Mechanical Properties of Fluids Test

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Mechanical Properties of Fluids Test - 39

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Question 1
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A small metal ball of mass '$$m$$' is dropped in a liquid contained in a vessel, attains a terminal velocity '$$V$$'. If a metal ball of same material but of mass $$'8m'$$ is dropped in same liquid then the terminal velocity will be.

A
$$V$$

B
$$2V$$

C
$$4V$$

D
$$8V$$

Solution

Terminal velocity of the ball $$V = \dfrac{2 gr^2(\rho - \sigma)}{9\eta}$$ ........(1)
where $$\rho$$ and $$\sigma$$ are the density of ball and liquid, respectively.
Mass of the metal ball $$m = \rho (\dfrac{4\pi}{3} r^3)$$
$$\implies$$ $$r \propto m^{1/3}$$
But from equation (1), we get $$V \propto r^2$$
$$\implies$$ $$V\propto m^{2/3}$$ ......(2)
Given : $$V_1 = V$$ $$m_1 = m$$ $$m_2 =8m$$
From (2) we get $$\dfrac{V_2}{V_1} = \bigg(\dfrac{m_2}{m_1}\bigg)^{2/3}$$
Or $$\dfrac{V_2}{V} = \bigg(\dfrac{8m}{m}\bigg)^{2/3} = 8^{2/3}$$
$$\implies$$ $$V_2= (2^3)^{2/3} V = 4V$$
Question 2
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For tap water and clean glass, the angle of contact is

A
$$0^o$$

B
$$90^o$$

C
$$140^o$$

D
$$8^o$$

Solution

The angle of contact between pure water and glass is found to be $${ 0 }^{ 0 }$$ as both cohesive forces and adhesive forces are equal. But due to presence of ions in tap water the angle of contact is acute from $${ 8 }^{ 0 }$$ to $${ 18 }^{ 0 }$$.
Question 3
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Water containing air bubbles flows without turbulence through a horizontal pipe which has a region of narrow cross-section. In this region the bubbles.

A
Move with greater speed and are smaller than in the rest of the pipe

B
Move with greater speed and are larger in size than in the rest of the pipe

C
Move with lesser speed and are smaller than in the rest of the pipe

D
Move with lesser speed and are of the same size as in the rest of the pipe

Solution

Since the water flow is assumed to be laminar, we can apply Bernoulli's equation between two points in the flow.

Bernoulli's equation is given by $$P+\frac{1}{2}\rho v^2 = \textrm{constant}$$

The continuity equation is given by $$Av = \textrm{constant}$$

At the narrow section, we have $$A$$ to be minimum. Thus, $$v$$ is maximum.
So, the bubbles travel faster at the narrow section.

Since, $$v$$ is maximum, by Bernoulli's equation, we have $$\frac{1}{2} \rho v^2$$ also maximum. Thus, $$P$$ is minimum.

If $$P$$ is minimum, we have the size of the bubbles as maximum.

Hence, at the minimum cross-section, the bubbles are larger in size and move with greater speed than rest of the pipe.
Question 4
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A particle released from rest is falling through a thick fluid under gravity. The fluid exerts a resistive force on the particle proportional to the square of its speed. Which one of the following graphs best depicts the variation of its speed $$v$$ with time $$t$$?

A

B

C

D

Solution

Let the effective $$g=g'$$
Now we know $$mg'-\alpha v^2=m\dfrac{dv}{dt}$$
Initially velocity will rise and later $$mg'=\alpha v^2$$ then velocity will become constant .
Question 5
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An open tank filled with water density $$\rho$$ has a narrow hole at a depth of $$h$$ below, then the velocity of water flowing out is:

A
$$h\rho g$$

B
$$2gh$$

C
$$\sqrt{2gh}$$

D
$$gh$$

Solution

From Bernoulli's equation, we can write:
$$\rho g h=\dfrac{1}{2}\rho v^2$$
So, $$v=\sqrt{2gh}$$
Question 6
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There are two identical small holes of area of cross section a on the either sides of a tank containing a liquid of density p (shown in figure). The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is :

A
$$\dfrac{2gh}{pa}$$

B
$$\dfrac{pgh}{a}$$

C
$$ghpa$$

D
$$2pagh$$

Solution

Net force (reaction)
$$F=F_B-F_A=\dfrac{dp_B}{dt}-\dfrac{dp_A}{dt}$$
$$=av_B p\times v_B-av_A p\times v_A$$
$$\therefore F = ap(v^2_B-v^2_A)$$
According to Bernaulli's theorem
$$P_A+\dfrac{1}{2}pv^2_A+pgh = P_B+\dfrac{1}{2}pv^2_B+0$$
$$\Rightarrow \dfrac{1}{2}p(v^2_B-v^2_A) = pgh$$
$$\Rightarrow v^2_B-v_A^2=2gh$$
From Eqs (i) and (ii), we get
$$F=ap(2gh)=2apgh$$
Question 7
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A rectangular vessel when full of water, takes $$10\ min$$ to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water?

A
$$9\ min$$

B
$$7\ min$$

C
$$5\ min$$

D
$$3\ min$$

Solution

If $$A_{0}$$ is the area of the orifice at the bottom below the free surface and $$A$$ that of vessel, time $$t$$ taken to be empited the tank,
$$t = \dfrac {A}{A_{0}}\sqrt {\dfrac {2H}{g}}$$
$$\therefore \dfrac {t_{1}}{t_{2}} = \sqrt {\dfrac {H_{1}}{H_{2}}}$$
$$\Rightarrow \dfrac {t}{t_{2}} = \sqrt {\dfrac {H_{1}}{H_{1}/2}}$$
$$\Rightarrow \dfrac {t}{t_{2}} = \sqrt {2}$$
$$\therefore t_{2} = \dfrac {t}{\sqrt {2}}$$
$$= \dfrac {10}{\sqrt {2}} = 5\sqrt {2}$$
$$\approx 7\ min$$
Question 8
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A uniform capillary tube of length $$l$$ and inner radius $$r$$ with its upper end sealed is submerged vertically into water. The outside pressure is $${p}_{0}$$ and surface tension of water is $$\gamma$$. When a length $$x$$ of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide, the value of $$x$$ is

A
$$\cfrac { l }{ \left( l+\cfrac { { p }_{ 0 }r }{ 4\gamma } \right) } $$

B
$$l\left( l-\cfrac { { p }_{ 0 }r }{ 4\gamma } \right) $$

C
$$l\left( l-\cfrac { { p }_{ 0 }r }{ 2\gamma } \right) $$

D
$$\cfrac { l }{ \left( l+\cfrac { { p }_{ 0 }r }{ 2\gamma } \right) } $$

Solution

For air inside capillary, $${ p }_{ 0 }(lA)=p'(l-x)A$$ where $$p'$$ is pressure in capillary after being submerged
$$\therefore p'=\cfrac { { p }_{ 0 }l }{ l-x } $$
Now since level of water inside capillary coincides with outside $$p'-{ p }_{ 0 }=\cfrac { 2\gamma }{ r } $$
$$\therefore \cfrac { { p }_{ 0 }l }{ l-x } -{ p }_{ 0 }=\cfrac { 2\gamma }{ r } \Rightarrow x=\cfrac { l }{ \left( l+\cfrac { { p }_{ 0 }r }{ 2\gamma } \right) } $$
Question 9
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Find the difference of air pressure between the inside and outside of a soap bubble $$5 mm$$ in diameter, if the surface tension is $$1.6 N/m$$.

A
$$2560 N/ m^2$$

B
$$3720 N /m^2$$

C
$$1208 N /m^2$$

D
$$950 N / m^2$$

Solution

$$\rho = \dfrac{4T}{R} = \dfrac{4\times 1.6}{2.5\times 10^{-3}}$$
$$=2560 N/m^2$$
Question 10
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The material of a wire has a density of 1.4 g/cm$$^3$$. If it is not wetted by a liquid of surface tension 44 dyne/cm, then the maximum radius of the wire which can float on the surface of liquid is :

A
$$\dfrac{10}{28} cm$$

B
$$\dfrac{10}{14} cm$$

C
$$\dfrac{10}{7} mm$$

D
0.7 cm

Solution

$$2 Tl = \pi r^2 Id \times g$$
$$r = \sqrt{\left(\dfrac{2T}{\pi dg} \right )} = \sqrt{\dfrac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$$
$$= \dfrac{1}{7} cm = \dfrac{10}{7} mm$$