Fluid Mechanics Multiple Choice Question And Answers

Fluid Mechanics

Question 1. Two nonmixing liquids of densities p and np (where n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats keeping its axis vertical so that a length of pL (where p < 1) is in the denser liquid. The density d is equal to

1. [2 + (n + 1)p]p
2. [2 + (n- 1)p]p
3. [1 + (n-1)p]p
4. [1 + (n + 1)p]p

Answer: 3. [1 + (n-1)p]p

Let A be the cross-sectional area of the cylindrical block. weight of the cylinder = ALdg (downward),

buoyant force due to the lower part of the liquid = ApLnpg (upward),

and that due to the upper part of the liquid =A(1 -p)Lpg (upward).

For equilibrium, when the cylinder floats,

ALdg= ApLnpg + A(1 -p)Lpg

=> d = pnp + (1-p)p = [1 + (n-1)p]p.

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Question 2. A candle of diameter d is floating in a liquid kept in a cylindrical container of diameter D (where D = d), as shown in the figure. It is burning at the rate of 2 cm h^-1. Then, the top of the candle will

1. Remain at the same height
2. Fall at the rate of 1 cm h^-1
3. Fall at the rate of 2 cm h^-1
4. Go up at the rate of 1 cm h^-1

Answer: 2. Fall at the rate of 1 cm h^-1

Let p_c and p_liq be the densities of the candle and liquid respectively. For the candle to float initially, weigh to f the candle- buoyant force

A.2L.p_cg = ALp_liqg

p_liq = 2p_c….(i)

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Thus, the candle’s length is equally distributed below and above foe liquid surface.

If 2 an of the candle is burnt out, for length above for liquid level will be

⇒ \(\frac{2 L-2 \mathrm{~cm}}{2}=L-1 \mathrm{~cm}\)

which is reduced by 1 cm.

So, for top footcandle will fall at a rate of 1 cm h^-1

Question 3. By sucking through a straw a person can reduce the pressure in his lungs to 750 mmHg (density of mercury being 13.6 g cm^-3). Using the straw, he can drink water from a glass up to a maximum depth of

1. 10 cm
2. 13.6 cm
3. 75 cm
4. 1.36 cm

Answer: 2. 13.6 cm

Standard atmospheric pressure = 760 mmHg.

The pressure in the lungs is reduced to 750 mmHg.

∴ pressure difference =10 mmHg.

Fewfoesame pressure difference, let ft = height of the water column.

∴ h (1 g cm^-33)g = (10 mm) (13.6 g cm^-3)g

= 136 mm Hg

= 13.6 cmHg.

Hence, for required depth is 13.6 cm

Question 4. A solid sphere of volume V and density p floats at the interface of two immiscible liquids of densities p₁, and p₂ respectively. If p₁ < p < p₂ then the ratio of the volume of the parts of the sphere in the upper and lower liquids is

1. \(\frac{\rho-\rho_1}{\rho_2-\rho}\)
2. \(\frac{\rho_2-\rho}{\rho-\rho_1}\)
3. \(\frac{\rho+\rho_1}{\rho+\rho_2}\)
4. \(\frac{\rho+\rho_2}{\rho+\rho_1}\)

Answer: 2. \(\frac{\rho_2-\rho}{\rho-\rho_1}\)

Let V₁ be the volume of the sphere immersed in the liquid of density p₁ and V₂ be the volume immersed in the liquid of density p₂.

Total weight of the sphere = Mg = (V₁ + V₂)pg (downward) and buoyant force due to the liquids displaced

= V₁p₁g + V₂p₂g

= (V₁ P₁ + V₂ P₂)g (upward)

For equilibrium,

(V₁ + V₂)pg = (V₁p₁ + V₂p₂)g

⇒  V₁(p-p₁) = V₂(p₂-p)

⇒ \(\frac{V_1}{V_2}=\frac{\rho_2-\rho}{\rho-\rho_1}\)

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Question 5. If water is used to construct a barometer, what would be the approximate height of the water column in the tube at the standard atmospheric pressure (76 cm Hg)?

1. 2m
2. 4 m
3. 6 m
4. 10 m

Answer: 4. 10 m

For the same pressure,

⇒ \(h_{\mathrm{w}} \rho_{\mathrm{w}} g=h_{\mathrm{Hg}} \rho_{\mathrm{Hg}} g\)

⇒ \(h_w=\frac{h_{\mathrm{Hg}} \rho_{\mathrm{Hg}}}{\rho_{\mathrm{w}}}\)

⇒ \(\frac{(76 \mathrm{~cm})\left(13.6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)}{1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}\)

= 1033.6 cm ≈ 10 m.

Question 6. A cube of ice floats partly in water and partly in an oil l as shown in the figure. If the relative densities of the oil, ice, and water are 0.8, 0.9, and 1.0 then the ratio of the volumes of ice immersed in water to that in the oil will be

1. 8:9
2. 9:8
3. 1:1
4. None of these

Answer: 3. 1:1

From the law of floatation, the weight of the ice block = the weight of the liquids displaced.

Now, the weight of the ice block = Vp_iceg = (V_w + V_o) p_iceg,

the weight of the displaced water = V_wp_wg,

and the weight of the displaced oil = V_op_og.

For equilibrium,

⇒ \(\left(V_{\mathrm{w}}+V_{\mathrm{o}}\right) \cdot \rho_{\mathrm{ice}}=V_{\mathrm{w}} \rho_{\mathrm{w}}+V_{\mathrm{o}} \rho_{\mathrm{o}}\)

V_w (P_ice – P_w) = V_o (p_o – P_ice)

⇒ \(\frac{V_w}{V_0}=\frac{\rho_0-\rho_{\text {ice }}}{\rho_{\text {ice }}-\rho_w}\)

= \(\frac{0.8-0.9}{0.9-1.0}=\frac{-0.1}{-0.1}=1\)

=> V_w:V_o =1:1.

Question 7. A thick spherical shell of inner and outer radii R and 2R respectively, floats half-submerged in a liquid of density p0. The density of the material of the spherical shell is

1. \(\frac{\rho_0}{2}\)
2. p₀
3. \(\frac{4}{3} \rho_0\)
4. \(\frac{4}{7} \rho_0\)

Answer: 4. \(\frac{4}{7} \rho_0\)

The volume of the material of the shell is

⇒ \(\frac{4}{3} \pi\left[(2 R)^3-R^3\right]=\frac{4}{3} \pi\left(7 R^3\right)\)

∴ its mass is \(M=\frac{4}{3} \pi\left(7 R^3\right) p\)

∴ the buoyant force (due to the displaced liquid) is

⇒ \(\frac{1}{2}\left[\frac{4}{3} \pi(2 R)^3\right] \rho_0 g\)

For equilibrium,

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⇒ \(\frac{4}{3} \pi\left(7 R^3\right) \rho g=\frac{1}{2} \cdot \frac{4}{3} \pi\left(8 R^3\right) \rho_0 \cdot g\)

⇒ \(\rho=\frac{4}{7} \rho_0 .\)

Question 8. A U-tube containing a liquid is accelerated horizontally with a constant acceleration a0. If the separation between the vertical arms is l then the difference in the heights of the liquid columns in the two arms will be

1. \(\frac{l g}{a_0}\)
2. \(\frac{l a_0}{g}\)
3. \(l\left(1+\frac{a_0}{g}\right)\)
4. \(i\left(1-\frac{a_0}{g}\right)\)

Answer: 2. \(\frac{l a_0}{g}\)

Let A be the area of the cross-section of the tube t and p be the density of the liquid. Consider section AB of the tube.

Mass of the liquid in AB = volume x density = Alp.

The pressures at A and B due to the liquid column are h₂pg and h₁pg respectively.

∴ the net force on the liquid in AB to the right is (h₂– h₁) pgA.

(h₂-h₁)pgA = Ma₀ = Alpa₀.

Hence, the difference in height is

(h₂– h₁) = la₀/g

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Question 9. A beaker containing a liquid of density p moves vertically upwards with a uniform acceleration a. The hydrostatic pressure due to the liquid column at a depth h below the free liquid surface is

1. hpg
2. hp (g – a)
3. hp (g + a)
4. \(h \rho g\left(\frac{g-a}{g+a}\right)\)

Answer: 3. hp (g + a)

Consider the forces acting on an element of the liquid of cross-section A and height h. The force on the upper section is F₁ = P₀A and that on the lower section is F₂ = (p₀ + p)A, where p = pressure due to the liquid column.

∵ weight = mg = Vpg = Ahpg,

∴ upward net force =F₂– F₁– mg = ma.

∴ (p₀+p)A – p₀A – Ahpg = Ahpa

Hence,p = hp(g + a)

Question 10. The area of the cross-section of two arms of a hydraulic press is A and nA (where n >1) respectively, as shown in the figure. A force F is applied to the liquid in the thinner arm. In order to maintain the equilibrium of the liquid, the force F’ to be applied on the liquid in the thicker arm is

1. \(\frac{F}{n}\)
2. nF
3. \(\frac{F}{n+1}\)
4. (n + 1)F

Answer: 2. nF

In equilibrium, the pressures at the two surfaces are equal as they lie in the same horizontal plane.

∴ \(p_{\mathrm{atm}}+\frac{F}{A}=p_{\mathrm{atm}}+\frac{F^{\prime}}{n A}\)

⇒ \(F=\frac{F^{\prime}}{n}\)

⇒ F’=nF

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Question 11. A wind blows parallel to the roof of a house at a speed of 40 m s^-1. The area of the roof is 250 m². Assuming the pressure inside the house to die standard atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (given that pair = 1.2 kg m³)

1. 4,8 x 10⁵ N (downward)
2. 4.8 x 10⁵ N (upward)
3. 2.4 x 10⁵ N (upward)
4. 2.4 x 10⁵ N (downward)

Answer: 3. 2.4 x 10⁵ N (upward)

According to Bernoulli’s theorem,

⇒ \(p_1+\frac{1}{2} \rho v_1^2+h \rho g=p_2+\frac{1}{2} \rho v_2^2+h \rho g\)

So, the pressure difference inside and outside the roof is

AP = P₂ ~ P₁

⇒ \(\frac{1}{2} \rho\left(v_2^2-v_1^2\right)\)

Substituting the values,

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⇒ \(\Delta p=\frac{1}{2}\left(1.2 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left[\left(40 \mathrm{~m} \mathrm{~s}^{-1}\right)^2-0\right]\)

= 960 N m^-2.

∴ the force acting on the roof upwards is

ΔpA = (960 N m^-2) (250 m²)

= 240000 N

= 2.4 X 10⁵N.

Question 12. The cylindrical tube of a spray pump has a radius of R. One end of the pump has n fine holes, each of radius r. If the speed of the liquid in the tube is v, the speed with which the liquid is ejected through the hole is

1. \(\frac{v R^2}{n^2 r^2}\)
2. \(\frac{v R^2}{n r^2}\)
3. \(\frac{v R^2}{n^3 r^2}\)
4. \(\frac{v^2 R}{n r}\)

Answer: 2. \(\frac{v R^2}{n r^2}\)

According to the equation of continuity,

A₁v₁ = A₂v₂

⇒ πR²v = n(πr²)v’

⇒ \(v^{\prime}=\frac{v R^2}{n r^2}\)

Question 13. Bernoulli’s theorem is a consequence of the conservation of

1. Energy
2. Linear momentum
3. Angular momentum
4. Mass

Answer: 1. \(\frac{v R^2}{n^2 r^2}\)

Bernoulli’s theorem is a consequence of the conservation of energy in fluids, which have their KE, PE, and pressure energy

Question 14. In old age, arteries carrying blood in the human body become narrow resulting in an increase in blood pressure. This follows from

1. Stokes’ law
2. Archimedes’ principle
3. Pascal’s law
4. Bernoulli’s principle

Answer: 4. Bernoulli’s principle

According to the equation of continuity,

av = constant.

So, an increase in area (a₂ >a₁ decreases the velocity (v₂ < v₁)

From Bernoulli’s theorem,

⇒ \(p_1+\frac{1}{2} \rho v_1^2=p_2+\frac{1}{2} \rho v_2^2\)

Hence, increase in pressure = \(p_2-p_1=\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\)

Question 15. A liquid kept in a cylindrical vessel is rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is r and the angular velocity of rotation is co, the difference in the heights of the liquid at the centre of the vessel and the edge is

1. \(\frac{r \omega}{2 g}\)
2. \(\frac{r^2 \omega^2}{2 g}\)
3. \(\sqrt{2 g \omega r}\)
4. \(\frac{\omega^2}{2 g r^2}\)

Answer: 2. \(\frac{r^2 \omega^2}{2 g}\)

At the point P, let

So, N cos θ = mg and N sinθ = \(\frac{m v^2}{x}\)

⇒ \(\tan \theta=\frac{v^2}{x g}=\frac{d y}{d x}\)

Integrating,

⇒ \(\int_0^h d y=\int_0^r \frac{\omega^2 x}{g} d x\) [∵ v = xwww]

⇒ \(h=\frac{r^2 \omega^2}{2 g}\)

Alternative method:

According to Bernoulli’s theorem,

⇒ \(p+\frac{1}{2} \rho v^2\) = constant.

At the sides, the speed (v = m) is large, so the pressure is low. However, the pressure at a given horizontal level in a liquid is constant. Hence, the liquid rises such that

⇒ \(\frac{1}{2} \rho v^2=\rho g h \text { and } h=\frac{r^2 \omega^2}{2 g}\)

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Question 16. The velocity of water flowing through a cylindrical tube of diameter Semis 4 m s^-1. It is connected to a pipe with an end tip of a diameter of 2 cm. The velocity of water at the free end is

1. 4 ms^-1
2. 8 ms^-1
3. 32 ms^-1
4. 64 ms^-1

Answer: 4. 64 ms^-1

From the equation of continuity,

A₁V₁ = A₂V₂

⇒ \(\pi R_1^2 v_1=\pi R_2^2 v_2\)

⇒ \(v_2=v_1\left(\frac{R_1}{R_2}\right)^2\)

= \(\left(4 \mathrm{~m} \mathrm{~s}^{-1}\right)\left(\frac{8 \mathrm{~cm}}{2 \mathrm{~cm}}\right)^2\)

= \(64 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 17. The pressure head in the Bernoulli equation is

1. \(\frac{p p}{g}\)
2. \(\frac{p}{pg}\)
3. ppg
4. Pg

Answer: 2. \(\frac{p}{pg}\)[/latex]

From Bernoulli’s theorem,

⇒ \(\frac{1}{2} \rho v^2+\rho g h+p=\text { constant }\)

⇒ \(\frac{v^2}{2 g}+h+\frac{p}{\rho g}=\text { constant. }\)

Hence, the pressure head is \(\frac{p}{p g}\)

Question 18. Water flows through a horizontal tube with a constriction at B, as shown in the figure. The pressure difference between the regions A and B is 600 Pa, while the cross-sectional areas are in the ratio 2:1. The speed of flow (v_A) in the region A is

1. 2ms^-1
2. √2 ms^-1
3. 4 ms^-1
4. √0.4 ms^-1

Answer: 4. √0.4 ms^-1

From the equation of continuity,

v_Aa_A = v_Ba_B

⇒ \(v_{\mathrm{B}}=v_{\mathrm{A}} \cdot \frac{a_{\mathrm{A}}}{a_{\mathrm{B}}}=v_{\mathrm{A}} \cdot \frac{2}{1}=2 v_{\mathrm{A}}\)

From Bernoulli’s theorem,

⇒ \(p_{\mathrm{A}}+\frac{1}{2} \rho v_{\mathrm{A}}^2=p_{\mathrm{B}}+\frac{1}{2} \rho v_{\mathrm{B}}^2\)

⇒ \(p_{\mathrm{A}}-p_{\mathrm{B}}=\frac{1}{2} \rho\left(v_{\mathrm{B}}^2-v_{\mathrm{A}}^2\right)\)

⇒ \(\frac{1}{2} \rho\left(4 v_{\mathrm{A}}^2-v_{\mathrm{A}}^2\right)=\frac{3}{2} \rho v_{\mathrm{A}}^2\)

⇒ \(v_{\mathrm{A}}=\sqrt{\frac{2}{3 \rho}\left(p_{\mathrm{A}}-p_{\mathrm{B}}\right)}\)

= \(\sqrt{\frac{2}{3} \cdot \frac{600 \mathrm{~N} \mathrm{~m}^{-2}}{10^3 \mathrm{~kg} \mathrm{~m}^{-3}}}\)

= \(\sqrt{0.4} \mathrm{~m} \mathrm{~s}^{-1}\)

Question 19. Water and mercury are filled in two cylindrical vessels up to the 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 v₁ and v₂ respectively, then

1. v₁ = 13.6v₂
2. \(v_1=\frac{v_2}{13.6}\)
3. v₁ = v₂
4. \(v_1=\sqrt{13.6} v_2\)

Answer: 3. v₁ = v₂

Efflux speed \(v=\sqrt{2 g h}\), which is independent of the density of the liquid. So, for the same height h, the efflux speed will always be the same.

Question 20. There is a small hole near the bottom of an open tank filled with a liquid. The speed of the liquid ejected out through the hole does not depend on the

1. Area of the hole
2. Density of the liquid
3. Acceleration due to gravity
4. Height of the liquid column from the hole

Answer: 1. Area of the hole

Efflux speed \(v=\sqrt{2 g h}\), which is independent of the area of the hole.

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Question 21. The efflux speed of the liquid coming out of a small hole of a large vessel containing two different liquids of densities 2p and p (as shown 2h in the figure) will be

1. \(\sqrt{g h}\)
2. \(2 \sqrt{g h}\)
3. \(\sqrt{8 g h}\)
4. \(\sqrt{6 g h}\)

Answer: 2. \(2 \sqrt{g h}\)

A liquid column of height 2h with a density of p is equivalent to a liquid column of height h with a density of 2p, as they produce the same pressure.- Hence, the equivalent height H of the top surface from the hole is

H = h + h = 2h,

and the efflux speed is \(v=\sqrt{2 g(2 h)}=2 \sqrt{g h}\)

Question 22. A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If the relative density of silver is 10, what will be the tension in the string?

1. 73 N
2. 21 N
3. 42.5 N
4. 37.2 N

Answer: 4. 37.2 N

Tension in the string = true weight- upthrust

⇒ \((4 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)-\frac{(4 \mathrm{~kg})\left(0.72 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\right)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right)}{10 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}\)

= 40 N – 2.8 N

= 37.2 N.

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Question 23. An incompressible fluid of density p is filled in two identical containers of base area A. On opening the l valve V connecting the containers, the levels become equal. The loss in the gravitational potential energy during the process is

1. \(\frac{A p g}{4}\left(x_1-x_2\right)^2\)
2. \(\frac{A \rho g}{4}\left(x_1+x_2\right)^2\)
3. \(\frac{A \rho g}{4}\left(x_1^2-x_2^2\right)\)
4. \(\frac{A \rho g}{4}\left(x_1^2+x_2^2\right)\)

Answer: 1. \(\frac{A p g}{4}\left(x_1-x_2\right)^2\)

The initial potential energy of the system is

U_i = m₁gh₁ + m₂gh₂

⇒ \(\left(\rho A x_1, \frac{x_1}{2}+\rho A x_2 \cdot \frac{x_2}{2}\right) g\)

⇒ \(\left(x_1^2+x_2^2\right)\left(\frac{A p g}{2}\right)\)

When the connecting valve is opened, the fluid attains the same height \(x=\frac{\left(x_1+x_2\right)}{2}\) in both containers.

Hence, the final potential energy is

⇒ \(U_f=2\left(\frac{x_1+x_2}{2}\right)^2\left(\frac{A p g}{2}\right)\)

⇒ \(\left(x_1+x_2\right)^2\left(\frac{A \rho g}{4}\right)\)

∴ loss in \(\mathrm{PE}=U_1-U_f=\frac{A \rho g}{2}\left[x_1^2+x_2^2-\frac{\left(x_1+x_2\right)^2}{2}\right]\)

⇒ \(\frac{A \rho g}{4}\left(x-x_2\right)^2\)

Question 24. A shell of relative density 27/9 relative to water is just submerged in water. If its inner and outer radii are r and R respectively, r/R is equal to

1. \(\left(\frac{1}{3}\right)^{\frac{1}{3}}\)
2. \(\left(\frac{2}{3}\right)^{\frac{1}{3}}\)
3. \(\left(\frac{3}{4}\right)^{\frac{1}{3}}\)
4. \(\left(\frac{5}{9}\right)^{\frac{1}{3}}\)

Answer: 2. \(\left(\frac{2}{3}\right)^{\frac{1}{3}}\)

For equilibrium,

mass of shell = buoyant force

⇒ \(\frac{4}{3} \pi\left(R^3-r^3\right) \rho_{\mathrm{sh}} g=\frac{4}{3} \pi R^3 \rho_{\mathrm{w}} g\)

⇒ \(\left(R^3-r^3\right) \cdot 3=R^3 \Rightarrow 1-\left(\frac{r}{R}\right)^3=\frac{1}{3}\)

⇒ \(\left(\frac{r}{R}\right)^3=\frac{2}{3} \Rightarrow \frac{r}{R}=\left(\frac{2}{3}\right)^{\frac{1}{3}}\)