Class 11th

11.11 The coefficient of volume expansion of glycerin is 49 * 10^–5 K^–1. What is the fractional change in its density for a 30 °C rise in temperature ?

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11.11 Coefficient of volume expansion of glycerin, $α_{V}$ = 49 $* 10^{- 5}$ /K

Rise in temperature, $? T$ = 30°C

Fractional change in volume = $\frac{? V}{V}$

We can write, $\frac{? V}{V}$ = $α_{V} * ? T$ = 49 $* 10^{- 5} * 30$ = 0.0147 ……(i)

If the final volume is $V_{2}$ and initial volume is $V_{1}$ , then

$\frac{? V}{V}$ = $\frac{V_{2} - V_{1}}{V_{1}}$

$V_{2} = \frac{m}{ρ_{2}}$ and $V_{1} = \frac{m}{ρ_{1}}$ where $ρ_{1}$ & $ρ_{2}$ are initial and final densities

$\frac{? V}{V}$ = $\frac{V_{2} - V_{1}}{V_{1}}$ = $\frac{ρ_{2} - ρ_{1}}{ρ_{1}}$ = fractional change in density = 0.0147 = 1.47 $*$ $10^{- 2}$

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10.4 Explain why

(a) To keep a piece of paper horizontal, you should blow over, not under, it

(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers

(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection

(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

(e) A spinning cricket ball in air does not follow a parabolic trajectory

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When air is blown under a paper, the velocity of air is more than the upper portion of the paper. As per Bernoulli's principle, atmospheric pressure reduces under the paper and makes it fall. To keep the paper horizontal, the air needs to be blown on the upper surface of the paper.

For a smaller opening, the flow of fluid is more than when it is bigger. When we try to close the tap with our fingers, water gushes through the small openings. Area and velocity are inversely proportional to each other.

Small opening of a syringe needle controls the velocity of the blood oozing out. At the constriction point of the syringe system, the flow ra

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10.3 Fill in the blanks using the word(s) from the list appended with each statement:

(a) Surface tension of liquids generally . . . with temperatures (increases / decreases)

(b) Viscosity of gases . . . with temperature, whereas viscosity of liquids . . . with temperature (increases / decreases)

(c) For solids with elastic modulus of rigidity, the shearing force is proportional to . . . , while for fluids it is proportional to. . (shear strain / rate of shear strain)

(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli's principle)

(e) For

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The surface tension of a liquid is inversely proportional to temperature. Decreases

Most fluids offer resistance to their motion. It is like internal mechanical friction, known as viscosity. Gas viscosity increases with temperature, whereas liquid viscosity decreases with temperature. Because, intermolecular forces weaken with temperature increase, viscosity decreases.

With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

For a steady-flowing fluid, an inc

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11.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a 'thermal stress' developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 * 10^–5 K^–1, steel = 1.2 * 10^–5 K^–1 ).

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11.10 Initial temperature, $T_{1}$ = 40.0°C, Final temperature, $T_{2}$ = 250°C, $?$ T = $T_{2}$ - $T_{1}$ = 210°C

Initial length of the brass rod at $T_{1}$ , $l_{b}$ = 50 cm, Initial diameter of the brass rod at $T_{1}$ , $d_{1}$ = 3 mm

Length of the steel rod $l_{s} = 50 c m$

For the expansion of the brass rod, we have:

$\frac{C h a n g e i n l e n g t h (? l_{b)}}{O r i g i n a l l e n g t h (l_{b)}}$ = $α_{b} ? T$ , then $? l_{b}$ = 50 $* 2.0 * 10^{- 5} * 210$ = 0.21 cm

For the expansion of the steel rod, we have:

$\frac{C h a n g e i n l e n g t h (? l_{s)}}{O r i g i n a l l e n g t h (l_{s})}$ = $α_{b} ? T$ , then $? l_{s}$ = 50 $* 1.2 * 10^{- 5} * 210$ = 0.126 cm

Total change in length = 0.21 + 0.126 = 0.336 cm

Since the rods are free at the end, no thermal stress developed.

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10.2 Explain why

(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute

(b) 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)

(c) Surface tension of a liquid is independent of the area of the surface

(d) Water with detergent dissolved in it should have small angles of contact

(e) A drop of liquid under no external forces is always spherical in shape

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The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact ( $θ)$ , as shown in the diagram

$S_{l a}$ = Interfacial tension between liquid-air interface

$S_{s l}$ = Interfacial tension between solid -liquid interface

$S_{s a}$ = Interfacial tension between solid-air interface

At the line of contact of contact, the surface forces between the three media must be in equilibrium. Hence

$\cos ? θ$ = $\frac{S_{s a} - S_{l a}}{S_{l a}}$

The angle of contact $θ$ is obtuse, if $S_{s a} < S_{l a}$ , as in the case of mercury on glass

This angle is acute if $S_{s l} < S_{l a}$ , as in the case of water on glass

Mercury molecules (which mak

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The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact ( $θ)$ , as shown in the diagram

$S_{l a}$ = Interfacial tension between liquid-air interface

$S_{s l}$ = Interfacial tension between solid -liquid interface

$S_{s a}$ = Interfacial tension between solid-air interface

At the line of contact of contact, the surface forces between the three media must be in equilibrium. Hence

$\cos ? θ$ = $\frac{S_{s a} - S_{l a}}{S_{l a}}$

The angle of contact $θ$ is obtuse, if $S_{s a} < S_{l a}$ , as in the case of mercury on glass

This angle is acute if $S_{s l} < S_{l a}$ , as in the case of water on glass

Mercury molecules (which makes an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction towards solids. Hence, they tend to form drops

Water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction towards solids. Hence, they tend to spread out

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 independent of the area of the liquid surface.

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

A liquid tend 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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11.9 A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 * 10^–5 K–1; Young's modulus of brass = 0.91 * 10¹¹ Pa.

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11.9 Initial temperature, $T_{1}$ = 27°C, Length of the wire $l_{1}$ at $T_{1}$ = 1.8 m

Final temperature, $T_{2}$ = -39°C

Diameter of the wire, d = 2.0 mm = 2 $* 10^{- 3}$ m

Coefficient of linear expansion of brass, $α = 2 * 10^{- 5}$ /K

Youngs' modulus of brass, Y = 0.91 $* 10^{11}$ Pa

Let the tension developed be F

We know Youngs' modulus = $\frac{S t r e s s}{S t r a i n}$ = $\frac{\frac{F}{A}}{\frac{? L}{L}} = Y$

Y $* \frac{? L}{L}$ = $\frac{F}{A}$ or F = $\frac{A Y ? L}{L}$

Here, A = cross-sectional area of the wire = $\frac{π}{4} d^{2}$ = $\frac{π}{4} {(2 * 10^{- 3})}^{2}$ $m^{2}$ = 3.1416 $* 10^{- 6} m^{2}$

Now $? L$ can be written as $? L$ = $α L (T_{2} - T_{1})$ = ( $2 * 10^{- 5}) * L * (- 39 - 27)$ = -1.32 $* 10^{- 3}$ L

Substituting all values, we get

F = (3.1416 $* 10^{- 6} * 0.91 * 10^{11} * (- 1.32 * 10^{- 3})$ = -377.37 N

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11.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C?

Coefficient of linear expansion of copper = 1.70 * 10^–5 K^–1.

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11.8 Initial temperature, $T_{1}$ = 27.0 °C, Initial diameter of the hole, $d_{1}$ = 4.24 cm

Final temperature, $T_{2}$ = 227.0 °C, Final diameter of the hole $= d_{2}$

Coefficient of linear expansion of copper, $α_{c o p p e r}$ = 1.70 $* 10^{- 5}$ K–1

We know

$\frac{C h a n g e i n a r e a (? A)}{O r i g i n a l a r e a (A)}$ = $β ?$ T where $β$ is the coefficient of superficial expansion, $β = 2 α_{c o p p e r}$

$\frac{(π \frac{d_{2}^{2}}{4} - π \frac{d_{1}^{2}}{4})}{π \frac{d_{1}^{2}}{4}}$ = $2 α_{c o p p e r} ?$ T

$\frac{d_{2}^{2} - d_{1}^{2}}{d_{1}^{2}}$ = $2 α_{c o p p e r} ?$ T

$\frac{d_{2}^{2}}{d_{1}^{2}}$ - 1= $2 α_{c o p p e r} (T_{2} - T) =$ 2 $*$ 1.70 $* 10^{- 5}$ (227-27) = 6.8 $* 10^{- 3}$

$d_{2}^{2} = 1.0068 * 4.24 * 4.24$

$d_{2}$ = 4.2544 cm

So change in diameter = 4.2544 – 4.24 = 0.01439 cm

Diameter increase by 1.44 $* 10^{- 2}$ cm

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10.1 Explain why

(a) The blood pressure in humans is greater at the feet than at the brain

(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km

(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area

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The pressure of a liquid is given by the following relation:

P = h $ρ$ g, where P = Pressure, h = height of the liquid column, $ρ$ = is the density of the liquid and g= acceleration due to gravity

From the above relation, because of the h factor (height of the human body), the pressure is more at the feet and less at the brain

The said phenomenon is due to the $ρ$ factor. Density of air is maximum at the sea level. At height, density decreases and pressure also decreases. At 6 km height, the density of air is nearly half of that of a t sea level

When pressure is applied on the liquid, the pressure is transmitted in all dir

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The pressure of a liquid is given by the following relation:

P = h $ρ$ g, where P = Pressure, h = height of the liquid column, $ρ$ = is the density of the liquid and g= acceleration due to gravity

From the above relation, because of the h factor (height of the human body), the pressure is more at the feet and less at the brain

When pressure is applied on the liquid, the pressure is transmitted in all directions in the liquid, Hence in absence of fixed direction, it is a scalar quantity

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11.7 A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : $α_{s t e e l}$ = 1.20* $10^{- 5}$ $K^{- 1}$

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11.7 Given, temperature $T_{1}$ = 27 °C = 27 + 273.16 K = 300.16 K

Outer dia of the shaft at temp $T_{1}$ , $d_{1}$ = 8.7 cm

Diameter of the central hole of the wheel, $d_{2}$ = 8.69 cm

The change in diameter, Δd= $d_{2} - d_{1} =$ 8.69 – 8.7 = $-$ 0.01 cm

After the shaft is cooled in dry ice, its temperature becomes $T_{2}$ . It can be calculated from the relation

Δd= $d_{1} * α_{s t e e l}$ ( $T_{2} - T_{1})$

-0.01 = 8.7 $* 1.20 * 10^{- 5} (T_{2} - 300)$

$T_{2} - 300$ = -95.78

$T_{2}$ = 204.22 K = -68.94 $?$

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11.6 A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0 °C ? Coefficient of linear expansion of steel = 1.20 * 10^–5 K^–1 .

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