflux is:
$$v= \sqrt{\dfrac{2\Delta P}{\rho}}$$
$$\Delta P= pressure$$ $$ at$$ $$ the$$ botton$$ - atmospheric$$ $$ pressure$$
So we get,$$v=\sqrt{\dfrac{2\times(3-1) \times 10^{5}}{10^{3}}}$$
Therefore $$v=\sqrt{400}ms^{-1}$$
hence Option A is correct
Question 7
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A tube $$1\ cm^2$$ in cross-section is attached to the top of a vessel $$1\ cm$$ high and cross- section $$100\ cm^2$$. Water fills the system upto a height of $$100\ cm$$ from the bottom of the vessel. The force exerted by the liquid at the bottom of the vessel is :

A
$$1000\ N$$

B
$$990\ N$$

C
$$900\ N$$

D
$$100\ N$$

Solution
Question 8
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A block of wood is floating in water in a closed vessel as shown in the figure. The vessel is connected to an air pump. When more air is pushed into the vessel, the block of wood floats with (neglect compressibility of water)

A
larger part in the water

B
smaller part in the water

C
same part in the water

D
at some instant it will sink

Solution

From Pascal's law, pressure is changing at every point by the same amount. Hence, buoyancy remains the same. So, the part of the block inside water remains the same.
Question 9
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For a fluid which is flowing steadily, the level in the vertical tubes is best represented by:

A

B

C

D

Solution
Question 10
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Directions For Questions

A uniform rod of length $$L$$ pivoted at the bottom of a pond of depth $$L/2$$ stays in stable equilibrium as shown in the figure.

...view full instructions

Find the force acting at the pivoted end of the rod in terms of mass $$m$$ of the rod.

A
$$\left(\sqrt{2}-1\right)mg$$

B
$$\left(\sqrt{2}+1\right)mg$$

C
$$\left(\sqrt{3}+1\right)mg$$

D
$$\left(\sqrt{2}+3\right)mg$$

Solution

Weight of the rod $$w=LA\rho{g}$$
(A=area of cross section, $$\rho$$=density of material)
Submerged length of the rod is given by $$OT=\dfrac{L}{2\sin\theta}$$

Buoyant force $$B=\left(\dfrac{L}{2\sin\theta}\right)A\rho_w{g}$$

from Archimedes' principle ($$\rho_w=$$density of water).

This force acts at $$P$$, where $$OP=\dfrac{OT}{2}=\dfrac{l}{4\sin\theta}$$

Balancing torque about $$O$$ is
$$W\ OQ\ \cos\theta=B\ OP\ \cos\theta$$

$$\left(LA\rho{g}\right){L/2}=\left(\dfrac{L}{2\sin\theta}\right)A\rho_w{g}\left(\dfrac{L}{4\sin\theta}\right)$$

$$\sin^2\theta=\dfrac{\rho_w}{4\rho}$$

$$\Rightarrow\sin^2\theta=1/2\Rightarrow\theta=45°$$

Also, balancing forces $$R+W=B$$
$$R=B-W$$

$$=\dfrac{\rho{A}\rho_w{g}}{2\sin\theta}-LA\rho{g}$$

$$=LA\rho{g}\left[\dfrac{1}{\sin\theta}-1\right]\left(\because\rho_w=2\rho\right)$$

$$=\left(\sqrt{2}-1\right)mg$$