Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 21

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Question 1
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A water tank standing on the floor has two small holes vertically one above the other punched on one side. The holes are $$h_{1}$$ cm and $$h_{2}$$ cm above the floor. How high does water stand in the tank, when the jets from the holes hit the floor at the same point?

A
$$(h_{2}-h_{1})$$

B
$$(h_{2}+h_{1})$$

C
$$(h^{2}_{2}-h^{2}_{1})$$

D
$$\dfrac{h_{2}}{h_{2}-h_{1}}$$

Solution

Time taken by water at (1) to reach the ground = $$\sqrt{\dfrac{2h_{2}}{g}}$$
velocity of water at (1) = $$\sqrt{2gH}$$
So,
$$d= \dfrac{\sqrt{2gH \times 2 P_{n}}}{g}$$-------------(1)
Similarly for point ,
Time taken = $$\sqrt{\dfrac{2h_{1}}{g}}$$
velocity = $$\sqrt{2g (H +h_{2}-h_{1})}$$
So, $$d= \sqrt{2g(H+h_{2}-h_{1}) \times \dfrac{2h_{1}}{g}}$$----------(2)
Equation (1) and (2),
$$\Rightarrow (H+h_{2}-h_{1})h_{1}=Hh_{2}$$
$$\Rightarrow Hh_{1}+h_{2}h_{1}-h_{1}^{2}=Hh_{2}$$
$$\Rightarrow H(h_{1}-h_{2})=h_{1}(h_{1}-h_{2})$$
So,$$\Rightarrow H=h_{1}$$
Height of water = $$h_{1}+h_{2}$$
Question 2
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A ball of mass m and radius r is released in the viscous liquid. The value of its terminal velocity is proportional to :

A
$$(\dfrac{1}{ r}) $$only

B
$$\dfrac{m}{ r}$$

C
$$(\dfrac{m}{ r})^{1/2}$$

D
$$m$$ only.

Solution
Question 3
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The cross-sectional area of a large tank is 0.5m$$^{2}$$. It has an opening near the bottom having cross-sectional area 1 cm$$^{2}$$. A load of 20 kg is applied on the water at the top. The velocity of water coming out of the opening, at the time when the height of water level is 50 cm above the bottom,is nearly (Take g $$=$$ 10 ms$$^{-2}$$)

A
$$1.3 ms^{-1}$$

B
$$3.3 ms^{-1}$$

C
$$5.3 ms^{-1}$$

D
$$7.3 ms^{-1}$$

Solution

Applying Bernoullis therom across point (1) and (2),
$$P_{0}+\dfrac{20 \times 10}{0.5}+ 1000 \times 10\times 0.5=P_{0}+\dfrac{1}{2}\times 1000 \times V^{2}$$
[$$A_{1}V_{1}=a_{1}v_{1}$$ from continuity equation]
$$\Rightarrow 400 + 5000 = 500 \times V^{2}$$
$$\Rightarrow \frac{5400}{500}= v^{2}\Rightarrow V^{2}= 10.8 m/s$$
$$\Rightarrow V \approx 3.3 m/s$$
Question 4
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There are two holes, each of cross-sectional area $$a$$, on the opposite sides of a wide rectangular tank containing a liquid of density $$\rho $$ . When the liquid flows out of the holes, the net force on the tank is :
( $$h$$ is the vertical distance between the two holes )

A
$$2a \rho gh$$

B
$$4a\rho gh$$

C
$$0.5a\rho gh$$

D
$$\rho gh$$

Solution

For 1st hole,
Velocity of water from P = $$\sqrt{2gH}=V_{1}$$
Q = AV
Force exerted by water on the tank = $$\rho aV_{1}^{2}$$
$$=2\rho agH$$-----------------(1)
Similarly for 2nd hole,
Velocity= $$\sqrt{2g(H+h)}=V_{2}$$
Q = $$aV_{2}^2$$
From exerted = $$2\rho ag(H+h)$$---------------(2)
Net force on the tank
$$=2\rho ag(H+h)-2\rho ag(H)$$
$$=2a\rho gh$$
Question 5
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A non-viscous liquid of constant density $$500 kg/m^{3}$$ flows in a variable cross-sectional tube. The area of cross-section of the tube at two points P and Q at heights of $$3 m $$ and $$6 m $$ are $$2\times 10^{-3} m^{3}$$ and $$4\times 10^{-3}m^{3}$$ respectively. The work done per unit volume by the forces of gravity as the fluid flows from point P to Q, is:

A
$$29.4 J/m^{3}$$

B
$$-1.47 \times 10^{4} J/m^{3}$$

C
$$-2.94 \times 10^{4} J/m^{3}$$

D
None of these

Solution

Difference in heigh of P and Q,$$=(6-3)=3m$$
work done by gravity per unit volume of the fluid
$$work\ done=\rho gh$$
$$=-3\times g\times 500$$
$$=-14700J/m^{3}$$ = $$-1.47\times 10^{4}J/^{3}$$
Question 6
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The cross-sectional area of a large tank is $$0.5m^{2}$$. It has an opening near the bottom having cross-sectional area $$1 cm^{2}$$. A load of $$20 kg$$ is applied on the water at the top. The velocity of water coming out of the opening, at the time when the height of water level is $$50 cm$$ above the bottom is nearly:(Take $$g = 10 ms^{-2}$$)

A
$$1.15$$

B
$$3.15$$

C
$$5.15$$

D
$$715$$

Solution

Applying Bernoullis theorem across the two points.
$$\displaystyle \dfrac{2000}{0.5}+1000\times 10\times \dfrac{50}{100}=\dfrac{1}{2}\times 1000\times v^{2}$$

$$\Rightarrow 400+5000=500v^{2}$$

$$\Rightarrow \dfrac{54}{5}=v^{2}\Rightarrow v=3.286m/s$$

$$\Rightarrow v\approx 3.3m/s$$ .....[ Neglecting the velocity of water at the top of beaker ]
Question 7
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A brick is kept in three different ways on a table as shown in Figure. The pressure exerted by the brick on the table will be

A
maximum in position A

B
maximum in position B

C
maximum in position C

D
equal in all cases.

Solution

$$ P = \frac{F}{A}$$
$$F= mg$$ is same in all the cases
P will maximum when Area (A) is minimum.
Question 8
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In the arrangement shown above, a block of mass 2700 kg is in equilibrium on applying a force F. The value of force F ? ($$d_{liquid}=0.75g cm^{-3} is$$)

A
$$147 N$$

B
$$1755.18 N$$

C
$$153 N$$

D
$$918 N$$

Solution

Total pressure acting on side of
$$A_1=\dfrac {F}{A_1}+h\rho g$$
$$=\dfrac {F}{6\times 10^{-4}}+(2\times 0.75\times 9.8\times 10^3)pa$$
pressure acting on side$$A_2=\dfrac {2700\times 9.8}{90\times 10^{-4}}pa$$

According to Pascal's law
Pressure acting on side of $$A_1=$$ pressure acting on side $$A_2$$
$$\dfrac {F}{6\times 10^{-4}}+(2\times 0.75\times 9.8\times 10^3)$$$$=\dfrac {2700\times 9.8}{90\times 10^{-4}}$$

$$\dfrac {F}{6\times 10^{-4}}=294\times 10^4-14.7\times 10^3$$

$$F=6\times 10^{-4}(292.53\times 10^4)$$
$$F=1755.18 N$$
Question 9
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A diver releases bubbles of gas from the bottom of a lake. The bubbles increase to 10 times of their original volume when they reach the surface. Assuming that the pressure exerted by a column of water of 5m height is double the atmospheric pressure, the depth of the lake is...

A
45m

B
90m

C
80m

D
22.5m
Question 10
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The force exerted by water on the base of a tank, of base area $$1.5 m^2$$ when filled with water up to a height of 1 m is (Density of water is $$1000 kg m^{-3}$$ and $$g=10 ms^{-2})$$ :

A
1500 N

B
15000 N

C
3000 N

D
30000 N

Solution

$$P = \rho g h \\ P = 10^{3} * 10 * 1 \\ P = 10^{4} N/m^2$$
$$ F = PA \\ F = 10^{4} * 1.5 \\ F = 1.5 * 10^{4} = 15000N$$