Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 18

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Question 1
1 / -0

Water is maintained at a height of 10m in a tank.The diameter of a circular aperture needed at the base of the tank to discharge water at the rate of 26.4m$$^{3}$$ min$$^{-1}$$ is(Given that g $$=$$ 9.8 m s$$^{-2})$$:

A
0.2 cm

B
2 m

C
0.2 m

D
2 cm

Solution

Velocity of water from the bottom operture $$=\sqrt{2gh}$$
$$=\sqrt{2 \times 9.8 \times 10}m/s$$
So,

$$Q=26.4 m^{3}=\pi r^{2}V$$
$$\Rightarrow 26.4 m^{3} = \pi \times r^{2} v$$
$$\Rightarrow \dfrac{26.4}{60}m^{3}/s = \pi r^{2} \times \sqrt{2 \times 9.8 \times 10}$$
$$\Rightarrow r^{2}=0.010004024$$
$$\Rightarrow r =0.1 m$$

So, Diameter of aperture $$=0.2\ m$$
Question 2
1 / -0

An aeroplane of mass $$6000 \ kg$$ is flying at an altitude of $$3 \ km$$. If the area of the wings is $$30 \ m^{2}$$ and the pressure at the lower surface of wings is $$0.6\times10^{5}\ pa$$, the pressure on the upper surface of wings is : ( in pascal) ( $$g= 10 \ ms^{-2}$$)

A
$$5.8 \times 10^{4}$$

B
$$6.8 \times 10^{4}$$

C
$$7.8 \times 10^{4}$$

D
$$8.8 \times 10^{4}$$

Solution

Pressure at lower surface of wing$$=0.6\times 10^{5}Pa$$
Let pressure at upper surface of wing$$=p$$
Weight of the airplane $$=5\times 10^{4}N$$
For equilibrium, Weight $$=$$upthrust due to air
$$\Rightarrow 5\times 10^{4}=\left ( 0.6\times 10^{5}-p \right )\times 50$$
$$\Rightarrow 10^{3}=0.6\times 10^{5}-p$$
$$\Rightarrow p=0.6\times 10^{5}-10^{3}$$
$$\Rightarrow p=59\times 10^{3}Pa$$
Question 3
1 / -0

At what speed will the velocity head of stream of water be equal to 40cm?

A
280 ms$$^{-1}$$

B
2.80 cm/s

C
280 cms$$^{-1}$$

D
0.280 m/s

Solution

Velocity head $$=\dfrac{V^{2}}{2g}=40 cm$$
$$\Rightarrow V^{2}=40 \times 980 \times 2$$
$$\Rightarrow V=280 cm/s$$
Question 4
1 / -0

The pressure that will be built up by a compressor in a paint-gun when a stream of liquid paint flows out with a velocity of $$25 ms^{-1}$$ ( density of paint is $$0.8 gm -cm^{-3}$$) is : ( in $$Nm^{-2}$$)

A
$$2.5 \times 10^{2}$$

B
$$2.5 \times 10^{3}$$

C
$$2.5 \times 10^{5}$$

D
$$5 \times 10^{5}$$

Solution

Applying Bernoulli's Theorem just inside and just outside the hole from which the point is flowing,
$$p_{1}+0=\dfrac{1}{2}\rho v^{2}$$
$$\Rightarrow p_{1}=\dfrac{1}{2}\times 800\times \left ( 25 \right )^{2}$$
$$p_{1}=2.5\times 10^{5}N/m^{2}$$
Question 5
1 / -0

A plane is in a level flight at a constant speed and each of its two wings has an area of $$25 \ m^{2}$$. If the speed of air is $$180 \ km h^{-1}$$ over the lower wing and $$234 \ km h^{-1}$$ over the upper wing surface, the plane's mass is : (Take density of air $$=1 kg m^{-3}$$.)

A
$$4.400 gm$$

B
$$4400 gm$$

C
$$44.0 kg$$

D
$$4400 kg$$

Solution

Applying bernouillis theorem above and below the wing,

$$p_{1}+\dfrac{1}{2}f v_{1}^{2}=p_{2}+\dfrac{1}{2} f v_{2}^{2}$$
$$\Rightarrow p_{1}-p_{2}=\dfrac{1}{2}\left ( 65^{2} -50^{2}\right )$$
$$\Rightarrow p_{1}-p_{2}=862.5pa$$

So, upthrust on wings=weight of plane

$$\Rightarrow \dfrac{862.5\times 2\times 25}{10}=$$mass of plane
$$\Rightarrow Mass=4312kg\approx 4400kg$$
Question 6
1 / -0

A tank containing water has an orifice on one vertical wall. If the centre of the orifice is 4.9 m below the surface of water in the tank, the velocity of discharge is :

A
0.98 m/s

B
9.8 m/s

C
98 m/s

D
9.8 cm/s

Solution

Velocity of discharge at the orifice = $$\sqrt{2gh}$$ [Toricelli's Theorem]
$$=\sqrt{2 \times 9.8 \times 4.9}$$
$$=9.8 m/s$$
Question 7
1 / -0

The level of water in a tank is 5m high. A hole of area of cross section 1 cm$$^{2}$$ is made at the bottom of the tank. The rate of leakage of water from the hole in $$m^{3}s^{-1}$$ is ( g $$=$$ 10ms$$^{-2}$$):

A
$$10^{-3}$$

B
$$10^{-4}$$

C
$$10$$

D
$$10^{-2}$$

Solution

Velocity of water from the hole= $$\sqrt{2 \times 5 \times 10}=10 m/s$$
So, $$Q=AV=1 \times 10^{-4} m^{2} \times 10m/s$$
$$= 10^{-3} m^{3}/s$$
Question 8
1 / -0

A cylindrical vessel contains a liquid of density $$\rho $$ upto a height $$h$$. The liquid is closed by a piston of mass $$m$$ and area of cross section $$A$$.There is a small hole at the bottom of the vessel.The speed with which the liquid comes out of the vessel is:

A
$$\sqrt{2gh}$$

B
$$\sqrt{2\left ( gh+\dfrac{mg}{\rho A} \right )}$$

C
$$\sqrt{2\left ( gh+\dfrac{mg}{A} \right )}$$

D
$$\sqrt{2gh+\dfrac{mg}{A}}$$

Solution

Applying Bernoulli theorem across 1 and 2,
$$\dfrac{2mg}{A\rho }+2\rho gh=\dfrac{1}{2}\rho v^{2}$$
$$v_{2}=\dfrac{2mg}{A\rho }+2gh$$
$$v^{2}=2g\left ( h+\dfrac{m}{\rho A} \right )$$
$$v = \sqrt{2\left ( gh+\dfrac{mg}{\rho A} \right )}$$
Question 9
1 / -0

Two hail stones with radii in the ratio of $$1:2$$ fall from a great height through the atmosphere. Then their terminal velocities are in the ratio of:

A
$$1:2$$

B
$$2:1$$

C
$$1:4$$

D
$$4:1$$

Solution

$$\textbf{Hint}$$: Use the formula of terminal velocity
$$\textbf{Step 1}$$:
We know
Terminal velocity $$V=\dfrac{2r^2(\rho-\rho_0)g}{9 \eta}$$
Here we can see that $$V \propto r^2$$
$$\textbf{Step 2}$$
We can write $$\dfrac{V_1}{V_2}=\dfrac{r_1^2}{r_2^2}$$
So, ratio of their terminal velocities of the given two hail stones is $$=\dfrac{r^2}{(2r)^2}$$
$$=\dfrac{r^2}{4r^2}=\dfrac{1}{4}$$
So, the ratio is $$1:4$$
Thus option C is correct.
Question 10
1 / -0

In a horizontal oil pipe line of constant cross sectional area the decrease of pressure between two points 100 km apart is 1500 pa. The loss of energy per unit volume per unit distance is ----Joule.

A
15

B
0.015

C
0.03

D
zero

Solution

Decreases in pressure$$10^{5}m$$\ apart$$=1500pa$$
energy cost per unit volume per unit distance$$=\dfrac{1500\times 10^{5}A}{10^{5}\times A\times 10^{5}}=\dfrac{work done}{lenght\times total volume}$$
$$=0.015\ joules$$

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Mechanical Properties of Fluids Test - 18

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Mechanical Properties of Fluids Test - 18

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Question 1

0.2 cm

2 m

0.2 m

2 cm

Solution

Question 2

$$5.8 \times 10^{4}$$

$$6.8 \times 10^{4}$$

$$7.8 \times 10^{4}$$

$$8.8 \times 10^{4}$$

Solution

Question 3

280 ms$$^{-1}$$

2.80 cm/s

280 cms$$^{-1}$$

0.280 m/s

Solution

Question 4

$$2.5 \times 10^{2}$$

$$2.5 \times 10^{3}$$

$$2.5 \times 10^{5}$$

$$5 \times 10^{5}$$

Solution

Question 5

$$4.400 gm$$

$$4400 gm$$

$$44.0 kg$$

$$4400 kg$$

Solution

Question 6

0.98 m/s

9.8 m/s

98 m/s

9.8 cm/s

Solution

Question 7

$$10^{-3}$$

$$10^{-4}$$

$$10$$

$$10^{-2}$$

Solution

Question 8

$$\sqrt{2gh}$$

$$\sqrt{2\left ( gh+\dfrac{mg}{\rho A} \right )}$$

$$\sqrt{2\left ( gh+\dfrac{mg}{A} \right )}$$

$$\sqrt{2gh+\dfrac{mg}{A}}$$

Solution

Question 9

$$1:2$$

$$2:1$$

$$1:4$$

$$4:1$$

Solution

Question 10

15

0.015

0.03

zero

Solution

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