
Chapter 23 – Heat and Temperature
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Question 1:
If two bodies are in thermal equilibrium in one frame, will they be in thermal equilibrium in all frames?
Answer:
If two bodies are in thermal equilibrium in one frame, they will be in thermal equilibrium in all the frames. In case there is any change in temperature of one body due to change in frame, the same change will be acquired by the other body.
Question 2:
Does the temperature of a body depend on the frame from which it is observed?
Answer:
No, the temperature of a body is not dependent on the frame from which it is observed. This is because atoms /molecules of matter move or vibrate in all possible directions. Increase in velocity at a particular direction of the container/ matter does not increase or decrease the overall velocity of the molecules/atoms because of the random collisions the entities suffer. So, there is no net rise in temperature of the system.
Question 3:
It is said that mercury is used in defining the temperature scale because it expands uniformly with temperature. If the temperature scale is not yet defined, is it logical to say that a substance expands uniformly with temperature?
Answer:
It is not illogical to say that mercury expands uniformly before temperature scale was defined. It’s uniform expansion can be studied by comparing the expansion of mercury with expansion of other substances (like alcohol water etc).
Question 4:
In defining the ideal gas temperature scale, it is assumed that the pressure of the gas at constant volume is proportional to the temperature T. How can we verify whether this is true or not? Do we have to apply the kinetic theory of gases? Do we have to depend on experimental result that the pressure is proportional to temperature?
Answer:
The ideal gas thermometer is based on the ideal gas equation, PV=nRT, where P is pressure of the gas at constant volume V with n number of moles at temperature T. Therefore, P = constant
×T. According to this relation, if the volume of the gas used is constant, the pressure will be directly proportional to the temperature of the gas. We need not use kinetic theory of gases or any experimental results.
Question 5:
Can the bulb of a thermometer be made of an adiabatic wall?
Answer:
The bulb of a thermometer plays an important role in measuring the temperature of the surrounding body. It is put in contact with the body whose temperature is to be measured. The bulb attains the temperature of the body, which allows calibration of temperature. If the bulb is made of an adiabatic wall, then no heat will be transferred through the wall and the bulb cannot attain thermal equilibrium with the surrounding body. Therefore, the bulb cannot be made of an adiabatic wall.
Question 6:
Why do marine animals live deep inside a lake when the surface of the lake freezes?
Answer:
Water possesses an anomalous behavour. The volume of a given amount of water decreases as it is cooled from room temperature, until its temperature reaches 4 °C. Below 4 °C, the volume increases, and therefore the density decreases.
When the temperature of the surface of lake falls in winter, the water at the surface becomes denser and sinks. As, the temperature reaches below 4^{ o}C , the density of the water at surface becomes less. Thus, it remains at surface and freezes. As, the ice is a bad conductor of heat, it traps the heat present in the lake’s water beneath itself. Hence, no further cooling of water takes place once the top layer of the lake is completely covered by ice. Thus the life of the marine animals inside the lake is possible.
Question 7:
The length of a brass rod is found to be less on a hot summer day than on a cold winter day as measured by the same aluminium scale. Can we conclude that brass shrinks on heating?
Answer:
On a hot summer day, metals tend to expand due to the heat. Different metals have different expansion coefficients. The coefficient of linear expansion of aluminium is more than that of brass. Therefore, it’ll expand more than brass, leading to an apparent decrease in length of the brass rod, as measured by the aluminium scale. So, we cannot conclude that brass shrinks on heating. Instead, aluminium expands more than brass on heating.
Question 8:
If mercury and glass had equal coefficients of volume expansion, could we make a mercury thermometer in a glass tube?
Answer:
Yes, we can make a mercury thermometer in a glass tube. Mercury and glass have equal coefficients of volume expansion. So, when temperature changes, the increase in the volume of the glass tube as which is equal to the real increase in volume minus the increase in the volume of the container, would be zero. Hence, it will give correct reading at every temperature.
Question 9:
The density of water at 4°C is supposed to be 1000 kg m^{–3}. Is it same at sea level and at high altitude?
Answer:
At sea level, the pressure is around 1 atmosphere and at high altitude, the density of air reduces.
Pressure of liquid,
P=hρg,where ρ=density of fluidThe above equation shows that pressure depends on density. Therefore at 4^{o}^{â€‹}C, the density of water will be less at high altitude, compared to the density at sea level.
Question 10:
A tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time. Explain.
Answer:
When a bottle with a tightlyclosed metal lid is put in hot water for sometime, its lid can be opened easily because metals have greater coefficient of expansion than glass. Therefore, when the metal lid comes in contact with hot water, it’ll expand more than the glass container. As a result, it will be easier to open the bottle.
Question 11:
If an automobile engine is overheated, it is cooled by pouring water on it. It is advised that the water should be poured slowly with the engine running. Explain the reason.
Answer:
In a hot engine the hot parts are expanded because of heat, if cold water is poured suddenly then there will be uneven thermal contraction in the parts. This will result in a stress to develop between the various parts of the engine and may let the engine to crack down.
Question 12:
Is it possible for two bodies to be in thermal equilibrium if they are not in contact?
Answer:
Two bodies are said to be in thermal equilibrium if they are at the same temperature. Consider two bodies A and B that are not in contact with each other but in contact with a heat reservoir. Since both the bodies will attain the temperature of the reservoir, they will be at the same temperature and, hence, in thermal equilibrium. Therefore, it is possible to have two bodies in thermal equilibrium even though they are not in contact.
Question 13:
A spherical shell is heated. The volume changes according to the equation V_{θ} = V_{0} (1 + γ^{θ}). Does the volume refer to the volume enclosed by the shell or the volume of the material making up the shell?
Answer:
When a spherical shell is heated, its volume changes according to the equation,
Vθ=V01+γ∆θ. The volume referred to here is the volume of the material used to make up the shell, as its volume expands with the rise of temperature with coefficient of expansion of volume,
γ.
Question 1:
A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z
(a) must be in thermal equilibrium
(b) cannot be in thermal equilibrium
(c) may be in thermal equilibrium
Answer:
(c) may be in thermal equilibrium
The given data in the question is insufficient to specify the relation between the physical conditions of systems Y and Z. As system X is not in thermal equilibrium with Y and Z, systems Y and Z may be at the same temperature or they may or may not be in thermal equilibrium with each other. So, the only possible option is (c).
Question 2:
Which of the curves in the figure (23Q1) represents the relation between Celsius and Fahrenheit temperatures?
Faigure
Answer:
(a)
Celsius and Fahrenheit temperatures are related in the following way:
C=59F1609Here, F = temperature in Fahrenheit
C = temperature in Celsius
If this equation is plotted on the graph, then the curve will be represented by curve ‘a’ lying in the fourth quadrant with slope 5/9.
So, the correct option is (a).
Question 3:
Which of the following pairs may give equal numerical values of the temperature of a body?
(a) Fahrenheit and Kelvin
(b) Celsius and Kelvin
(c) Kelvin and Platinum
Answer:
(a) Fahrenheit and Kelvin
Let θ be the temperature in Fahrenheit and Kelvin scales.
We know that the relation between the temperature in Fahrenheit and Kelvin scales is given by
TF32180=TK273.15100T_{F} = T_{K} = θ
Therefore,
θ32180=θ273.151005θ160=9θ2458.54θ=2298.35θ=574.59 oIf we consider the same for Celsius and Kelvin scales
TC0100=TK273.15100Let the temperature be t
t0100=t273.15100t=t273.15Thus, t does not exist.
The Kelvin scale uses mercury as thermometric substance, whereas the platinum scale uses platinum as thermometric substance. The scale depends on the properties of the thermometric substance used to define the scale. The platinum and Kelvin scales do not agree with each other. Therefore, there is no such temperature that has same numerical value in the platinum and Kelvin scale.
Question 4:
For a constantvolume gas thermometer, one should fill the gas at
(a) low temperature and low pressure
(b) low temperature and high pressure
(c) high temperature and low pressure
(d) high temperature and high pressure
Answer:
(c) high temperature and low pressure.
A constantvolume gas thermometer should be filled with an ideal gas in which particles don’t interact with each other and are free to move anywhere, so that the thermometer functions properly. An ideal gas is only a theoretical possibility. Therefore, the gas that is filled in the thermometer should be at high temperature and low pressure, as under these conditions, a gas behaves as an ideal gas.
Question 5:
Consider the following statements.
(A) The coefficient of linear expansion has dimension K^{–1}.
(B) The coefficient of volume expansion has dimension K^{–1}.
(b) A is correct but B is wrong.
(c) B is correct but A is wrong.
(d) A and B are wrong.
Answer:
(a) A and B are correct.
The coefficient of linear expansion,
α=1L△L△T
=LLT=K1Here, L = initial length
△L = change in length
△T = change in temperature
On the other hand, the coefficient of volume expansion,
γ=1V△V△T=L3L3T=K1Here, V = initial volume
△V = change in volume
△T = change in temperature
K = kelvin, the S.I. unit of temperature
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Question 6:
A metal sheet with a circular hole is heated. The hole
(a) gets larger
(b) gets smaller
(c) retains its size
(d) is deformed
Answer:
(a) gets larger
When a metal sheet is heated, it starts expanding and its surface area will start increasing, which will lead to an increase in the radius of the hole. Hence, the circular hole will become larger.
Question 7:
Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a bimetallic strip (α_{copper} > α_{steel}). On heating, this strip will
(a) remain straight
(b) bend with copper on convex side
(c) bend with steel on convex side
(d) get twisted
Answer:
(b) bend with copper on convex side
We are provided with two metal strips of copper and steel. On heating, both of them will expand. Expansion coefficient of copper is more than that of steel. So, the copper metal strip will expand more, causing the bimetallic strip to bend with copper at the convex side, as it’ll have more surface area compared to the steel sheet, which will be on the concave side.
Question 8:
If the temperature of a uniform rod is slightly increased by âˆ†t, its moment of inertia I about a perpendicular bisector increases by
(a) zero
(b) αIâˆ†t
(c) 2αIâˆ†t
(d) 3αIâˆ†t.
Answer:
(c) 2αIâˆ†t
The change in moment of inertia of uniform rod with change in temperature is given by,
I’=I(1+2αΔt)Here, I = initial moment of inertia
I’ = new moment of inertia due to change in temperature
α= expansion coefficient
∆t = change in temperature
So,
I’I=2αIΔt
Question 9:
If the temperature of a uniform rod is slightly increased by âˆ†t, its moment of inertia I about a line parallel to itself will increase by
(a) zero
(b) αIâˆ†t
(c) 2αIâˆ†t
(d) 3αIâˆ†t.
Answer:
(c) 2αIâˆ†t
The moment of inertia of a solid body of any shape changes with temperature as
I’=I1+2αΔtHere, I = initial moment of inertia
I’ = new moment of inertia due to change in temperature
α = expansion coefficient
Δt = change in temperature
So,
I’I=2αIΔtI−I0=2αΔt
Question 10:
The temperature of water at the surface of a deep lake is 2°C. The temperature expected at the bottom is
(a) 0 °C
(b) 2 °C
(c) 4 °C
(d) 6 °C
Answer:
(c) 4 ^{o}C
The density of water is maximum at 4 ^{o}C, and the water at the bottom of the lake is most dense, compared to the layers of water above. Therefore, the temperature expected at the bottom is 4^{o}C.
Question 11:
An aluminium sphere is dipped into water at 10°C. If the temperature is increased, the force of buoyancy
(a) will increase
(b) will decrease
(c) will remain constant
(d) may increase or decrease depending on the radius of the sphere
Answer:
(b) will decrease
When an aluminium sphere is dipped in water and the temperature of water is increased, the aluminium will start expanding leading to increase in its volume. This will lead to increase in the surface area of the shell and it’ll exert less pressure on the water such that the volume of the sphere submerged in water will decrease and it’ll start float easily on water. Now, the volume of water displaced will be less compared to what was displaced initially. Therefore, the force of buoyancy will decrease, as it is directly proportional to the volume of water displaced.
Question 1:
A spinning wheel is brought in contact with an identical wheel spinning at identical speed. The wheels slow down under the action of friction. Which of the following energies of the first wheel decreases?
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal
Answer:
(a) Kinetic
(c) Mechanical
The kinetic energy of a body depends on its speed. Since when a spinning wheel is slowed down, its speed decreases leading to reduction in its kinetic energy. The mechanical energy of a body is defined as the sum of its potential and kinetic energies. Since the kinetic energy of the wheel has been decreased, it’ll lead to decrease in its mechanical energy. When the wheel slows down due to friction, its mechanical energy gets converted into heat energy, leading to increase in internal energy, which increases with increase in temperature.
Question 2:
A spinning wheel A is brought in contact with another wheel B, initially at rest. Because of the friction at contact, the second wheel also starts spinning. Which of the following energies of the wheel B increases?
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal
Answer:
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal
When the wheel B starts spinning because of the friction at contact, it will gain kinetic energy and, hence, mechanical energy (kinetic + potential energies). Also, internal energy will increase, which increases with rise in temperature. Along with it, the generation of heat energy due to friction will lead to increase in the net sum of all the energies, i.e. total energy.
Question 3:
A body A is placed on a railway platform and an identical body B in a moving train. Which of the following energies of B are greater than those of A, as seen from the ground?
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal
Answer:
(a) Kinetic
(b) Total
(c) Mechanical
As body A is at rest on the ground, it possesses only potential energy, whereas body B, being placed inside a moving train, possesses kinetic energy due to its motion along with the train. Therefore, body B will have greater kinetic, mechanical (energy possessed by the body by virtue of its position and motion = kinetic energy+potential energy) energy and, hence, total (sum of all the energies) energy. No information is given about the temperature of the body so we can not say wheather body B^{‘} s internal energy will be or will not be greater than that of body A.
Question 4:
In which of the following pairs of temperature scales, the size of a degree is identical?
(a) Mercury scale and ideal gas scale
(b) Celsius scale and mercury scale
(c) Celsius scale and ideal gas scale
(d) Ideal gas scale and absolute scale
Answer:
(c) Celsius scale and ideal gas scale
(d) Ideal gas scale and absolute scale
Celsius scale and ideal gas scale measure temperature in kelvin (K) and the ideal gas scale is sometimes also called the absolute scale. A mercury scale gives reading in degrees and its size of degree, which depends on length of mercury column, doesn’t match any of the abovementioned scales.
Question 5:
A solid object is placed in water contained in an adiabatic container for some time. The temperature of water falls during this period and there is no appreciable change in the shape of the object. The temperature of the solid object
(a) must have increased
(b) must have decreased
(c) may have increased
(d) may have remained constant
Answer:
(a) must have increased.
The whole system (water + solid object) is enclosed in an adiabatic container from which no heat can escape. After some time, the temperature of water falls, which implies that the heat from the water has been transferred to the object, leading to increase in its temperature.
Question 6:
As the temperature is increased, the time period of a pendulum
(a) increases proportionately with temperature
(b) increases
(c) decreases
(d) remains constant
Answer:
(b) increases
In general, the time period of a pendulum,t, is given by
t=12πlg.
When the temperature (T) is increased, the length of the pendulum (l) is given by,
l=l0(1+αT),
where l_{0} = length at 0 ^{o}C
α= linear coefficient of expansion.
Therefore, the time period of a pendulum will be
t=12πl0(1+αT)g
Hence, time period of a pendulum will increase with increase in temperature.
Question 1:
The steam point and the ice point of a mercury thermometer are marked as 80° and 20°. What will be the temperature on a centigrade mercury scale when this thermometer reads 32°?
Answer:
Given:
Ice point of a mercury thermometer, T_{0} = 20° C
Steam point of a mercury thermometer, T_{100} = 80° C
Temperature on thermometer that is to be calculated in centigrade scale, T_{1} = 32° Câ€‹
Temperature on a centigrade mercury scale, T, is given as:
T=T1T0T100T0×100⇒T =32208020×100⇒T =1260×100⇒T=1206⇒T=20° CTherefore, the temperature on a centigrade mercury scale will be 20^{o} C.
Question 2:
A constantvolume thermometer registers a pressure of 1.500 × 10^{4} Pa at the triple point of water and a pressure of 2.050 × 10^{4 }Pa at the normal boiling point. What is the temperature at the normal boiling point?
Answer:
Given:
Pressure registered by a constantvolume thermometer at the triple point, P_{tr} = 1.500 × 10^{4 } Pa
Pressure registered by the thermometer at the normal boiling point, P = 2.050 × 10^{4} Pa
We know that for a constantvolume gas thermometer, temperature (T) at the normal boiling point is given as:
T=PPtr×273.16 K⇒T=2.050×1041.500×104×273.16 K⇒T=373.31 KTherefore, the temperature at the normal point (T) is 373.31 K.
Question 3:
A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water, find the melting point of lead.
Answer:
Given:
In a gas thermometer, the pressure measured at the melting point of lead, P = 2.20 × Pressure at triple point(P_{â€‹tr})
So the melting point of lead,(T) is given as:
T=PPtr×273.16 K⇒T=2.20×PtrPtr×273.16 K⇒T=2.20×273.16 K⇒T=600.952 K⇒T ≃601 KTherefore, the melting point of lead is 601 K.
Question 4:
The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100°C)?
Answer:
Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, P_{tr} = 40 kPa = 40 × 10^{3} Pa
Boiling point of water, T = 100°C = 373.16 K
Let the pressure measured at the boiling point of water be P.
â€‹For a constant volume gas thermometer, temperaturepressure relation is given below:
T=PPtr×273.16 K⇒P=T×Ptr273.16⇒P=373.16×40×103273.16⇒P=54643 Pa⇒P=54.6×103 Pa ⇒P≃55 kPaTherefore, the pressure measured at the boiling point of water is 55 kPa.
Question 5:
The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point.
Answer:
Given:
â€‹Temperature of ice point, T_{1} = 273.15 K
Temperature of steam point, T_{2} = 373.15 K
Pressure of the gas in a constant volume thermometer at the ice point, P_{1}_{â€‹} = 70 kPa,
Let P_{tr} be the pressure at the triple point and P_{2} be the pressure at the steam point.
The temperaturepressure relations for ice point and steam point are given below:
For ice point,
T1=P1Ptr×273.16 K
⇒273.15=70Ptr×103×273.16⇒Ptr=70×273.16×103273.15 PaFor steam point,
T2=P2×273.16Ptr KOn substituting the value of P_{tr} ,we get:
373.15=P2×273.15×273.1670×273.16×103 ⇒P2=373.15×70×103273.15⇒P2=95.626×103 Pa⇒P2≃96 kPaTherefore, the pressure at steam point is 96 kPa.
Question 6:
The pressures of the gas in a constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath, respectively. Find the temperature of the wax bath.
Answer:
Given:
In a constant volume gas thermometer,
Pressure of the gas at the ice point, P_{â€‹â€‹â€‹0} = 80 cm of Hgâ€‹
Pressure of the gas at the steam point, P_{100} = 90 cm of Hg
â€‹Pressure of the gas in a heated wax bath, P = 100 cm of Hg
The temperature of the wax bath
Tis given by:
T=PP0P100P0×100° C⇒ T=100809080×100⇒ T =2010×100⇒ T =200° CTherefore, the temperature of the wax bath is 200^{o} C.
Question 7:
In a Callender’s compensated constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel.
Answer:
Given:
Volume of the bulb in a Callender’s compensated constant pressure air thermometer, (V) =
1800 cc
Volume of mercury that has to be poured out, V’ = 200 cc
Temperature of ice bath, T_{o} = 273.15 K
â€‹So the temperature of the vessel(T’) is given by:
T’=VVV’×T0⇒T’=18001600×273.15 K⇒T’=307.293⇒T’≃307 KTherefore, the temperature of the vessel is 307 K.
Question 8:
A platinum resistance thermometer reads 0° when its resistance is 80 Ω and 100° when its resistance is 90 Ω.
Find the temperature at the platinum scale at which the resistance is 86 Ω.
Answer:
Given:
Resistance at 0^{o}C, R_{0} = 80
Ω
Resistance at 100^{o}C, R_{100} = 90
ΩLet t be the temperature at which the resistance (R_{t}) is 86
Ω.
t=RtR0R100R0×100⇒t=86809080×100⇒t=610×100⇒t=60°Therefore, the resistance is 86
Ωat 60^{o}C.
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Question 9:
A resistance thermometer reads R = 20.0 Ω, 27.5 Ω, and 50.0 Ω at the ice point (0°C), the steam point (100°C) and the zinc point (420°C), respectively. Assuming that the resistance varies with temperature as R_{θ} = R_{0} (1 + αθ + βθ^{2}), find the values of R_{0}, α and β. Here θ represents the temperature on the Celsius scale.
Answer:
Given:
Reading on resistance thermometer at ice point, R_{0} = 20 Ω
Reading on resistance thermometer at steam point, R_{100} = 27.5 Ω
Reading on resistance thermometer at zinc point, R_{420} = 50 Ω
The variation of resistance with temperature in Celsius scale,θ, is given as:
R100=R01+α θ+βθ2⇒R100=R0+R0αθ+R0βθ2⇒R100=R0+R0αθ+R0βθ2⇒R100R0R0=αθ+βθ2⇒27.52020=αθ+βθ2⇒7.520=α×100+β×10000 …iAlso,R420=R01+αθ+βθ2⇒R420=R0+R0αθ+R0βθ2⇒R420R0R0=αθ+βθ2⇒502020=420α+176400 β⇒32=420α+176400 β …iiSolving (i) and (ii), we get:
α = 3.8 ×10^{–3}°C^{â€‹1}
β = –5.6 ×10^{–7}°C^{1}
Therefore, resistance R_{0 }is 20 Ω and the value of α is 3.8 ×10^{–3}°C^{â€‹1 }and that of β is –5.6 ×10^{–7}°C^{1}.
Question 10:
A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concrete is 1.0 × 10^{–5} °C^{–1}.
Answer:
Given:
Length of the slab when the temperature is 0°C, L_{0} = 10 m
Temperature on the summer day, t = 35 °C
Let L_{1} be the length of the slab on a summer day when the temperature is 35°C.
The coefficient of linear expansion of concrete, α = 1 ×10^{–5} °C^{â€‹}^{1}
L1=L01+αt
⇒L1= 10 (1 + 10^{–5} × 35)
⇒L1 = 10 + 35 × 10^{–4}
⇒L1 = 10.0035 m
â€‹So, the length of the slab on summer day when the temperature is 35^{o}C is 10.0035 m.
Question 11:
A metre scale made of steel is calibrated at 20°C to give correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10^{–5} °C^{–1}.
Answer:
Given:
Temperature at which the steel metre scale is calibrated, t_{1} = 20^{o}C
Temperature at which the scale is used, t_{2} = 10^{o}C
So, the change in temperature,
Δt = (20^{o}
10^{o}) C
The distance to be measured by the metre scale, L_{o} = (51
50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel,
αsteel= 1.1 × 10^{–5} °C^{–1}
Let the new length measured by the scale due to expansion of steel be L_{â€‹2}, Change in length is given by,
∆L=L1αsteel Δt⇒∆L=1×1.1×10–5×10⇒∆L=0.00011 cmAs the temperature is decreasing, therefore length will decrease by
∆L.
Therefore, â€‹the new length measured by the scale due to expansion of steel (L_{2}) will be,
L_{2} = 1 cm
0.00011 cm = 0.99989 cm
Question 12:
A railway track (made of iron) is laid in winter when the average temperature is 18°C. The track consists of sections of 12.0 m placed one after the other. How much gap should be left between two such sections, so that there is no compression during summer when the maximum temperature rises to 48°C? Coefficient of linear expansion of iron = 11 × 10^{–6} °C^{–1}.
Answer:
Given:
Length of the iron sections when there’s no effect of temperature on them, L_{o} = 12.0 m
â€‹Temperature at which the iron track is laid in winter, t_{â€‹w}â€‹ = 18^{ o}C
Maximum temperature during summers, t_{s} = 48^{ o}C
Coefficient of linear expansion of iron,
α= 11 × 10^{–6} °C^{–1}
Let the new lengths attained by each section due to expansion of iron in winter and summer be L_{w} and L_{s}_{,} respectively, which can be calculated as follows:
Lw=L01+αtw⇒Lw=12 1+11×106×18⇒Lw=12.00237 m Ls=L0 1+α ts⇒Ls=12 1+11×106×48⇒Ls=12.006336 m∴ ∆L=LsLw⇒ΔL=12.00633612.002376⇒ΔL=0.00396 m⇒ΔL≈0.4 cmTherefore, the gap (
ΔL) that should be left between two iron sections, so that there is no compression during summer, is 0.4 cm.
Question 13:
A circular hole of diameter 2.00 cm is made in an aluminium plate at 0°C. What will be the diameter at 100°C? α for aluminium = 2.3 × 10^{–5} °C^{–1}.
Answer:
Given:
Diameter of a circular hole in an aluminium plate at 0°C, d_{1} = 2 cm = 2 × 10^{–2} m
Initial temperature, t_{1} = 0 °C
Final temperature, t_{2} = 100 °C
So, the change in temperature, (
Δt) = 100°C – 0°C = 100°C
The linear expansion coefficient of aluminium, α_{al}_{â€‹} = 2.3 × 10^{–5} °C^{–1}
Let the diameter of the circular hole in the plate at 100^{o}C be d_{2} , which can be written as:
d2=d11+αΔt
⇒d2= 2 × 10^{–2 }(1 + 2.3 × 10^{–5 }× 10^{2})
⇒d2= 2 × 10^{–2} (1 + 2.3 × 10^{–3})
⇒d2= 2 × 10^{–2} + 2.3 × 2 × 10^{–5}
⇒d2= 0.02 + 0.000046
⇒d2= 0.020046 m
⇒d2≈ 2.0046 cm
Therefore, the diameter of the circular hole in the aluminium plate at 100^{o}C is â€‹2.0046 cm.
Question 14:
Two metre scales, one of steel and the other of aluminium, agree at 20°C. Calculate the ratio aluminiumcentimetre/steelcentimetre at (a) 0°C, (b) 40°C and (c) 100°C. α for steel = 1.1 × 10^{–5} °C^{–1} and for aluminium = 2.3 × 10^{–5}°C^{–1}.
Answer:
Given:
At 20°C, length of the metre scale made up of steel, L_{st}= length of the metre scale made up of aluminium, L_{al}â€‹
Coefficient of linear expansion for aluminium, α_{al} = 2.3 × 10^{–5 }°C^{1}
Coefficient of linear expansion for steel, α_{st} = 1.1 × 10^{–5}^{ }°C^{1}
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L_{0al}â€‹,_{ }Lâ€‹_{4}_{0al}_{ }and L_{10}_{â€‹0al}.
And let the length of the steel scale at 0°C, 40°C and 100°C be L_{0st}â€‹,_{ }Lâ€‹_{4}_{0st }and L_{10}_{â€‹0st}.
(a) So, L_{0st}(1 – α_{st} × 20) = L_{0al}(1 – α_{al} × 20)
L0stL0al=1αal×201αst×20⇒L0stL0al=12.3×105×2011.1×105×20⇒L0stL0al=0.999540.99978⇒L0stL0al=0.999759(b)
L40 alL40 st=L0 al 1+α al×40L0 st 1+α st×40⇒L40 alL40 st=L0 alL0 st×1+2.3×105×401+1.1×105×40⇒L40 alL40 st=0.99977×1.000921.00044⇒L40 alL40 st=1.0002496(c)
L100 alL100 st=L0 al 1+α al×100L0 st 1+αst×100⇒ L100 alL100 st=0.99977×1.00231.0023⇒ L100 alL100 st=1.00096
Question 15:
A metre scale is made up of steel and measures correct length at 16°C. What will be the percentage error if this scale is used (a) on a summer day when the temperature is 46°C and (b) on a winter day when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10^{–6} °C^{–1}.
Answer:
(a) Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t_{1} = 16 °C
Temperature on a hot summer day, t_{2} = 46 °C
â€‹So, change in temperature, Δθ = t_{2}_{ }
–t_{1}_{ } = 30 °C
Coefficient of linear expansion of steel,
α= 1.1 × 10^{–5 }°C^{â€‹1}
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10^{–5} × 30
% of error =∆LL×100% =Lα ΔθL×100% =1.1×105×30×100% =3.3×102%(b) Temperature on a winter day, t_{2} = 6 °Câ€‹
â€‹So, change in temperature, Δθ = t_{1}
–t_{2} = 10 °Câ€‹
ΔL = L_{â€‹2 }
–L_{1 }= L αΔθ = L × 1.1 × 10^{–5} × 10â€‹
% of error =∆LL×100% =Lα ΔθL×100% =1.1×105×10×100% = 1.1×102
Question 16:
A metre scale made of steel reads accurately at 20°C. In a sensitive experiment, distances accurate up to 0.055 mm in 1 m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel = 11 × 10^{–6} °C^{–1}.
Answer:
Given:
Temperature at which a metre scale gives an accurate reading, T_{1} = 20 °C
The value of variation admissible, ΔL = 0.055 mm = 0.055 × 10^{–3} m, in the length, L_{0} = 1 m
Coefficient of linear expansion of steel, α = 11 × 10^{–6} °C^{–1}
Let the range of temperature in which the experiment can be performed be T_{2}.
We know: ΔL = L_{0} αΔT
⇒0.055×103=1×11×106×T1±T2⇒5×103=20±T2×103⇒20 ± T2=5⇒Either T2=20+5=25 °C or T2=205=15 °CHence, the experiment can be performed in the temperature range of 15 °C to 25 °C .
Question 17:
The density of water at 0°C is 0.998 g cm^{–3 }and at 4°C is 1.000 g cm^{–1}. Calculate the average coefficient of volume expansion of water in the temperature range of 0 to 4°C.
Answer:
Given:
Density of water at 0°C, ( f_{0})= 0.998 g cm^{3}
Density of water at 4°C, â€‹(f_{4}) = 1.000 g cm3
Change in temperature, (Δt) = 4^{o}C
Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be γ.
We know: f4=f01+γ∆t⇒ f0=f41+γ∆t⇒0.998=11+γ.4⇒1+4γ=10.998⇒4γ=10.9981⇒γ=0.0005=5×104 oC1As the density decreases,
γ=5×104 oC1
Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to 4°C will be
γ=5×104^{o}C^{1}.
Question 18:
Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature. Coefficients of linear expansion of iron and aluminium are 12 × 10^{–6} °C^{–1} and 23 × 10^{–6} °C^{–1} respectively.
Answer:
Let the original length of iron rod be L_{Fe} and L^{‘}^{â€‹}^{â€‹â€‹}_{Fe }be its length when temperature is increased by ΔT.
Let the original length of aluminium rod be L_{Al} and L^{‘}^{â€‹}^{â€‹â€‹}_{Al }be its length when temperature is increased by ΔT.
Coefficient of linear expansion of iron,
αFe = 12 × 10^{–6} °Câ€‹
1
Coefficient of linear expansion of aluminium, α_{Al} = 23 × 10^{–6} °C^{â€‹}â€‹
1
Since the difference in length is independent of temperature, the difference is always constant.
L’Fe=LFe 1+αFe×∆Tand L’Al=LAl 1+αAl×∆T⇒L’FeL’Al=LFeLAl+LFe×αFe∆TLAl×αAl×∆T(1)Given:L’FeL’Al=LFeLAlHence, LFeαFe=LAl αAl [using (1)]⇒LFeLAl=2312The ratio of the lengths of the iron to the aluminium rod is 23:12.
Question 19:
A pendulum clock shows correct time at 20°C at a place where g = 9.800 m s^{–2}. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m s^{–1}. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12 × 10^{–6} °C^{–1}.
Answer:
Given:
The temperature at which the pendulum shows the correct time, T_{1} = 20 °C
Coefficient of linear expansion of steel,
α= 12 × 10^{–6} °C^{–1}
Let T_{2} be the temperature at which the value of g is 9.788 ms^{–2 }and
ΔT be the change in temperature.
^{â€‹}So, the time periods of pendulum at different values of g will be t_{1} and t_{2} , such that
t1=2πl1g1t2=2πl2g2 =2πl11+αΔTg2 ∵l2=l11+α∆TGiven, t1=t2⇒2πl1g1=2πl11+α∆Tg2⇒l1g1=l11+α∆Tg2⇒19.8=1+12×106×∆T9.788⇒9.7889.8=1+12×106×∆T ⇒9.7889.81=12×106×∆T⇒∆T=0.0012212×106⇒T220=102.4⇒T2=102.4+20 =82.4⇒T2≈82 °CTherefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms^{–2} should be
82 ^{o}C.
Question 20:
An aluminium plate fixed in a horizontal position has a hole of diameter 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of 10 °C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is 23 × 10^{–6} °C^{–1} and that of steel is 11 × 10^{–6} °C^{–1}.
Answer:
Given:
Diameter of the steel sphere at temperature (T_{1} = 10 °C) , d_{st} = 2.005 cm
Diameter of the aluminium sphere, d_{Al} = 2.000 cm
Coefficient of linear expansion of steel, α_{st} = 11 × 10
6 °C
1
Coefficient of linear expansion of aluminium, α_{Al} = 23 × 10
6 °C
1^{â€‹}
Let the temperature at which the ball will fall be T_{2}_{ , }so that change in temperature be ΔTâ€‹.
d‘_{st} = 2.005(1 + α_{st} ΔT)
⇒d’st=2.005+2.005×11×106×∆T d’Al=21+αAl×∆T⇒ d’Al=2+2×23×106×∆TThe steel ball will fall when both the diameters become equal.
So, d‘_{st}_{ }= d‘_{Al}
⇒2.005+2.005×11×106∆T=2+2×23×106∆T⇒4622.055×106 ∆T=0.005⇒∆T=0.005×10623.945=208.81Now, ∆T=T2T1=T210 °C ⇒T2=∆T+T1=208.81+10⇒T2=218.8≅219 °CTherefore, â€‹the temperature at which the ball will fall is 219 °C.
Question 21:
A glass window is to be fit in an aluminium frame. The temperature on the working day is 40°C and the glass window measures exactly 20 cm × 30 cm. What should be the size of the aluminium frame so that there is no stress on the glass in winter even if the temperature drops to 0°C? Coefficients of linear expansion for glass and aluminium are 9.0 × 10^{–6} °C^{–1} and 24 ×100^{–6}°C^{–1} , respectively.
Answer:
Given:
At 40^{o}C, the length and breadth of the glass window are 20 cm and 30 cm, respectively.
Coefficient of linear expansion of glass,
αg= 9.0 × 10^{–6} °C^{–1}
Coefficient of linear expansion for aluminium,
αAl= 24 ×100^{–6} °C^{–1}
The final length of aluminium should be equal to the final length of glass so that there is no stress on the glass in winter, even if the temperature drops to 0 °C.
â€‹Change in temperature,
Δθ = 40 °C
Let the initial length of aluminium be l.
l1αAl ∆θ = 201αg∆θ⇒l124×106×40=2019×106×40⇒l10.00096=2010.00036⇒l=20×0.9996410.00096 =20×0.999640.99904⇒l=20.012 cmLet the initial breadth of aluminium be b.
b1αAl∆θ=301αg∆θ⇒b=30×19×106×40124×106×40 =30×0.999640.99904⇒b=30.018 cmTherefore, the size of the aluminium frame should be 20.012 cm × 30.018 cm.
Question 22:
The volume of a glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature? Coefficients of cubical expansion of mercury and glass are 1.8 × 10^{–6} °C^{–1} and 9.0 × 10^{–6} °C^{–1} , respectively.
Answer:
At T = 20°C, the volume of the glass vessel, V_{g}_{ }= 1000 cc.
Let the volume of mercury be V_{Hg} .
Coefficient of cubical expansion of mercury, γ_{Hg} = 1.8 × 10^{–4} /°C
Coefficient of cubical expansion of glass, γ_{g} = 9 × 10^{–6} /°C
â€‹Change in temperature, ΔT, is same for glass and mercury.
Let the volume of glass and mercury after rise in temperature be V’_{g} and V’_{Hg} respectively.
Volume of remaining space after change in temperature,(V’_{g} – V’_{Hg}) = Volume of the remaining space (initial),(V_{g}â€‹â€‹ – V_{Hg})
We know: V’_{g} = V_{g} (1 + γ_{g} ΔT) …(1)
V’_{Hg} = V_{Hg} (1 + γ _{Hg } ΔT) …(2)
Subtracting (2) from (1), we get:
V’gV’Hg = VgVHg+Vgγg∆TVHgγHg∆T⇒Vgγg∆TVHgγHg∆T=0⇒VgVHg=γHgγg⇒1000VHg=1.8×1049×106⇒VHg=9×1031.8×104⇒VHg=50 ccTherefore, the volume of mercury that should be poured into the glass vessel is 50 cc.
Question 23:
An aluminium can of cylindrical shape contains 500 cm^{3} of water. The area of the inner cross section of the can is 125 cm^{2}. All measurements refer to 10°C.
Find the rise in the water level if the temperature increases to 80°C. The coefficient of linear expansion of aluminium is 23 × 10^{–6} °C^{–1} and the average coefficient of the volume expansion of water is 3.2 × 10^{–4} °C^{–1}.
Answer:
Given:
Volume of water contained in the aluminium can, V_{0} = 500 cm^{3}
Area of inner crosssection of the can, A = 125 cm^{2}
^{â€‹}Coefficient of volume expansion of water, γâ€‹ = 3.2 × 10^{–4} °C^{–1}
Coefficient of linear expansion of aluminium,
αAL= 23 × 10^{–6} °C^{–1}
If
∆θis the change in temperature, then final volume of water
Vdue to expansion,
V = V_{0}(1 + γΔθ)
= 500 [1 + 3.2 × 10^{–4} × (80 – 10)]
= 500 [1 + 3.2 × 10^{–4 }× 70]
= 511.2 cm^{3}
The aluminium vessel expands in its length only.
So, area of expansion of the base can be neglected.
Increase in volume of water = 11.2 cm^{3}
Consider a cylinder of volume 11.2 cm^{3}
∴ Increase in height of the water
=11.2125= 0.0896
= 0.089 cm
Question 24:
A glass vessel measures exactly 10 cm × 10 cm × 10 cm at 0°C. It is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1.6 cm^{3} of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5 × 10^{–1 }°C^{–1}.
Answer:
Given: At 0^{o}^{â€‹}C, volume of glass vessel, V_{g} = 10 × 10 × 10 = 1000 cc = volume of mercury, V_{Hg}
Let the volume of mercury at 10°C be V’_{Hg }and that of glass be V’_{g}.
At 10^{â€‹o}C, the additional volume of mercury than glass, due to heating, V’_{Hg} – V’_{g} = 1.6 cm^{3}
So change in temperature, ΔT = 10°C
Coefficient of linear expansion of glass, α_{g} = 6.5 × 10^{–6 }°C^{–1}
Therefore, the coefficient of volume expansion of glass, γ_{g} = 3 × 6.5 × 10^{–6}°C^{–1}â€‹
Let the coefficient of volume expansion of mercury be γ_{Hg}.
We know that
V’_{Hg} = V_{Hg} (1 + γ _{Hg } ΔT) …(1)
V’_{g} = V_{g} (1 + γ_{g} ΔT) …(2)
Subtracting (2) from (1) we get,
V’_{Hg} – V’_{g} = V_{H}_{g} – V_{g} + V_{Hg} γ_{Hg} ΔT – V_{g} γ_{g} ΔT (as V_{Hg} = V_{g})
⇒1.6=1000×γHg×101000 × 6.5×3×106×10⇒γHg=1.6+19.5×10210000⇒γHg=1.6+0.19510000⇒γHg=1.79510000⇒γHg=1.795×104⇒γHg≅1.8×104°C1Therefore, the coefficient of volume expansion of mercury is 1.8× 10^{–4 }°C^{–1}.
Question 25:
The densities of wood and benzene at 0°C are 880 kg m^{3} and 900 kg m^{–3} , respectively. The coefficients of volume expansion are 1.2 × 10^{–3 }°C^{–1} for wood and 1.5 × 10^{–3 }°C^{–1} for benzene. At what temperature will a piece of wood just sink in benzene?
Answer:
Given:
Density of wood at 0 °C, f_{w} = 880 kgm^{â€‹3}
Density of benzene at 0 °C, f_{b} = 900 kgm^{–}^{â€‹3}
Coefficient of volume expansion for wood, γ_{w} = Coefficient of volume expansion for benzene, γ_{b} = 1.5 × 10^{–3 }°C^{–1}
So, initial temperature, T_{1} = 0 °C
Let T_{2} be the temperature at which the piece of wood will just sink in benzene and
ΔT = T_{2}
–T_{1}.
The piece of wood begins to sink when its weight is equal to the weight of the benzene displaced.
Mass = volume
×density
Therefore, Vf’wg = Vf’bg⇒fw1+γw ΔT=fb1+γb ΔT⇒8801+1.2×103 ΔT=9001+1.5×103 ΔT⇒880+880×1.5×103ΔT =900+900×1.2×103 ΔT⇒13201080×103 ΔT=20⇒ΔT=83.3°C⇒T2T1≅83o ⇒T20°≅83o⇒T2≅83°CTherefore, the piece of wood will just sink in benzene at 83 ^{o}C.
Question 26:
A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed?
Answer:
The steel rod is resting on a horizontal base at 0 °C. When the temperature is increased to 100 °C, it will lead to an increase in the length of the steel due to expansion on heating. Since, there is no opposition in expansion of length, no longitudinal strain will be developed.
Question 27:
A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C.
Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10^{–5} °C^{–1}.
Answer:
Given:
Temperature at which rod is resting on a fixed horizontal base without any strain, T_{1}=20 °C. Then the rod is heated to temperature, T_{2} = 50 °C
â€‹So change in temperature,
ΔT=T2T1=30 oCCoefficient of linear expansion of steel, α = 1.2 × 10^{–5} °C
1â€‹
Let L be the length of the rod without heating and L’ be the length of the rod on heating.
Let longitudinal strain developed in the rod be S.
We know that
L’=L(1+αΔT)⇒ΔL=LαΔTStrain, S =ΔLL =LαΔTL =αΔT ⇒S =1.2×105×5020 =1.2×105×30 =36×105 S =3.6×104The strain of 3.6 × 10
4 will be opposite to the direction of expansion.
Question 28:
A steel wire of crosssectional area 0.5 mm^{2} is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10^{–5} °C^{–1} and its Young’s modulus is 2.0 × 10^{–11} Nm^{–2}.
Answer:
Given:
Crosssectional area of the steel wire, A = 0.5 mm^{2} = 0.5 × 10^{–6} m^{2}
The wire is taut at a temperature, T_{1} = 20 °C,
After this, the temperature is reduced to T_{2} = 0 °C
â€‹So, the change in temperature, Δθ = T_{1}
–T_{2} = 20 °C
Coefficient of linear expansion of steel, α = 1.2 ×10^{–5} °C^{â€‹1}
Young’s modulus, γ = 2 ×10^{11} Nmâ€‹
2
Let L be the initial length of the steel wire and L‘ be the length of the steel wire when temperature is reduced to 0°C.
Decrease in length due to compression, ΔL = L’
L= LαΔθ …(1)
Let the tension applied be F.
γ=stressstrain=FAΔLL⇒γ=FA×LΔL⇒ΔL=FLAY …(2)
Change in length due to tension produced is given by (1) and (2).
So, on equating (1) and (2), we get:
LαΔθ=FLAY⇒F=αΔ θAY =1.2×105×200×0.5×106×2×1011 =1.2×20⇒F=24 NTherefore, the tension produced when the temperature falls to 0°C is 24 N.
Question 29:
A steel rod is rigidly clamped at its two ends. The rod is under zero tension at 20°C. If the temperature rises to 100°C, what force will the rod exert on one of the clamps? Area of crosssection of the rod is 2.00 mm^{2}. Coefficient of linear expansion of steel is 12.0 × 10^{–6} °C^{–1} and Young’s modulus of steel is 2.00 × 10^{11} Nm^{–2}.
Answer:
Given:
Temperature of the rod at zero tension, T_{1} = 20 °C
Final temperature, T_{2} = 100 °C
â€‹Change in temperature,
Δθ = 80 °C
Crosssectional area of the rod, A = 2 mm^{2} = 2 × 10
6 m^{2}
Coefficient of linear expansion of steel, α_{ }= 12 ×10^{–6 }°C^{â€‹}
1
Young’s modulus of steel, Y_{ }= 2 × 10^{11 }Nm
2
Let L be the length of the steel rod at 20 °C and L’ be the length of steel rod at 100 °C.
Change of length of the rod,
∆L= L’
–L
If F be the force exerted by the rod on one of the clamps due to rise in temperature, then
Y=stressstrain=F/AΔLL⇒F=Y×ΔLL×A ΔL=LαΔθ⇒F=YLαΔθAL⇒F=YAαΔθ =2×1011×2×106×12×106×80 =48×80×101So, F=384 NTherefore, the rod will exert a force of 384 N on one of the clamps.
Page No 14:
Question 30:
Two steel rods and an aluminium rod of equal length l_{0} and equal crosssection are joined rigidly at their ends, as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to θ. Coefficient of linear expansion of aluminium and steel are α_{a} and α_{s}_{,} respectively. Young’s modulus of aluminium is Y_{a} and of steel is Y_{s}.
Steel 
Aluminium 
Steel 
Figure 23E1
Answer:
Given:
Initial length of the system (two identical steel rods + an aluminium rod) at 0 °C = l_{0}
Coefficient of linear expansion of steel and aluminium are α_{s}_{ }and α_{Al}â€‹, respectively.
Temperature is raised by θ.
_{â€‹}So, the change in temperature,
∆θ= θ
0 °C = θ
Young’s modulus of steel and aluminium are γ_{s} and γ_{Al}_{, }respectively.
If l be the final length of the system at temperature θ,
strain on the system
=ll0l0 â€‹ …(1)
Young’s modulus = Stress/ Strain
Therefore, the total strain on the system
=Total stress on the systemTotal Young’s modulus of the systemNow, total stress = stress due to the two steel rods + stress due to the aluminium rod
Stress=FA=αY∆θ Total stress = γ_{s}α_{s}θ + γ_{s}α_{s}θ + γ_{Al}α_{Al}θ
= 2γ_{s}α_{s}θ + γ_{Al}α_{Al}θ …(2)
Young’s modulus of the system,
Y = γ_{s} + γ_{s} + γ_{Al} = 2γ_{s} + γ_{Al}_{ }…(3)
Using (1), (2) and (3), we get:
Strain on the system =2γsαsθ+γAlαAl θ2γs+γAl⇒ll0l0=2γsαsθ+γAlαAlθ2γs+γAl⇒l=l0 1+2γsαsθ+γAlαAlθ2γs+γAlTherefore, the final length of the system will be
l0 1+2γsαsθ+γAlαAlθ2γs+γAl, where l_{0} is its initial length.
Question 31:
A steel ball that is initially at a pressure of 1.0 × 10^{5 }Pa is heated from 20°C to 120°C, keeping its volume constant.
Find the pressure inside the ball. Coefficient of linear expansion of steel = 12 × 10^{–6} °C^{–1} and bulk modulus of steel = 1.6 × 10^{11} Nm^{–2}.
Answer:
Given:
Initial pressure on the steel ball = 1.0 × 10^{5 }Pa
The ball is heated from 20 °C to 120 °C.
So, change in temperature,
Δθ = 100 °C.
Coefficient of linear expansion of steel,
α= 12 × 10^{–6} °C^{–1}
Bulk modulus of steel, B = 1.6 × 10^{11} Nm^{–2}
Pressure is given as,
⇒P=B×γ∆θ⇒P=B×3α∆θ ∵γ=3α⇒P=1.6×1011×3×12×106×12020 =1.6×3×12×1011×106×102 =57.6×107⇒P=5.8×108 PaTherefore, the pressure inside the ball is 5.8 × 10^{8 }Pa.
Question 32:
Show that the moment of inertia of a solid body of any shape changes with temperature as I = I_{0} (1 + 2αθ), where I_{0} is the moment of inertia at 0°C and α is the coefficient of linear expansion of the solid.
Answer:
Given:
Coefficient of linear expansion of solid = α
Moment of inertia at 0 °C = I_{0}
If temperature changes to θ from 0 °C, then change in temperature,
∆T=
θLet I be the new moment of inertia attained due to rise in temperature.
Let R_{0} be the radius of gyration at 0 °C.
We know that on heating, radius of gyration will change as
R_{ }= R_{0}(1 + αθ)
Here, R is the radius of gyration after heating.
I_{0} = MR_{0}^{2} , where M = mass of the body
Now, I = MR^{2} = MR_{0}^{2}(1 + αθ)^{2}
Expanding binomially and neglecting the higher terms of order (αθ) that will be very small, we get
I = MR_{0}^{2}(1 + 2 αθ)
So, I = I_{0}(1 + 2 αθ)
Hence, proved.
Question 33:
A torsional pendulum consists of a solid disc connected to a thin wire (α = 2.4 × 10^{–5 }°C^{–1}) at its centre. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).
Answer:
Given:
Coefficient of linear expansion of the wire, α = 2.4 × 10^{–5 }°C^{–1}
Let I_{0} be the moment of inertia of the torsional pendulum at 0 °C.
If K is the torsional constant of the wire, then time period of torsional pendulum
T:
T=2πIK …1Here, I = moment of inertia after change in temperature
When the temperature is changed by
∆θ, moment of inertia
I,
I = I_{0}(1+2
α ∆θ)
On substituting the value of I in equation(1), we get:
T=2πI01+2α∆θKIn winter,
∆θ= 5 °C
∴Time period
T1
=2πI01+2α×5KIn summer,
∆θ= 45 °C
Time period
T2
=2πI01+2α×45K
So,T2T1=1+90α1+10α =1+90×2.4×1051+10×2.4×105⇒T2T1=1.002161.00024%change =T2T11×100 =0.0959%⇒%change in time period≈9.6×102%Therefore, the percentage change in time period of a torsional pendulum between peak winters and peak summers is 9.6 × 10^{–2 }% .
Question 34:
A circular disc made of iron is rotated about its axis at a constant velocity ω. Calculate the percentage change in the linear speed of a particle of the rim as the disc is slowly heated from 20°C to 50°C, keeping the angular velocity constant. Coefficient of linear expansion of iron = 1.2 × 10^{–5} °C^{–1}.
Answer:
Let initial radius of the circular disc at 20 ^{â‚’}C =
r20Let final radius of the circular disc at 50 ^{â‚’}C =
r50Coefficient of linear expansion of iron,
α= 1.2 × 10^{–5} °C^{–1}.
Change in temperature,
∆T= 30 ^{â‚€}C
Let R‘ and R be the radius of the paricle at 50 ^{â‚’}C and 20 ^{â‚’}C respectively.
If v and v‘ be the linear speed of the particle at 50 ^{â‚’}C and 20 ^{â‚’}C respectively, as the angular velocity remains(
ω) constant.
Therefore,
ω=vR=v’R’ ….1Now,
R‘ = R(1+
α∆T)
⇒R’ = R + R
×1.2 × 10^{–5} °C^{–1}
×∆T.
⇒R‘ = 1.00036R
Using equation(1) we have,
vR=v’R’⇒vR=v’1.00036R⇒v’=1.00036vPercentage change in linear speed will be,
=v’vv×100=1.00036vvv×100=3.6×102
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity