Class 11th

11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^–1 K^–1; heat of fusion of water = 335 J g^–1 ).

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11.13 Mass of the copper block, m = 2.5 kg = 2.5 $* 10^{3}$ gm

Rise in temperature of the copper block, $?$ T = 500°C

Specific heat of the copper, C = 0.39 g–1 K–1

Heat of fusion of water, L = 335 J g–1

The maximum heat the copper block can lose, Q = mc $?$ T = 2.5 $* 10^{3} * 0.39 * 500$ = 487500 J

Let $m_{1}$ gm be the mass of the ice, which will melt because of the copper block.

Heat gained by ice block = Q = $m_{1} L$

$m_{1} = \frac{Q}{L}$ = $\frac{487500}{335}$ g = 1455.22 gm = 1.45 kg

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10.6 Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^–3. Determine the height of the wine column for normal atmospheric pressure.

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Density of mercury, $ρ_{1}$ = 13.6 $* 10^{3}$ kg/ $m^{3}$

Density of wine, $ρ_{2}$ = 9.84 $* 10^{2}$ kg/ $m^{3}$

Height of the mercury column for atmospheric pressure, $h_{1}$ = 760 mm = 0.76 m

Height of the mercury column for atmospheric pressure = $h_{2}$

From the relation, P = $ρ g h$ , since the pressure on both the system are equal

$ρ_{1} g h_{1}$ = $ρ_{2} g h_{2}$ , we get $h_{2}$ = $\frac{ρ_{1} g h_{1}}{ρ_{2} g}$ = $\frac{13.6 * 10^{3} * 0.76}{9.84 * 10^{2}}$ = 10.5 m

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11.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g^–1 K^–1.

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11.12 Power of the drilling machine, P = 10 kW= 10 $* 10^{3}$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $* 10^{3}$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $* 60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $δ$ T

Total energy consumed by the drilling machine= P $* t$ = 10 $* 10^{3} * 150$ J = 1.5 $* 10^{6}$ J

It is given 50% of energy is useful.

So useful energy, $? Q$ = 50% of Pt= 0.5 $*$ 1.5 $* 10^{6} =$ 7.5 $* 10^{5}$ J

We know, $? Q$ = mc $?$ T or T = $\frac{? Q}{m c}$ = $\frac{7.5 * 10^{5}}{8 * 10^{3} * 0.91}$ = 103 $?$

Therefore 2.5 minute drilli

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11.12 Power of the drilling machine, P = 10 kW= 10 $* 10^{3}$ W

Mass of the Aluminium block, m = 8.0 kg = 8 $* 10^{3}$ gm

Time for which the machine is used, t = 2.5 minute = 2.5 $* 60$ s = 150 s

Specific heat of Aluminium, c = 0.91 J/gK

Let the rise of temperature in the block after drilling be $δ$ T

Total energy consumed by the drilling machine= P $* t$ = 10 $* 10^{3} * 150$ J = 1.5 $* 10^{6}$ J

It is given 50% of energy is useful.

So useful energy, $? Q$ = 50% of Pt= 0.5 $*$ 1.5 $* 10^{6} =$ 7.5 $* 10^{5}$ J

We know, $? Q$ = mc $?$ T or T = $\frac{? Q}{m c}$ = $\frac{7.5 * 10^{5}}{8 * 10^{3} * 0.91}$ = 103 $?$

Therefore 2.5 minute drilling will raise the block temperature by 103 $?$

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10.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

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Mass of the girl, m = 50 kg

Diameter of the heel, d = 1.0 cm = 0.01 m, Radius of the heel, r = d/2 = 0.005m

Area of the heel, $π r^{2}$ = 7.85 $* 10^{- 5}$ $m^{2}$

Force exerted by heel on the floor, F = mg = 50 $* 9.8$ N = 490 N

Pressure exerted by heel on the ground, p = F/A = 490/ (7.85 $* 10^{- 5})$ N/ $m^{2}$

= 6.24 $* 10^{6}$ N/ $m^{2}$

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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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