Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 11 Thermal Properties Of Matter are provided here with simple step-by-step explanations. These solutions for Thermal Properties Of Matter are extremely popular among class 11 Science students for Physics Thermal Properties Of Matter Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 11 Science Physics Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 77:

Question 11.1:

A bimetallic strip is made of aluminium and steel (αAl > αsteel). On heating, the strip will
(a) remain straight.
(b) get twisted.
(c) will bend with aluminium on concave side.
(d) will bend with steel on concave side.

Answer:

On heating the metallic strip with higher coefficient of linear expansion (αAl) will expand more.
According to the question αAl >αsteel , so aluminium will expand more. So it should have larger radius of curvature. Hence aluminium is on the convex side. The metal of smaller α bends on the inner side.

Hence the correct answer is option (d).

Page No 77:

Question 11.2:

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:

When the rod is heated uniformly to raise its temperature slightly high , it expands. So the moment of inertia of the rod will increase. As in this process , no external torque is being acted. Hence the Angular momentum of the system will remain constant:

(Iω)initial=(Iω)finalAs rod expands, Iinitial<Ifinal ωinitial>ωfinalTherefore the angular speed of the rod will decrease.

Hence the correct answer is option (b). 

Page No 77:

Question 11.3:

The graph between two temperature scales A and B is shown in the Figure. 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) tA-180100=tB150

(b) tA-30150=tB100

(c) tB-180150=tA100

(d) tB-40100=tA180

Answer:

Temp on one scale can be converted in to another by using the following identity:
Reading on any scale-LFPUFP-LFP= constantwhere LFP: Lower fixed pointUFP: Upper fixed pointFrom the graph it is clear that the lowest point for the scale will be 30° and the highest point will be 180°Now, TA-LFPA(UFP)A-(LFP)A=TB-LFPB(UFP)B-(LFP)BTA-30150=TB-0100-0TA-30150=TB100

Hence the correct answer is option (b).



Page No 78:

Question 11.4:

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:

Liquids generally increase the volume or expands while heating. but in case of water , it expands on heating if the temperature is greater than 4° C. The density of the water reaches a maximum value of 1g/cm3 at 4° C. This behaviour of the water is know as anomalous behaviour of the water.
Force of buoyancy
 F=VρgFrom this , Fρ               FBρ0°              at 0°C                             ...eq1FB'ρ4°             at 4°C                             ...eq2Divide eq(1)and (2), we getFBFB'=ρ0°ρ4°>1Hence,  F at 4°C >F at 0°C

Hence the correct answer is option (a).

Page No 78:

Question 11.5:

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:

According to the relation,

T=2πlgTl

As we increase the temperature , the length of the pendulum will also increase , hence the time period of the pendulum will increase too. As the expansion of the length of the pendulum is uniform hence, center of mass of the system will be intact.

Hence the correct answer is option (a).

Page No 78:

Question 11.6:

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:

When liquids and gases are heated, then there is vibration of molecules about their mean position increases, hence kinetic energy associated with random motion of molecules increases.
Hence, the correct answer is option (a)
 



Page No 79:

Question 11.7:

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 ΔT so that its new temperature is T + ΔT. The increase in the volume of the sphere is approximately
(a) 2πR α ΔT
(b) πR2 α ΔT
(c) 4πR3 α ΔT/3
(d) 4πR3 α ΔT

Answer:

Initial volume of the sphere = 43πR3
Coefficient of linear expansion = α
∴ Coefficient of volume expansion = 3α
1VdVdT=3α

Now change in volume of the sphere after heating,
dV=3VαdT    =4πR3αT

Hence, the correct answer is option (d).

Page No 79:

Question 11.8:

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 is directly proportional to the surface area of the object exposed to the surrounding. As the thickness of the circular plate is least, so it has maximum surface area and sphere has minimum surface area. Therefore, plate will cool fastest and sphere the slowest.
Hence, the correct answer is option (c).

Page No 79:

Question 11.9:

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:

The Zeroth Law clearly suggests that when two systems A and B, are in thermal equilibrium, there must be a physical quantity that has the same value for both.
According to this law we can conclude that, if a system is in thermal equilibrium with Y  but not with then the​ systems and are not in thermal equilibrium with each other. 
Also. if a system is neither in thermal equilibrium with nor with Z then the system and may be in thermal equilibrium with each other.
Hence, the correct options are (b) and (d).

Page No 79:

Question 11.10:

‘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:

In case of an object with smaller surface area, heat radiation will be less and it will be heated first. Therefore, smaller gulab jamuns are heated before bigger ones. Similarly, smaller pizzas are heated before bigger ones.

Hence, the correct options are (b) and (c).



Page No 80:

Question 11.11:

Refer to the plot of temperature versus time 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:

Temperature of the system remains constant during the phase change process. During the process AB, ice transforms into water while temperature of the system remains 0 oC. Thus, the region AB represents ice and water in thermal equilibrium.
BC represents temperature rise of water from 0 oC to 100 oC. During the process CD, water and steam are in equilibrium at boiling point.

Hence, the correct options are (a) and (d).

Page No 80:

Question 11.12:

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:


The heat from hot milk spread on the table is transferred to the surroundings by conduction, convection and radiation. According to Newton's law of cooling the temperature of milk falls exponentially with time.
Hence, the correct options are (b), (c) and (d).

Page No 80:

Question 11.13:

Is the bulb of a thermometer made of diathermic or adiabatic wall?

Answer:

Bulb of a thermometer is made of diathermic walls, as they allow conduction of heat.

Page No 80:

Question 11.14:

A student records the initial length l, change in temperature ΔT and change in length Δl of a rod as follows:
 

S.No. l (m) ΔT (C) Δl (m)
1. 2 10 4 × 10−4
2. 1 10 4 × 10−4
3. 2 20 2 × 10−4
4. 3 10 6 × 10−4

If the first observation is correct, what can you say about observations 2, 3 and 4.

Answer:

If the first observation is correct, the coefficient of linear expansion is given as
α=ll×T  =4×10-42×10  =2×10-5 °C-1

For second observation,
l=αlT    =2×10-5×1×10    =2×10-4 m4×10-4 m
So, the second observation is incorrect.

For third observation,
l=αlT    =2×10-5×2×20    =8×10-4 m2×10-4 m
So, the third observation is incorrect.

For fourth observation,
l=αlT    =2×10-5×3×10    =6×10-4 m
So, the fourth observation is correct.



Page No 81:

Question 11.15:

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:

Metals have high conductivity compared to wood. The rate of heat transfer in metal is faster than in the wood. Thus, a metal bar appears hotter than a wooden bar at the same temperature and it also appears cooler than wooden bar if they are both colder than room temperature.

Page No 81:

Question 11.16:

Calculate the temperature which has same numeral value on Celsius and Fahrenheit scale.

Answer:

The relation between temperature readings of the Celsius(C) and Fahrenheit(F) scale is given as:
C-0100=F-32212-32

Let T be the temperature which has same numeral value on Celsius and Fahrenheit scale.
T-0100=T-32212-32180T=100T-3200T=-320080=-40 °C=-40 F

Page No 81:

Question 11.17:

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:

Conductivity of Cu is high as compared to steel. The junction of Cu and steel gets heated quickly and steel allows food inside to get heated uniformly as it does not conduct as quickly as that of Cu.

Page No 81:

Question 11.18:

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:

If M is the mass of the rod and L is the length of the rod initially, then its moment of inertia is given as:
I=ML212
When temperature is increased, the length of rod will become:
L'=L1+αT
Thus, moment of inertia is given by
I'=ML'212  =ML21+αT212  =ML2121+2αT+α2T2  =I1+2αT                       

I'-I=I1+2αT-I       =2αIT

 

Page No 81:

Question 11.19:

During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured 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:

P-T diagram of water is shown below:

Increasing pressure at 0 °C and 1 atm takes ice into liquid state and decreasing pressure in liquid state at 0 °C and 1 atm takes water to ice state. On squeezing the crushed ice, some of it melts, which fills up the gap between ice flakes. When pressure is released, this water freezes which binds all ice flakes and thus, this makes the ball more stable.

Page No 81:

Question 11.20:

100 g of water is supercooled to –10 °C. At this point, due to some disturbance mechanised 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=1cal/g/°C and LFusionw=80cal/g

Answer:

Mass of water (m) = 100 g
Change in temperature (T) = 0°C − (−10 °C) = 10 °C
Latent heat fusion LFusion = 80 cal/g
Amount of heat required to change the temperature of 100 g of water at −10 oC to 0 oC, Q = mSwT 100×1×10 = 1000 cal
If m gram ice melts,
m=QL=100080=12.5 g
Therefore, the temperature of the resultant mixture will be 0 oC and it consists of 12.5 g of ice and rest of water.

Page No 81:

Question 11.21:

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:

According to Newton’s law of cooling, the rate of loss of heat is directly proportional to the difference of temperature of the body and the surrounding. The difference of temperature is less in the first case, therefore rate of loss of heat will also be less in the first case. Hence, water remains warmer when Ramesh fill the remaining bucket completely by cold water and then attend to the work.



Page No 82:

Question 11.22:

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 × 10 –5 /K and αbrass = 1.8 × 10–5 /K, what should we take as length of each strip?

Answer:

According to the question, at all temperatures,
liron-lbrass=10 cmAlso, liron-lbrass=10 cmliron(1+αiront)-lbrass(1+αbrasst)=10 cmlironαiront=lbrassαbrasst

lironlbrass=αbrassαiron=1.8×10-51.2×10-5lironlbrass=32

liron-lbrass=10 cm32lbrass-lbrass=10 cm12lbrass=10 cmlbrass=20 cmHence,liron-lbrass=10 cmliron-20 cm=10 cmliron=30 cm

 

Page No 82:

Question 11.23:

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:

Consider an iron vessel with a brass rod inside it.
So, according to the question, at all temperatures,
Viron-Vbrass=100 cm3Also, Viron-Vbrass=100 cm3Viron(1+βviront)-Vbrass(1+βvbrasst)=100 cm3Vironβviront=Vbrassβvbrasst

VironVbrass=βvbrassβviron=6×10-53.55×10-5VironVbrass=63.55

Viron-Vbrass=100 cm363.55Vbrass-Vbrass=100 cm3Vbrass=144.9 cm3Viron=244.9 cm3






 

Page No 82:

Question 11.24:

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57°C is drunk. You can take body (tooth) temperature to be 37°C and α = 1.7 × 10–5/°C, bulk modulus for copper = 140 × 109 N/m2.

Answer:

Bulk Modulus, B=StressVolume strain

Stress=B×volume strain         =B×VV         =B×γt         =B×3αt

Substituting all the given values, 

Stress=140×109×3×1.7×10-5×20         =1.428×108 N/m2

Page No 82:

Question 11.25:

A rail track made of steel having length 10 m is clamped on a raillway line at its two ends. On a summer day due to rise in temperature by 20°C, it is deformed as shown in figure. Find x (displacement of the centre) if αsteel = 1.2× 10–5 /°C.

Answer:

Using Pythagoras theorem in the right-angled triangle,

x=12L+L2-L22
Neglecting the term L2, we get
x=122L×L
Now, using L=Lαt
x=L22α×t  =1022×1.2×10-5×20  =0.11 m=11 cm
 

Page No 82:

Question 11.26:

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:

If temperature of a rod varies linearly, then the average temperature of the rod can be taken as mean of temperatures at the two endsθ1 and θ2. i.e.,
θ=θ1+θ22
Now, new length will become,
L=L01+αθ  =L01+αθ1+θ22

Page No 82:

Question 11.27:

According to Stefan’s law of radiation, a black body radiates energy σT4from its unit surface area every second where T is the surface temperature of the black body and σ = 5.67 × 10–8 W/m2K4is 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 10K 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 1s?
[Sw = 4186.0 J/kgK and Lv = 22.6 × 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: 
σ = 5.67 × 10–8 W/m2K4 
Radius, R = 0.5 m
Temperature, T = 10

(a) According to Stefan's law, power radiated is given as:

P=σAT4  =5.67×10-8×4π×0.52×1064  =1.78×1017 J/s1.8×1017 J/s

(b) Energy available per second, U = 1.8 × 1017 J/s
Energy needed for evaporation, U' = 10 % of 1.8 × 1017 J/s = 1.8 × 1016 J/s
Energy required to raise the temperature of m kg of water from 30 oC to 100 oC and then into vapour at 100 oC, U' = mswθ+mLv
U'=m×4186×100-30+m×22.6×105    =25.53×105 m J/s 

Now,
      1.8×1016=25.53×105 mm=1.8×101625.53×105   =7.0×109 kg

(c) Momentum per unit time,
p=Uc=1.8×10173×108=6×108 kg-m/s2

Momentum per unit time per unit area = p4πR2=6×1084×3.14×1032 = 47.7 N/m2

 



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