Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 36

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Question 1
1 / -0

The velocity of the water flowing from the inlet pipe is less than the velocity of water flowing out from the spin pipe B.

A
variation of water level in vessel will be irregular.

B
water level will remains constant.

C
the water level will perform periodic oscillation motions.

D
none of the above.

Solution

When the water level increases in the vessel, then pressure will also increase, which will lead to increase in flow rate at pipe B and when again level of water decreases, velocity will also decrease. This will follow a periodic oscillatory motion.
Question 2
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Two spheres of the same material, but of radii $$R$$ and $$3R$$ are allowed to fall vertically downwards through a liquid of density $$\rho$$. The ratio of their terminal velocities is:

A
$$1:3$$

B
$$1:6$$

C
$$1:9$$

D
$$1:1$$

Solution

Terminal velocity of the sphere $$v = \dfrac{2 gr^2 (\sigma - \rho)}{9 \eta}$$
where $$\eta$$ is the viscosity of the liquid, $$\sigma$$ is the density of material of sphere, $$\rho$$ is the density of liquid and $$r$$ is the radius of the sphere.
$$\implies$$ $$v \propto r^2$$
Given : $$r_1 = R$$ and $$r_2 = 3R$$
Ratio of terminal velocities $$v_1 : v_2 = r_1^2 : r_2^2$$
Or $$v_1 : v_2 = R^2 : (3R)^2$$
$$\implies$$ $$v_1:v_2 = 1:9$$
Question 3
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A tank is filled with water to a height $$H$$. Two holes are made on its side wall, one at a height of $$h$$ from the bottom and other at a depth $$h$$ from the top. The horizontal jets starting from the two holes meet the ground or side (in level with the bottom of the tank) at the same point. This distance of this point from the side of the tank is

A

B

C

D

Solution

Range of water jet coming out of a hole in a tank if given as $$\boxed { Range=2\sqrt { h(H-h) } } $$
Here H is height of liquid & h is height of hole from bottom.
Question 4
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In a streamline flow of a liquid

A
Every particle has its own velocity, different from others.

B
All particles move with a constant velocity, even if the path is curvilinear.

C
At a point on the streamline, particle can have two velocities.

D
At a point on the streamline, particle can have only one velocity along the tangent.

Solution

In a stream line flow, all the particles have a common velocity along the tangent to the streamline drawn at that point.
Question 5
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A square hole of side length $$l$$ is made at a depth of $$y$$ and a circular hole is made at a depth of $$4y$$ from the surface of water in a water tank kept on a horizontal surface. If equal amount of water comes out of the vessel through the holes per second then the radius of the circular hole is equal to $$(r, l << y)$$ :

A
$$ l / \sqrt2$$

B
$$l / 2$$

C
$$l / \sqrt {2 \pi}$$

D
$$ \sqrt2\pi$$

Solution

Given  $${ A }_{ 1 }{ v }_{ 1 }={ A }_{ 2 }{ v }_{ 2 }$$
here  $${ A }_{ 1 }={ l }^{ 2 }\quad \quad \& \quad { A }_{ 2 }=\Pi { r }^{ 2 }$$
$${ v }_{ 1 }=\sqrt { 2gh } \quad \quad { v }_{ 2 }=\sqrt { 2g4h } $$
So  $$\sqrt { 2gh } \times { l }^{ 2 }=2\sqrt { 2gh } \times \Pi { r }^{ 2 }$$
So $$\boxed { r=\dfrac { l }{ \sqrt { 2\Pi } } } $$
Question 6
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Three capillaries of lengths $$L$$, $$\dfrac{L}{2}$$ and $$\dfrac{L}{3}$$ are connected in series. Their radii are $$r$$, $$\dfrac{r}{2}$$ and $$\dfrac{r}{3}$$ respectively. Then if stream line flow is to be maintained and the pressure across the first capillary is $$P$$, then

A
the pressure difference across the ends of the second capillary is $$8 P$$

B
the pressure difference across the third capillary is $$43 P$$

C
the pressure difference across the ends of the second capillary is $$16 P$$

D
the pressure difference across the ends of the second capillary is $$59 P$$

Solution
Question 7
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A water tank placed on the floor has two small holes, pinched in the vertical wall, one above the other. The holes are $$3.3 cm$$ and $$4.7 cm$$ above the floor. If the jets of water issuing out from the holes hit the floor at the same point on the floor, then the height of water in the tank is

A
$$3 cm$$

B
$$6 cm$$

C
$$8 cm$$

D
$$9 cm$$

Solution

Since Range = $$2\sqrt { (H-h)h } $$
Given $${ h }_{ 1 }=3.3\quad \& \quad { h }_{ 2 }=4.4$$
and $${ R }_{ 1 }={ R }_{ 2 }$$
$$2\sqrt { (H-{ h }_{ 1 }){ h }_{ 1 } } =2\sqrt { (H-{ h }_{ 2 }){ h }_{ 2 } } $$
$$\left( H-{ h }_{ 1 } \right) { h }_{ 1 }=(H-{ h }_{ 2 }){ h }_{ 2 }$$
$$(H-3.3)\dfrac { 3.3 }{ 4.7 } =H-4.7$$
0.702H - 2.317 = H - 4.7
0.297H = 2.3829
$$\boxed { H=8cm } $$
Question 8
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Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are $$v_1$$ and $$v_2$$ respectively. Thus

A
$$v_1 = v_2$$

B
$$v_1 = 13.6 v_2$$

C
$$v_1 = \dfrac{v_2}{13.6}$$

D
$$v_1 = \sqrt {13.6} \ v_2$$

Solution

Since velocity of liquid coming out of tank is irrespective of density of material & given as velocity = $$\sqrt { 2gh } $$. Hence both will come out at same velocity.
Question 9
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A tube of length L and radius $$R$$ is joined to another tube of length $$\dfrac{L}{3}$$ and radius $$\dfrac{R}{2}$$ . A fluid is flowing through this tube. If the pressure difference across the first tube is $$P$$, then the pressure difference across the second tube is

A
$$\dfrac{16P}{3}$$

B
$$\dfrac{4P}{3}$$

C
$$P$$

D
$$\dfrac{3P}{16}$$

Solution

Since we know that flow rate remains constant, then
flow rate in 1 = flow rate in 2.
$$\dfrac { \pi { P }_{ 1 }{ r }_{ 1 }^{ 4 } }{ 8g{ l }_{ 1 } } =\dfrac { \pi { P }_{ 1 }{ r }_{ 2 }^{ 4 } }{ 8g{ l }_{ 2 } } $$
$$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { l }_{ 1 } }{ { l }_{ 2 } } \times { \left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }^{ 4 }\quad \quad \quad \left[ now\quad since\ { r }_{ 2 }=\dfrac { { r }_{ 1 } }{ 2 } \ \& \quad { l }_{ 2 }=\dfrac { { L }_{ 1 } }{ 3 } \right] $$
So,
$$\dfrac { P }{ { P }_{ 2 } } =\dfrac { 3{ L }_{ 1 } }{ { L }_{ 1 } } ={ \left( \dfrac { { r }_{ 1 } }{ { 2r }_{ 1 } } \right) }^{ 4 }=\dfrac { 3 }{ 16 } $$
$$\boxed { { P }_{ 2 }=\dfrac { 16 }{ 3 } P } $$
Question 10
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For a liquid, which is rising in a capillary tube, the angle of contact is

A
$$90^o$$

B
$$180^o$$

C
Acute

D
Obtuse

Solution

If a liquid is rising in a capillary then adhesive forces are greater than cohesive forces. Hence the angle of contact is acute.