Thermal Properties Of Matter
Class 11th Physics NCERT Exemplar Solution
Mcq I
Question 1.A bimetallic strip is made of aluminium and steel (αAl > αsteel) . On heating, the strip will
A. remain straight.
C. will bend with aluminium on concave side.
D. will bend with steel on concave side.
Answer:
First of all understand the bimetallic strip, when two strips of equal length but of different materials, means different coefficient of linear expansion, join together then this is known as bimetallic strip. This strip has the property of bending on heating due to unequal linear expansion of two metals.
Both strips of Al and steel are fixed together initially in bimetallic strip. When both are heated then expansion in steel will be smaller than aluminium.
So, Al strip will be convex side and steel on concave side.
Therefore option (d) is correct.
Question 2.
A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
A. its speed of rotation increases.
B. its speed of rotation decreases.
C. its speed of rotation remains same.
D. its speed increases because its moment of inertia increases.
Answer:
As we know moment of inertia of a rod about its perpendicular bisector is 
so from this formula we can say easily that if L increases then I also increases, so on heating a uniform metallic rod its length will increase so moment of inertia of rod increased from I1 to I2 an obviously I2 > I1 or 
.
Now also we will apply conservation of angular momentum i.e.
I1ω1= I2ω2 or 
so from here ω1>ω2 which means angular speed decreases.
Therefore option (b) is correct.
Question 3.
The graph between two temperature scales A and B is shown in Fig. 11.1. Between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by
A. 
= 
B. 
= 
C. 
= 
D. 
= 
Answer:
As we know the identity by which temperature on one scale can be converted into another scale i.e.
where LFP & UFP stands for Lower fixed point & upper fixed point respectively.
Now from above graph we can say Lower fixed point is
and Upper fixed point 
Hence applying the above formula for A and B
i.e. 
Therefore option (b) is correct.
Question 4.
An aluminium sphere is dipped into water. Which of the following is true?
A. Buoyancy will be less in water at 0°C than that in water at 4°C.
B. Buoyancy will be more in water at 0°C than that in water at 4°C.
C. Buoyancy in water at 0°C will be same as that in water at 4°C.
D. Buoyancy may be more or less in water at 4°C depending on the radius of the sphere.
Answer:
As we know that buoyant force on a body of volume V and density of ρ, when immersed in liquid is FB
.
So at 0℃ assume the volume be V and density be ρ0 then buoyant force F0℃
and at 4℃ F4℃
also density i.e. ρ4 will be maximum all over the range of temperature so
Now ratio of buoyant force 
So 
Therefore option (a) is correct.
Question 5.
As the temperature is increased, the time period of a pendulum
A. increases as its effective length increases even though its centre of mass still remains at the centre of the bob.
B. decreases as its effective length increases even though its centre of mass still remains at the centre of the bob.
C. increases as its effective length increases due to shifting of centre of mass below the centre of the bob.
D. decreases as its effective length remains same but the centre of mass shifts above the centre of the bob.
Answer:
As we know time period of a pendulum is given by 
so from this we can say 
Now also we know that as the temperature increased the length L increases due to linear expansion i.e. 
.
So from above we can say on increasing temperature, its effective length increases hence T also increases.
Therefore option (a) is correct.
Question 6.
Heat is associated with
A. kinetic energy of random motion of molecules.
B. kinetic energy of orderly motion of molecules.
C. total kinetic energy of random and orderly motion of molecules.
D. kinetic energy of random motion in some cases and kinetic energy of orderly motion in other.
Answer:
We know that vibration of molecules about their mean position increases as the temperature increases or the body get heated.
Hence kinetic energy associated with random motion of molecule increases.
Therefore option (a) is correct.
Question 7.
The radius of a metal sphere at room temperature T is R, and the coefficient of linear expansion of the metal is α. The sphere is heated a little by a temperature ∆T so that its new temperature T is (T + ∆T). The increase in the volume of the sphere is approximately
A. 2π R α ∆ T
B. π R2 α ∆T
C. 4 π R3 α ∆T / 3
D. 4πR3 α ∆T
Answer:
As we know the volume of sphere 
Also given in question that coefficient of linear expansion is 
also coefficient of cubical expansion
So 
Option (d) is matching hence correct option is (d).
Question 8.
A sphere, a cube and a thin circular plate, all of same material and same mass are initially heated to same high temperature.
A. Plate will cool fastest and cube the slowest
B. Sphere will cool fastest and cube the slowest
C. Plate will cool fastest and sphere the slowest
D. Cube will cool fastest and plate the slowest.
Answer:
loss of heat on cooling is directly proportional to the following factors
(i) surface area exposed to surrounding
(ii) temperature difference between body and surrounding
(iii) material of object
Also we know that surface area of sphere is minimum and of circular plate is maximum.
So option (c) s correct and rests are wrong.
Mcq Ii
Question 1.Mark the correct options:
A. A system X is in thermal equilibrium with Y but not with Z. System Y and Z may be in thermal equilibrium with each other.
B. A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other.
C. A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other.
D. A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other.
Answer:
In this question we will apply Zeroth law of thermodynamics which state if two thermodynamics systems are each in thermal equilibrium with third one, then they are in thermal equilibrium with each other.
So look option (a) here Tx=Ty but Tx ≠ Tz hence Ty ≠ Tz which means Y and Z never in thermal equilibrium so option (a) is wrong.
In option (b) Tx=Ty but Tx ≠ Tz hence Ty ≠ Tz so this option is correct.
In option (c) Tx≠ Ty also Tx ≠ Tz hence Ty ≠ Tz so this option is also wrong.
And at last in option (d) Tx≠ Ty and Tx ≠ Tz hence Ty may be equal Tz so option (d) is correct.
Question 2.
‘Gulab Jamuns’ (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice big (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following:
A. Both size gulab jamuns will get heated in the same time.
B. Smaller gulab jamuns are heated before bigger ones.
C. Smaller pizzas are heated before bigger ones.
D. Bigger pizzas are heated before smaller ones.
Answer:
As we know, material whose surface area less will get heated before than the more surface area.
Hence options (b) and (c) are correct and rest options are wrong.
Question 3.
Refer to the plot of temperature versus time (Fig. 11.2) showing the changes in the state of ice on heating (not to scale).
Which of the following is correct?
A. The region AB represents ice and water in thermal equilibrium.
B. At B water starts boiling.
C. At C all the water gets converted into steam.
D. C to D represents water and steam in equilibrium at boiling point.
Answer:
As we know whenever any substance get heated continuously but its temperature not changing, then obviously the state changes i.e. solid to liquid or liquid to gas or similarly.
Now look carefully the above graph, we are seeing that part AB (0℃) and CD (100℃) represents the conversion of solid into liquid and liquid into gas respectively. That means part AB represents ice and water both till B, and part CD represents water and steam.
So from above we can say that option (b) is wrong and also option (c) is wrong because at D all waters get converted into steam. Hence options (a) & (d) are correct.
Question 4.
A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?
A. The rate of cooling is constant till milk attains the temperature of the surrounding.
B. The temperature of milk falls off exponentially with time.
C. While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools.
D. All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.
Answer:
Lost of heat is directly proportional to the temperature difference with surrounding and body. So as milk cool, temperature difference decreases so rate of cooling decreases with time so from here we can say option (a) is wrong.
According to Newton’s law of cooling the heat of milk falls exponentially so option (b) is correct.
While cooling very small amount of heat also flows from surrounding to milk as compared to heat lost by milk to surrounding so option (c) is also correct.
By conduction, convection and radiation, when hot milk spread on table it transfer heat to surroundings. Hence option (d) is correct.
Vsa
Question 1.Is the bulb of a thermometer made of diathermic or adiabatic wall?
Answer:
As we know diathermic walls allows to conducts heat through it while adiabatic walls doesn’t allow. Hence bulb of a thermometer made of diathermic walls.
Question 2.
A student records the initial length l, change in temperature ∆T and change in length ∆l of a rod as follows:
If the first observation is correct, what can you say about observations 2, 3 and 4.
Answer:
As we know, 
so from 1st observation we can say that linear expansion of this rod will be
. Now this linear expansion will be same for all observations. Now we will check for rest of observations, so for 2nd observation
but in question
which is not equal to our calculated value hence this observation is wrong.
Similarly for 3rd observation
but given 
hence this observation is also wrong. Now last observation
which is correct.
Question 3.
Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.
Answer:
The law of conduction of heat depends on the conductivity of the material.
Here, “H” is the rate of heat transfer, “K” is the thermal conductivity constant, A is the area of cross-section, T2-T1 is the temperature difference.
Metals have higher conductivity than wood. This means when we touch hot metals or wood, the heat from the metals flow after towards our hands as compared to that of wood. So, they appear hotter. The same theory goes when the two bars are cooled. Here, the heat from our body has a least resistive path when we touch metals as they have higher conductivity. So, heat is dissipated faster in metals than in wood and we feel them cooler.
Question 4.
Calculate the temperature which has same numeral value on Celsius and Fahrenheit scale.
Answer:
The relation between Celsius and Fahrenheit is as follows:
Here, temperature on Celsius is same as Fahrenheit. This means
Here, x is any variable.
So, at -40°C we have the same temperature in Fahrenheit scale i.e. -40°F.
Question 5.
These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor.
Answer:
The use of copper with steel in utensils has to do with their different conductivities. The heat transfer depends on the conductivity of the material and is given as:
Here, “H” is the rate of heat transfer, “K” is the thermal conductivity constant, A is the area of cross-section, T2-T1 is the temperature difference.
Copper has higher conductivity and so is heated immediately. Steel on the other hand has lower conductivity. Because of this, the transfer of heat at the intersection of steel and copper is a gradual process. This ensures uniform heating of food.
Sa
Question 1.Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion α) about its perpendicular bisector when its temperature is slightly increased by ∆T.
Answer:
The moment of inertia of a uniform rod (along perpendicular bisector) is given by:
————— (1)
where, I is the inertia, M is the mass of the rod and l is its length.
The thermal expansion ΔL is given by:
————— (2)
Here, L’ is the new length, L was the original length, 𝛂 is the thermal expansion coefficient and ΔT is the temperature difference.
————— (3)
Substituting (3) in (1)
We know the change in length, ΔL is very small. So, we can neglect square terms like (𝛂ΔT)2. Thus, we get,
Now,
Therefore, the increase in moment of inertia is 2I𝛂ΔT.
Question 2.
During summers in India, one of the common practices to keep cool is to make ice balls of crushed ice, dip it in flavored sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of P–T diagram of water.
Answer:
The process of making an ice ball deals with interconversion of the states of water.
While making an ice ball, we apply pressure on the crushed ice. The temperature is 0°C and the atmospheric pressure is 1 atm. Increasing this pressure shift the state from solid to liquid as indicated by the circle. Due to this, a part of the crushed ice melts and fills any gaps present in between ice flakes. When we release pressure, the water again solidifies to ice thus making the ice ball.
Question 3.
100 g of water is supercooled to –10° C. At this point, due to some disturbance mechanized or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?
[Sw = 1 cal/g/° C and Lwfusion = 80 cal/ g]
Answer:
Here, it’s given
Mass of water, m=100g
Cooled temperature, T1=-10°C
Specific heat capacity of water, Sw= 1 Cal/g/°C
Latent heat of fusion, Lw=80 Cal/g
Let the mass which would freeze be m’.
Ice forms at 0°C.
So, the temperature of the resultant mixture will be T2 = 0°C
We know,
Hence, the temperature of the resultant mixture is 0°C and the mass which would freeze is 12.5g.
Question 4.
One day in the morning, Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 minutes before he could take bath. Now he had two options: (i) fill the remaining bucket completely by cold water and then attend to the work, (ii) first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain.
Answer:
The rate of heat energy loss is directly proportional to the difference in temperature. In the first scenario, mixing the cold water with the hot water overall brings the water to an intermediate temperature which isn’t too hot. The difference between this intermediate temperature and the surrounding temperature is less. Hence the heat loss would be less. In the second scenario, the temperature difference between the hot water and the surrounding is considerable and thus, the heat loss will occur rapidly. This will cool the water down.
So, the first option will keep the water warmer.
La
Question 1.We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If αiron 1.2 X 10-5/ K and αbrass 1.8 X 10-5 /K , what should we take as length of each strip?
Answer:
Let the length of iron strip be LI and that of brass be LB. It is given that the length of the scale should be 10 cm. So,
————— (1)
Both the strips expand at different rates due to different expansion coefficient. But the net expansion in the strips needs to be equal to maintain 10cm length. This means,
————— (2)
————— (3)
————— (4)
(as, αiron 1.2 X 10-5/ K and αbrass 1.8 X 10-5 /K)
—————(5)
Substituting (5) in (1)
The length of the brass rod should be 20 cm and that of the iron rod be 30 cm to maintain 10 cm length at all temperatures.
Question 2.
We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron (βvbrass = 6 × 10-5/K and βviron = 3.55 ×10-5 / K) to create a volume of 100 cc. How do you think you can achieve this?
Answer:
Here, we can have a bimetallic vessel with iron on the outside and brass on the inside as brass expands more than iron.
The change in volume of both the metals should be same.
Now, 
So, the volume of the brass vessel should be 144.9 cc and that of iron vessel should be 244.9 cc.
Question 3.
Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57o C is drunk. You can take body (tooth) temperature to be 37o C and α = 1.7× 10-5 /o C, bulk modulus for copper = 140 × 10 9 N/m2.
Answer:
Given,
Coefficient of linear expansion, α = 1.7× 10-5 /°C
Therefore, Coefficient of volumetric expansion(𝛾)
𝛾 = 3α = 1.7× 10-5 /°C = 5.1× 10-5 /°C
Bulk modulus, B= 140 × 10 9 N/m2.
Body temperature, T1 = 37°C
Temperature of tea, T2 = 57 °C
Temperature difference, ΔT = 57 - 37 = 20 °C
We know,
Now, stress, 
Atmospheric pressure is 105 N/m2.
So, the stress is about 103 times atmospheric pressure.
Question 4.
A rail track made of steel having length 10 m is clamped on a railway line at its two ends (Fig 11.3). On a summer day due to rise in temperature by 20° C, it is deformed as shown in figure. Find x (displacement of the center) if αsteel = 1.2× 10-5 /°C.
Answer:
Here, we apply Pythagoras theorem to find x in terms of L.
From the figure, it is clear that
As, ΔL2 is a very small quantity, we can neglect it.
(here, we multiply and divide 2 in the numerator and denominator)
We know,
Here,
αsteel = 1.2× 10-5 /°C and ΔT = 20 °C. Therefore,
So, the displacement is about 11 cm.
Question 5.
A thin rod having length L0 at 0° C and coefficient of linear expansion α has its two ends maintained at temperatures θ1 and θ2, respectively. Find its new length.
Answer:
The length of the rod is L0. And the coefficient of linear expansion is α. Therefore, temperature (𝜃) at any length x is:
Where 𝜃1 and 𝜃2 are temperatures at the two ends.
We know,
For a small change dx in length dx’ for temperature 𝜃, we have
Integrating the equation,
Now, 
(original length) and 
(NEW LENGTH)
Therefore,
(ANS)
Question 6.
According to Stefan’s law of radiation, a black body radiates energy σT 4 from its unit surface area every second where T is the surface temperature of the black body and σ = 5.67 × 10-8 W/ m2K4 is known as Stefan’s constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When detonated, it reaches temperature of 106K and can be treated as a black body.
(a) Estimate the power it radiates.
(b) If surrounding has water at 30 C°, how much water can 10% of the energy produced evaporate in 1 s?
[Sw = 4186.0 J/ kgK and Lv = 22.6 X 105 J / kg]
(c) If all this energy U is in the form of radiation, corresponding momentum is p = U/c. How much momentum per unit time does it impart on unit area at a distance of 1 km?
Answer:
Given,
The radius of the weapon, r=0.5 m
Therefore, Surface area, A=4𝜋r2 = 𝜋
Temperature after detonation, T=106 K
Stefan’s constant, σ = 5.67 × 10-8 W/ m2K4
(a) Therefore, power radiated (P) is given by
(ANS)
(b) Here, 10% of the energy(E) is to evaporate water which is at 30°C. This means the energy heats the water up to 100°C and then evaporates.
So, 0.1E = msΔT +mLv
Here, the first term is due to the contribution of energy in heating the liquid to 100°C and the second term is due to vaporization and m is the mass of the water used.
Energy, 
Here, t is the time which is 1 sec.
Given, Heat capacity, Sw = 4186.0 J/ kgK
Latent heat of vaporization, Lv = 22.6 X 105 J / kg
Temperature difference, ΔT = 100°C - 30°C = 70°C
So, 7x109 kg of water would be evaporated.
(c) Here, energy, 
Therefore, momentum, 
Now, momentum at a distance 1km (per unit time) is
Therefore, the momentum imparted at 1km is 47.7 N/m2