2 Marks Questions

Question 12 Marks

Explain why To keep a piece of paper horizontal, you should blow over, not under, it.

Answer

When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according to Beroulli's Theorem), where as pressure of air blow the paper is atmospheric. Hence, the paper stays horizontal.

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Question 22 Marks

Figures(a) and (b) refer to the steady flow of a (non viscous) liquid. Which of the two figures is incorrect? Why?

Answer

Fig. (a) is incorrect. According to equation of continuity, i.e., av = Constant, where area of cross-section of tube is less, the velocity of liquid flow is more. So the velocity of liquid flow at a constriction of tube is more than the other portion of tube. According to Bernoulli’s Theorem, $\text{P}+\frac{1}{2}\rho\text{v}^2$= Constant, where v is more, P is less and vice versa.

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Question 32 Marks

Explain why Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)

Answer

Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops.
On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.

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Question 42 Marks

Explain why Water with detergent disolved in it should have small angles of contact.

Answer

Water with detergent dissolved in it has small angles of contact $(\theta).$ This is because for a small $\theta$, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact $(\theta).$ If $(\theta)$ is small, then $\cos\theta$ will be large and the rise of the detergent water in the cloth will be fast.

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Question 52 Marks

Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.

Answer

No, it does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation. The two points where Bernoulli’s equation is applied should have significantly different atmospheric pressures.

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Question 62 Marks

Explain why A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.

Answer

When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity: Area × Velocity = Constant. According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.

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Question 72 Marks

Explain why The blood pressure in humans is greater at the feet than at the brain.

Answer

Blood pressure is the pressure exerted by the column of blood on the blood vessel. Blood pressure depends on the force at which heart pumps the blood. The pressure of Liquid is given by: P = hdg where h = height of liquid column(blood in this case) d = density of the liquid g = Acceleration due to gravity. Hence we can say that height of blood column at feet is more than it is at brain and also at the feet the blood has to be pumped to the heart against the force of gravity and has to travel a greater distance than at the brain. so we can conclude, by saying that "The blood pressure in humans is greater at the feet than at the brain".

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Question 82 Marks

Explain why A drop of liquid under no external forces is always spherical in shape.

Answer

A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

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Question 92 Marks

Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Answer

Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamline flow.

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Question 102 Marks

Explain why Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer

When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.

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Question 112 Marks

Explain why The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.

Answer

The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of the equation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.

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Question 122 Marks

A manometer reads the pressure of a gas in an enclosure as shown in Fig.(a) When a pump removes some of the gas, the manometer reads as in Fig. (b) The liquid used in the manometers is mercury and the atmospheric pressure is $76cm$ of mercury. Give the absolute and gauge pressure of the gas in the enclosure for cases (i) and (ii) in units of cm of mercury. How would the levels change in case (i) if $13.6cm$ of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in volume of the gas.)

Answer

Given : Atmospheric pressure,

$P_0=6 \mathrm{~cm} \text { of } \mathrm{Hg} .$
In figure (i) pressure head,
$h_1=+20 \mathrm{~cm} \text { of } \mathrm{Hg}$
Absolute pressure $(P)$ of the gas is greater than the $\mathrm{P}_0$ i.e.,
$P=P_0+h_1 p g$
$=76 \mathrm{~cm} \text { of } \mathrm{Hg}+20 \mathrm{~cm} \text { of } \mathrm{Hg}$
$=96 \mathrm{~cm} \text { of } \mathrm{Hg} .$
Gauge pressure is the difference between the absolute pressure and the atmospheric pressure. $\frac{1}{2}$ It means,
$\text { Gauge pressure }=P-P_0$
$=96 \mathrm{~cm} \text { of } \mathrm{Hg}-76 \mathrm{~cm} \text { of } \mathrm{Hg}$
$=20 \mathrm{~cm} \text { of } \mathrm{Hg} .$
In figure (ii), pressure head,
$\mathrm{h}_2=-18 \mathrm{~cm} \text { of } \mathrm{Hg} .$
$\therefore$ The absolute pressure of the gas is lesser than the atmospheric pressure is given by
$P=P_0+h_2 p g$
$=76 \mathrm{~cm} \text { of } \mathrm{Hg}+(-18 \mathrm{~cm}) \text { of } \mathrm{Hg}$
$=58 \mathrm{~cm} \text { of } \mathrm{Hg}$
$\text { Gauge pressure }=\text { Absolute pressure }- \text { Atmospheric pressure }$
$=58 \mathrm{~cm} \text { of } \mathrm{Hg}-76 \mathrm{~cm} \text { of } \mathrm{Hg}$
$=-18 \mathrm{~cm} \text { of } \mathrm{Hg}$
It means, Gauge pressure is simply equal to h cm of Hg .
b. Given : 13.6 cm of water added in the right limb is equivalent to $\frac{13.6}{13.6}=1 \mathrm{~cm}$ of Hg column i.e., $\mathrm{h}=1 \mathrm{~cm}$ of Hg column, which can be calculated as follows

Given : 13.6cm of water added in the right limb is equivalent to $\frac{13.6}{13.6}=1\text{cm}$ of Hg column i.e., h = 1cm of Hg column, which can be calculated as follows

$\text{h}_{\omega}=13.6\text{Cm}$ of water
Suppose $h_m$ = height of Hg column equivalent to 13.6cm of water, thus equilibrium.
$\text{h}_{\text{m}}\rho_{\text{m}}\text{g}=\text{h}_{\omega}\rho_{\omega}\text{g}.$
$\text{h}_{\text{m}}=\text{h}_{\omega}\frac{\rho_{\omega}}{\rho_{\text{m}}}=\frac{\text{h}_{\omega}}{\Big(\frac{\rho_{\text{m}}}{\rho_{\omega}}\Big)}$
$=\frac{13.6}{13.6}=1\text{cm}$ of hg
The mercury will rise in the left limb such that the difference in the height of Hg column in the two limbs.
= 20cm - 1m
= 19 cm of Hg column.

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Question 132 Marks

Explain why Surface tension of a liquid is independent of the area of the surface.

Answer

Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

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Question 142 Marks

A piece of an alloy of mass 96gm is composed of two metals whose specific gravities are 11.4 and 7.4. If the weight of the alloy is 86gm in water, find the mass of each metal in the alloy.

Answer

Suppose the mass of the metal of specific gravity 11.4 be m. Now the mass of the second metal of specific gravity 7.4 will be (96 - m). Volume of first metal $=\frac{\text{m}}{11.4}\text{cm}^3$ Volume of second metal $=\frac{96-\text{m}}{7.4}\text{cm}^3$ Total volume $=\frac{\text{m}}{11.4}+\frac{96-\text{m}}{7.4}$ Apparent loss of wt. in water $=\Big(\frac{\text{m}}{11.4}+\frac{96-\text{m}}{7.4}\Big)\text{gm wt.}$ According wt. in water $=96-\Big[\Big(\frac{\text{m}}{11.4}\Big)+\frac{(96-\text{m})}{7.4}\Big]$ According to the given peoblem,$96-\Big[\Big(\frac{\text{m}}{11.4}\Big)+\frac{(96-\text{m})}{7.4}\Big]=86$ or $\frac{\text{m}}{11.4}+\frac{(96-\text{m})}{7.4}=10$
Solving we get, m = 62.7gm.$\therefore$ Mass of second metal = 96 - 62.7 = 33.3gm

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Question 152 Marks

A vertical barometer tube $100cm$ in length and dipping into mercury contains a small quantity of air. When the open end is $15cm$ below the surface of mercury the meniscus is $30cm$ from the upper closed end. When the tube is pushed into the mercury so that the open end is $35cm/ s$ below the surface, the meniscus is $20cm$ from the closed end. Calculate the atmospheric pressure.

Answer

Let the atmospheric pressure be $P$ and area of cross section of the tube be $a$. Case (i): Length of air column $=30 \mathrm{~cm}$
$\therefore$ Volume of mercury column $=100-(30+15)=55 \mathrm{~cm}$.
Pressure $\mathrm{P}_1$ of air in tube $=(P-55) \mathrm{cm}$ Case (ii): Length of colum $=20 \mathrm{~cm} . \therefore$ Volume of air, $\mathrm{V}_2=(20 \times$ a)c.c.
Length of mercury column $=100-(20+30)=45 \mathrm{~cm}$. Pressure $P_2$ of air in tube $=(P-45) \mathrm{cm}$ Applying Boyle's law, $P_1 V_1=P_2 V_2(P-55) \times 30 a=(P-45) \times 20$ a Solving, we get, $P=75 \mathrm{~cm}$

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Question 162 Marks

Explain why To keep a piece of paper horizontal, you should blow over, not under, it.

Answer

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Question 172 Marks

Figures(a) and (b) refer to the steady flow of a (non viscous) liquid. Which of the two figures is incorrect? Why?

Answer

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Question 182 Marks

Derive an expression for the excess of pressure inside an air bubble.

Answer

Consider a bubble of radius R with $\sigma$ the surface tension of liquid. Excess pressure inside the bubble, $P = P_i - P_0$ ($\because$ air bubble has only one free surface)$\delta\text{R}$ = Small increase in radius of bubble due to excess pressure
Work done, W = Force × Displacement. = (Excess pressure × Area) × Increase in radius$=\text{P}\times4\pi\text{R}^2\times\delta\text{R}$
Increase in surface area of bubble = Final surface area - Initial surface area$=4\pi(\text{R}+\delta\text{R})^2-4\pi\text{R}^2$
$=8\pi\text{R}(\delta\text{R})(\text{Neglecting }\delta\text{R}^2)$
$\therefore\text{P}\times4\pi\text{R}^2\times\delta\text{R}=8\pi\text{ R}(\delta\text{R})\times\sigma$
Increase in P.E. = increase in surface area × Surface tension$=8\pi\text{R}(\delta\text{R})\times\sigma$
Since the drop is in equilibrium.$\therefore\text{P}\times4\pi\text{R}^2\times\sigma\text{R}=8\pi\text{ R}(\delta\text{R})\times\sigma$
$\text{P}=\frac{2\sigma}{\text{R}}$

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Question 192 Marks

Explain why Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)

Answer

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Question 202 Marks

Two soap bubbles of radii 6cm and 8cm coalesce to form a single bubble. Find the radius of the new bubble.

Answer

Surface energy of first soap bubble = Surface tension × surface area$=2\times4\pi\text{R}^2_1\text{S}=8\pi\text{R}^2_1\text{S}$
Surface energy of second soap bubble $=8\pi\text{R}^2_2\text{S}$ Let the radius of the new shoap bubble is R, so the surface energy of new bubble $=8\pi\text{R}^2\text{S}$ By the law of conservation of enrgy,$8\pi\text{R}62\text{S}=8\pi\text{R}^2_1\text{S}+8\pi\text{R}^2_2\text{S}$
$\text{R}^2=\text{r}^2_1+\text{R}^2_2=36+64$
$\therefore\text{R}^2=100\text{cm}^2$
$\Rightarrow\text{R}=10\text{cm}$

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Question 212 Marks

The density of atmosphere is $1.29 \mathrm{~kg} \mathrm{~m}^{-3}$ at sea level, where atmospheric pressure is $1.013 \times 10^5 \mathrm{~Pa}$. If we assume that atmospheric density does not change with altitude, then what should be the height of the atmosphere?

Answer

Here, Pa = 1.013 × 105 Pa and $\rho=1.29\text{kg m}^{-3}$ If height of air column, assuming its density to be constant, be h, then,$\text{P}\text{a}=\text{h}\rho\text{g}$
$\Rightarrow\text{h}=\frac{\text{P}\text{a}}{\rho\text{g}}$
$=\frac{1.013\times10^5}{1.29\times9.8}=8013\text{m}\simeq8\text{km.}$

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Question 222 Marks

A body of mass 6kg is floating in a liquid with $\frac23$ of its volume inside the liquid. Find ratio between the density of the body and density of liquid. Take $g = 10m/ s^2$.

Answer

As we know that, for a floating body Buoyant force = Weight of liquid displaced Let V be the volume of the body, $\frac23\text{V}\rho_\text{l}\text{g}=\text{V}\rho_\text{l}\text{g}$ Where, $\rho_\text{b}=$ density of floating body and $\rho_\text{l}=$ density of liquid$\therefore\frac{\text{}\rho_\text{b}}{\rho_\text{l}}=\frac23$

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Question 232 Marks

A liquid drop breaks into 27 small drops. If surface tension of the liquid is S, then find the energy released.

Answer

Let the radius of larger drop = R and radius of each small drop = r Volume of 27 small drops = Volume of the large drop$=27\times\frac43\times\pi\text{r}^3=\frac43\pi\text{R}^3$
So, $\text{r}=\frac{\text{R}}{3}$ Surface area of large drop $=4\pi\text{R}^2$ Surface area of 27 small drops $=27\times4\pi\text{r}^2$$=27\times4\pi\Big(\frac{\text{R}}{2}\Big)^3=12\pi\text{R}^2$
$\therefore$ Increase in surface area $=12\pi\text{R}^2-4\pi\text{R}^2=8\pi\text{R}^2$
Increase in energy = increase in surface area × Surface tention$=8\pi\text{R}^2\times\text{S}$

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Question 242 Marks

In a horizontal pipe line of uniform area of cross section, the pressure falls by $8N-m^2$ between two points separated by a distance of $1km$. What is the change in kinetic energy per kg of the oil flowing at these points? Density of oil is $806kg-m^{-3}$.

Answer

According to Bernoulli's theorem,$\text{P}_1+\frac12\rho\upsilon_1^2=\text{P}_2+\frac12\rho\upsilon^2_2$ (pipe is horizontal)
$\text{P}_1-\text{P}_2=\frac12\rho\ (\upsilon^2_2-\upsilon^2_1)=$ change in K.E. per kg mass.
$\therefore$ Change in K.E. per kg. mass of oil $=\frac{\text{P}_1-\text{P}_2}{\rho}$
Substituting the given values, we have Change in K.E. per kg mass $=\frac{8}{800}=10^{-2}\text{J/kg.}$

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Question 252 Marks

As shown in figure, water flows from P to Q. Explain why height h, of column AB of water is greater than height h, of column CD of water.

Answer

Height in the column of water depends on difference of pressure. Since $P_1 - P_2$ greater than in arm CD. So, $h_1 > h_2$.$\text{P}_1-\text{P}_2=\text{h}\rho_\text{m}\text{g}$

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Question 262 Marks

At, what speed will the velocity head of stream of water be 40cm?

Answer

Given, h = 40cm, $g = 980cm/ s^2$ We know that velocity head, $\text{h}=\frac{\text{v}^2}{2\text{g}}$$\therefore\text{v}=\sqrt{2\text{gh}}=\sqrt{2\times980\times40}$
$=280\text{cm s}^{-1}$

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Question 272 Marks

The antiseptics used for cuts and wounds in the human flesh have low surface tensions. Why?

Answer

For wound to heal quickly, the antiseptic used should be able to spread itself into a thin layer on the wound. A liquid spreads more on a surface if its own surface tension is less. Therefore, the antiseptic should possess a low surface tension.

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Question 282 Marks

A liquid has a definite volume but no shape of its own. Explain.

Answer

Liquids take the shape of the container in which they are placed and do not possess a shape of their own. It takes the same volume irrespective the shape of the container.

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Question 292 Marks

What should be the maximum average velocity of water in a tube of diameter $2cm$ so that flow is laminar? The viscosity of water is $0.001Nm^{-2}s$.

Answer

D = 2cm = 0.02m$\rho=10^3\text{kg m}^{-3}$
$\eta=0.001\text{ Nm}^{-2}\text{s}=10^{-3}\text{ Nm}^{-2}\text{s}$
Flow of water will be laminar if, $N_R = 1000$ where $N_R$ is Reynold number Let $\upsilon=$ maximum average velocity
$\therefore$ Using the relation,
$\text{N}_\text{R}=\frac{\rho\upsilon\text{D}}{\eta}$ or $\upsilon=\frac{\text{N}_\text{R}\eta}{\rho\text{D}}$
$=\frac{1000\times0.001}{1000\times0.02}=0.05\text{ms}^{-1}$

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Question 302 Marks

A liquid drop of radius 4mm breaks into 1000 identical drops. Find the change in surface energy. $S=0.07 \mathrm{Nm}^{-1}$.

Answer

Volume of 1000 small drops = Volume of a large drop$1000\times\frac43\pi\text{r}^3=\frac43\pi\text{R}^3$
$\text{r}=\frac{\text{R}}{10}$
Surface area of large drop $=4\pi\text{R}^2$ Surface area of 1000 drop $4\pi\times1000\text{r}^2=40\pi\text{R}^2$$\therefore$ Increase in surface area $=(40-4)\pi\text{R}^2=36\pi\text{R}^2$
The increase in surface energy = Surface tension × increase in suraface area$=36\pi\text{R}^2\times0.07=36\times3.14\times(4\times10^{-3})^2\times0.07$
$=1.26\times10^{-4}\text{J}$

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Question 312 Marks

A balloon with hydrogen in it rises up but a balloon with air comes down, Why?

Answer

The density of hydrogen is less than air. So, the buoyant force on the balloon will be more than its weight in case of the hydrogen. So, in this case the balloon rises up. In case of air, the weight of balloon is more than the buoyant force acting on it, so balloon will come down.

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Question 322 Marks

What should be the maximum average velocity of water in a tube of diameter 0.5cm. So that the flow is laminar? The viscosity of water is $0.00125Nm^{-2}s$.

Answer

Here D = 0.5cm = 0.005m$\rho=10^3\text{kg m}^{-3}$
$\eta=0.00125\text{Nm}^{-2}\text{s}$
For laminar flow, the Reynold number for water, $N_R$ = 2000 Let $\text{v}$ b the maximum average velocity,$\therefore\text{N}_\text{R}=\frac{\rho\upsilon\text{D}}{\eta}$ or $\text{v}=\frac{\text{N}_\text{R}-\eta}{\rho-\text{D}}$
$=\frac{2000\times0.00125}{1000\times0.005}=0.5\text{m/s}.$

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Question 332 Marks

Explain why Water with detergent disolved in it should have small angles of contact.

Answer

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Question 342 Marks

If work required to blow a soap bubble of radius r is W, then what additional work is required to be done to blow it to a radius 3r?

Answer

Increase in surface area $=2[4\pi(3\text{r})^2-4\pi\text{r}^2]$ Increase in surface energy $=\sigma\times2\times4\pi\times8\text{r}^2=\text{8W}$ Addditional work done = 8W.

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Question 352 Marks

In rising from the bottom of a lake to the top, the temperature of an air bubble remains unchanged, but its diameter gets doubled. What is the depth of the lake? Given h is the barometric height in metres of mercury of relative density ρ at the surface of the lake.

Answer

At the surface, $\text{P}_1\text{h}\rho\text{g};\text{V}_1=\frac43\pi(2\text{r})^3$ At the bottom of depth x, $\text{P}_2=(\text{h}\rho+\text{xg})$ and $\text{V}_2=\frac{4}{3}\pi\text{r}^3$ Using Boyle's law, $\text{P}_1\text{V}_1=\text{P}_2\text{V}_2$$\therefore\text{h}\rho\text{g}\times\frac{4}{3}\pi(2\text{r})^3(\text{h}\rho\text{g}+\text{xg})\times\frac43\pi\text{r}^3$ or $\text{x}=8\text{h}\rho-\text{h}\rho=7\text{h}\rho\text{ metres}.$

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Question 362 Marks

A cylindrical jar of cross-sectional area $0.01m^2$ is filled with water to a height of 50cm (given figure). It carries a tight fitting piston of negligible mass. Calculate the pressure at the bottom of the jar when a mass of 5kg is placed on the piston.

Answer

Total force acting on the base $=\text{hg}\rho\text{A}+\text{mg}$
$\text{F}=0.5\times9.8\times1000\times0.01+5\times9.8$
$=5\times9.8+5\times9.8=98.0\text{N}$
$\therefore\text{Pressure}=\frac{\text{Force}}{\text{Area}}=\frac{98.0}{0.01}=9800\text{Nm}^{-2}$

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Question 372 Marks

Find the work done required to make a soap bubble of radius 0.02m. Given surface tension of soap 0.03N/ m.

Answer

Given, S = 0.03N/ m Work done = surface area × surface tension$=2\times4\pi\text{r}^2\times\text{S}$
$=2\times4\times3.14\times(2.02)^2\times0.03$
$=3\times10^{-4}\text{J}$

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Question 382 Marks

On what factors does the critical velocity of the liquid depend?

Answer

Critical velocity ($v_c$) of a liquid is:

Directly proportional to the coefficient of viscosity of the liquid.
Inversely proportional to the density of the liquid i.e., $\text{V}_\text{c}\propto\frac{1}{\rho}.$
Inversely proportional to the diameter of the tube through which it flows i.e., $\text{V}_\text{c}\propto\frac1{\text{D}}.$

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Question 392 Marks

A hole of area $4cm^2$ is formed in the side of a ship $2.4m$ below the water level. What minimum force is required to hold on a patch covering the hole from the inside of the ship? Given that density of sea water = $1.03 \times 103kg m^{-3}$.

Answer

Here, depth of hole below the water level $h = 2.4m$, Density of sea water $\rho=1.03\times10^3\text{kg m}^{-3}$ and surface area of hole $A = 4cm^2 = 4 \times 10^{-4} m^2$.
$\therefore$ Minimum force required to hold on a patch covering the hole from inside the ship
F = Pressure at height h of sea water column × Surface area of hole $=\text{h}\rho\text{g A}=2.4\times1.03\times10^3\times9.8\times4\times10^{-4}$
$=9.69\text{N}.$

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Question 402 Marks

Calculate the work done in blowing a soap bubble from a radius of 2cm to 3cm. The surface tension of the soap solution is 30 dynes cm.

Answer

$\sigma=30$ dynes/ cm, $r_1 = 2cm, r_2 = 3cm$,Since, bubble has two surface, initail surface area of bubble,
$=2\times4\pi\text{r}^2_1=2\times4\pi\times(2)^2$
$=32\pi\text{ cm}^2$
Final surface area of bubble,
$=2\times4\pi\text{r}^2_2=2\times4\pi\times(3)^2$
$=72\pi\text{r}^3$
Increase in surface area,
$=72\pi-34\pi=40\pi\text{ cm}^2$
Work done $=\sigma\times$ Increase in surface area,
$=30\times40\pi=3768\text{ ergs}$

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Question 412 Marks

Explain why “A drop of liquid under no external force is always spherical in shape”.

Answer

To reduce the P.E., the surface area for a given volume is reduced by forming a spherical shape.

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Question 422 Marks

What is the excess pressure inside a soap bubble that is 5 cm in diameter, assuming $0.026 \mathrm{Nm}^{-1}$ as the surface tension of the soap solution?

Answer

Excess pressure inside a soap bubble is$(\text{P}-\text{P}_\text{a})=\frac{\text{4T}}{\text{r}}$
Here $\text{T}=0.026\text{Nm}^{-1}$$\text{r}=\frac52=2.5\text{cm}=2.5\times10^{-2}\text{m}$
$\therefore\text{P}-\text{P}_\text{a}=\frac{4\times0.026\text{Nm}^{-1}}{2.5\times10^{-2}\text{m}}=4.16\text{Nm}^{-2}$

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Question 432 Marks

Two equal drops of water are falling through air with a steady velocity v. If the drops coalesce, what will be the new steady velocity?

Answer

The terminal velocity v of a drop of radius r, density $\rho$ while falling through a viscous medium of viscosity $\eta$ and density $\rho_0$ is,$\text{v}=\frac29\frac{2\text{r}^2(\rho-\rho_0)\text{g}}{\eta}\dots\text{(i)}$
If R is the radius of the new drop formed when two drops coalesce, then$\therefore$ New terminal velocity of drop of radius R is,
$\text{v}'=\frac{2[(2)^{\frac13}\text{r}](\rho-\text{}\rho_0)\text{g}}{\eta}=(2)^{\frac13}\text{v}$ [From (i)]

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Question 442 Marks

It is advised not to stand near a running train. Why?

Answer

When fast moving train passes on a rail, then the velocity of air streams in between the rail and the person standing near rail will be very large as compared to the velocity of air streams on the other side of person away from the rail. According to Bernoulli's theorem, the pressure of air will become low in between person and rail and is high on the other side of person. As a result of the pressure difference, a thrust acts on the person which may push the person towards rail side and the person may meet with an accident.

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Question 452 Marks

What height of water column produces the same pressure as a 760mm high column of mercury?

Answer

At sea level atmospheric pressure is the pressure exerted by 0.76m of mercury column i.e., h = 0.76m Density of mercury,$\rho=13.6\times10^3\text{kg/m}^3$
$g = 9.8 m/s^2$ Atmospheric pressure,$\text{P}=\text{h}\rho\text{g}=0.76\times(13.6\times10^3)\times9.8$
$=1.015\times10^5\text{Pa}$
Now atmosphere on earth exerts a pressure of $1.01 \times 10^5Nm^{-2}$ on the surface of earth Consider the water to be of uniform density $\rho=10^3\text{kg/m}^3.$ Then height of water is,$\text{h}=\frac{\text{P}}{\rho\text{g}}=\frac{1.01\times10^3}{10^3\times9.8}=10.3\text{m}$

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Question 462 Marks

Derive an expression for the excess pressure inside a liquid drop.

Answer

Consider a liquid drop of radius r. If one tries to enhance the radius by a small amount ‘dr’ work has to be done to overcome the excess pressure (P).

Work done $=\text{dW}=\text{p}\times4\pi\text{r}^2\times\text{dr}$ Due to surface tension $'\sigma',$ the excess pressure exists. The work done to change the area is also written as,$\text{dW}=\sigma \times\text{change in area}$
$=\sigma\times4\pi\{(\text{r}+\text{dr})^2-\text{r}^2\}=\sigma8\pi\text{rdr}$
$\therefore\text{P}4\pi\text{r}^2\text{dr}=\sigma8\pi\text{r}\text{ dr}$
$\therefore\text{P}=\frac{2\sigma}{\text{r}}$

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Question 472 Marks

Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.

Answer

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Question 482 Marks

Two syringes of different cross$-$sections $($without needles$)$ filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are $1.0\ cm$ and $3.0\ cm$ respectively.

Find the force exerted on the larger piston when a force of $10N$ is applied to the smaller piston.
If the smaller piston is pushed in through $6.0\ cm$, how much does the larger piston move out?

Answer

Here, $d_1 = 1.0\ cm, d_2 = 3.0\ cm$, Force on smaller piston $F_1 = 10N$.

According to Pascal's law of transmission of pressure $\text{P}=\frac{\text{F}_1}{\text{A}_1}=\frac{\text{F}_2}{\text{A}_2}.$

$\therefore\frac{\text{F}_2}{\text{F}_1}=\frac{\text{A}_2}{\text{A}_1}$
$=\frac{\text{r}^2_2}{\text{r}^2_1}=\Big(\frac{\text{d}_2}{\text{d}_1}\Big)^2$
$=\Big(\frac{3.00\text{cm}}{1.0\text{cm}}\big)^2=9$
$\Rightarrow\text{F}_2=\text{F}_1\times9$
$=10\times9$
$=90\text{N}.$

Water is considered to be completely incompressible. Therefore, volume covered by the movement of smaller piston inwards is exactly equal to volume moved outwards due to movement of larger piston, distance through which smaller piston is pushed $L_1 = 6.0\ cm$. Let larger piston is pushed by a distance $L_2$ then,

$\text{V}=\text{L}_1\text{A}_1=\text{L}_2\text{A}_2$
$\therefore\text{L}_2=\text{L}_1\Big(\frac{\text{A}_1}{\text{A}_2}\Big)=\text{L}_2\Big(\frac{\text{d}_1}{\text{d}_2}\Big)^2$
$=6.0\text{cm}\times\Big(\frac{1.0\text{cm}}{3.0\text{cm}}\Big)^2$
$=\frac69\text{cm}$
$=0.67\text{cm}$

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Question 492 Marks

If the excess pressure inside a spherical soap bubble of radius 1cm is balanced by that due to a column of oil of specific gravity 0.9, 1.36mm high. Calculate the surface tension.

Answer

Redius, r = 1cm; $\rho=0.9\text{g cm}^{-3}$$\text{h}=1.36\text{mm}=0.136\text{cm}$
Pressure, $\text{P}=\text{h}\rho\text{g}=0.136\times0.9\times980$$=119.95\text{ dyne cm}^{-2}$
Let T be surfac4e tension of shoap solution$\therefore$ Excess pressure, $\text{P}=\frac{4\text{T}}{\text{r}}$ or $\text{T}=\frac{\text{Pr}}{\text{4}}=\frac{119.95\times1}{4}$
$= 29.988\text{ dynecm}^{-1}$

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Question 502 Marks

A piece of copper having an internal cavity weight 264 g in air and 221 g in water. Find the volume of the cavity. The density of copper $=8.8 \mathrm{gcm}^{-3}$.

Answer

Mass of copper piece in air $=264 \mathrm{~g}$ Mass of copper piece in water $=221 \mathrm{~g}$ Apparent loss of mass $=264-221=43 \mathrm{~g}$ This is the mass of water displaced by the copper piece when immersed in water. Volume of copper piece with cavity $=43 \mathrm{~cm}^3$ Volume of copper only $=\frac{m}{\rho}=\frac{264}{8.8}=30 \mathrm{~cm}^3$ Volume of the cavity $=43-30=13 \mathrm{~cm}^3$.

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