Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A marble of mass $$ x $$ and diameter $$ 2r $$ is gently released a tall cylinder containing honey .If the marble displaces mass $$y(<\, x) of the liquid , then the terminal velocity is proportional to

A
$$ (x+y) $$

B
$$ (x-y) $$

C
$$(\dfrac{x+y}{r})$$

D
$$(\dfrac{x-y}{r})$$

Solution

If $$ v $$ is the terminal velocity , then
$$ xg-yg=6 \pi\,\eta\, r\,v$$
$$v=\dfrac{(x-y)}{r}.\dfrac{g}{6 \pi \eta}$$
$$v\,\alpha \dfrac{x-y}{r}$$
Question 2
1 / -0

The pressure at depth $$h$$ below the surface of a liquid of density $$\rho $$ open to the atmosphere is

A
Greater than the atmospheric pressure by $$\rho gh$$

B
Less than the atmospheric pressure by $$\rho gh$$

C
Equal to the atmospheric pressure

D
Decreases exponentially with depth

E
Increases exponentially with depth

Solution

Pressure at depth $$h$$ to the open surface
$$p={ p }_{ 0 }+h\rho g$$.
i.e., the pressure at depth $$h$$ below the surface of a liquid of density $$\rho $$ open to the atmosphere is greater than the atmosphere pressure by $$\rho gh$$.
Question 3
1 / -0

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference p. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled, is

A
$$\rho$$

B
$$\displaystyle\frac{3\rho}{4}$$

C
$$\displaystyle\frac{\rho}{2}$$

D
$$\displaystyle\frac{\rho}{4}$$

Solution

$$\because$$ Rate of flow, $$Q=\displaystyle\frac{v}{t}=\frac{\pi pr^4}{8\eta l}$$
and rate of flow, $$2Q=\displaystyle\frac{\pi p'(2r)^4}{8\eta (2l)}=\frac{8\pi p'\pi^4}{8\eta l}$$
$$\displaystyle\frac{Q}{2Q}=\frac{p}{8p'}$$
$$\Rightarrow p'=\frac{p}{4}$$.
Question 4
1 / -0

Water flows along a horizontal pipe whose cross.section is not constant. The pressure is 1 cm of Hg, where the velocity is $$35\, cms^{-1}$$. At a point where the velocity is $$65\, cms^{-1}$$, the pressure will be

A
0.89 cm Hg

B
8.9 cm of Hg

C
0.5 cm of Hg

D
1 cm of Kg

Solution

Using Bernoulli's theorem in horizontal pipe,
$$P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 +\dfrac{1}{2}\rho v_2 ^2$$

Here, $$P_1 =\rho_mgh_1= 13600 \times 9.8 \times 10^{-2}$$

$$P_2 = 13600 \times 9.80 \, h$$

$$\rho_w= 1000\, kg/m^3$$

$$v_1 = 35\times 10^{-2}\, m/s$$

$$v_2 = 65\times 10^{-2}\, m/s$$

So, we have
$$ 13600 \times 9.8 \times 10^{-2}+\dfrac{1}{2}\times 1000 \times (0.35)^2$$ $$= 13600 \times 9.8 \times h+\dfrac{1}{2}\times 1000 \times (0.65)^2$$
After solving, we get $$h=0.98 \ cm$$ of Hg.
Question 5
1 / -0

A liquid is filled upto a height of $$20\ m$$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take, $$g = 10\ ms^{-2}$$).

A
$$1.2\ ms^{-1}$$

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

C
$$2\ ms^{-1}$$

D
$$3.2\ ms^{-1}$$

E
$$1.4\ ms^{-1}$$

Solution

Given, $$h = 20\ cm, = 0.2\ m$$
and $$g = 10\ ms^{-2}$$
We know that,
$$v = \sqrt {2gh}$$
Speed of liquid coming out of small hole,
$$\Rightarrow v = \sqrt {2\times 10\times 0.2} = 2\ ms^{-1}$$.
Question 6
1 / -0

A cylindrical vessel of base radius R and height H has a narrow neck of height h and radius r at one end(see figure). The vessel is filled with water(density $$\rho_w$$) and its neck is filled with immiscible oil (density $$\rho_o$$). Then the pressure at :

A
M is $$g(h\rho_o+H\rho_w)$$

B
N is $$g(h\rho_o+H\rho_w)\displaystyle\frac{r^2}{R^2}$$

C
M is $$gH\rho_w$$

D
N is $$g\displaystyle\frac{\rho_wHR^2+\rho_ohr^2}{R^2+r^2}$$

Solution

Pressure at M equals pressure at N (same depth) which is given by:
$$P_M = P_N = \rho_w g H + \rho_o g h$$
Question 7
1 / -0

Equal volumes of two immisible liquids of densities $$\rho$$ and $$2\rho$$ are filled in a vessel as shown in figure. Two small holes are punched at depths $$h/2$$ and $$3h/2$$ from the surface of lighter liquid. If $$v_1$$ and $$v_2$$ are the velocities of efflux at these two holes, then $$v_1/v_2$$ is?

A
$$\displaystyle\frac{1}{2\sqrt{2}}$$

B
$$1/2$$

C
$$1/4$$

D
$$\displaystyle\frac{1}{\sqrt{2}}$$

Solution

$$v_1=\sqrt{2g\left(\displaystyle\frac{h}{2}\right)}=\sqrt{gh}$$ .......(i)
From Bernoulli's theorem
$$\rho gh+2\rho g\left(\displaystyle\frac{h}{2}\right)=\displaystyle \frac{1}{2}(2\rho)v^2_2$$
$$\therefore \displaystyle v_2=\sqrt{2gh}$$ .........(ii)
$$\therefore \displaystyle\frac{v_1}{v_2}=\frac{1}{\sqrt{2}}$$.
Question 8
1 / -0

A tank of height $$5\ m$$ is full of water. There is a hole of cross sectional area $$1\ cm^2$$ in its bottom. The initial volume of water that will come out from this hole per second is

A
$$10^{-3}m^3/s$$

B
$$10^{-4}m^3/s$$

C
$$10m^3/s$$

D
$$10^{-2}m^3/s$$

Solution

Since velocity of water coming out = $$\sqrt { 2gh } $$
= $$\sqrt { 2\times 10\times 5 } $$
= 10 m/s
So volume of water = velocity $$\times $$ area
= $$10\times { 10 }^{ -4 }{ m }^{ 3 }/s$$
volume flow rate = $${ 10 }^{ -3 }m/s$$
Question 9
1 / -0

Two mercury drops(each of radius r) merge to form a bigger drop. The surface energy of the bigger drop if T is the surface tensions

A
$$2^{5/3}\pi r^2T$$

B
$$4\pi r^2T$$

C
$$2\pi r^2T$$

D
$$2^{8/3}\pi r^2T$$

Solution

Since final volume = 2 initial volume
$${ R }^{ 3 }=2{ r }^{ 3 }$$
$$R={ 2 }^{ 1/3 }r$$
So final surface area = $$4\pi { R }^{ 2 }$$
= $$4\pi { 2 }^{ 2/3 }{ r }^{ 2 }$$
Area = $${ 2 }^{ 8/3 }\pi { r }^{ 2 }$$
So surface energy = T $$\times $$ area
= $${ 2 }^{ 8/3 }\pi { r }^{ 2 }T$$
Question 10
1 / -0

In football sport a banana kick is an off-centre kick that makes the ball curve or suddenly change direction mid-air.The player achieves this by imparting revolution on the ball while kicking .The physical principle that best describes the banana kick is

A
Hooke's law

B
Bernoulli's law

C
Torricelli's law

D
Newton law of motion

Solution

A banana kick is caused by putting a spin on the ball while kicking in one specific direction. Due to the "Bernoulli effect", the ball will change direction in mid-air. This is because there is a difference in velocity on different portions of the ball. This difference causes there to be a force moving the ball sideways.

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Re-Attempt Weekly Quiz Competition

Mechanical Properties of Fluids Test - 41

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Mechanical Properties of Fluids Test - 41

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SHARING IS CARING

Question 1

$$ (x+y) $$

$$ (x-y) $$

$$(\dfrac{x+y}{r})$$

$$(\dfrac{x-y}{r})$$

Solution

Question 2

Greater than the atmospheric pressure by $$\rho gh$$

Less than the atmospheric pressure by $$\rho gh$$

Equal to the atmospheric pressure

Decreases exponentially with depth

Increases exponentially with depth

Solution

Question 3

$$\rho$$

$$\displaystyle\frac{3\rho}{4}$$

$$\displaystyle\frac{\rho}{2}$$

$$\displaystyle\frac{\rho}{4}$$

Solution

Question 4

0.89 cm Hg

8.9 cm of Hg

0.5 cm of Hg

1 cm of Kg

Solution

Question 5

$$1.2\ ms^{-1}$$

$$1\ ms^{-1}$$

$$2\ ms^{-1}$$

$$3.2\ ms^{-1}$$

$$1.4\ ms^{-1}$$

Solution

Question 6

M is $$g(h\rho_o+H\rho_w)$$

N is $$g(h\rho_o+H\rho_w)\displaystyle\frac{r^2}{R^2}$$

M is $$gH\rho_w$$

N is $$g\displaystyle\frac{\rho_wHR^2+\rho_ohr^2}{R^2+r^2}$$

Solution

Question 7

$$\displaystyle\frac{1}{2\sqrt{2}}$$

$$1/2$$

$$1/4$$

$$\displaystyle\frac{1}{\sqrt{2}}$$

Solution

Question 8

$$10^{-3}m^3/s$$

$$10^{-4}m^3/s$$

$$10m^3/s$$

$$10^{-2}m^3/s$$

Solution

Question 9

$$2^{5/3}\pi r^2T$$

$$4\pi r^2T$$

$$2\pi r^2T$$

$$2^{8/3}\pi r^2T$$

Solution

Question 10

Hooke's law

Bernoulli's law

Torricelli's law

Newton law of motion

Solution

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