Chapter 10 - Thermal Properties of Matter

NCERT Solutions Class 11 Physics Chapter 10 – Thermal Properties of Matter

NCERT Solutions for Class 11 Physics Chapter 10 (Thermal Properties of Matter) explores the fascinating behavior of matter when subjected to heat energy. While previous chapters dealt primarily with the mechanical motion of bodies, this chapter transitions into the study of internal energy, temperature, and the physical changes—such as expansion and phase transitions—that occur when substances absorb or lose thermal energy.

NCERT Solutions for Class 11 Physics Chapter 10 by ALLEN are designed by expert faculty to make complex thermodynamic concepts simple with clear theory and detailed numerical examples. The solutions are prepared with a structured approach based on the exam patterns of CBSE, JEE and NEET and hence, ensure a strong conceptual foundation for every student. The principles in this chapter are fundamental to engineering and meteorology and underpin the design of bridges with expansion gaps and the cooling effect of a sea breeze. These solutions provide a rigorous logical basis for learning the laws of calorimetry, thermal conductivity and Newton’s law of cooling.

1.0Class 11 Physics Chapter 10 Thermal Properties of Matter: Key Concepts

This chapter examines how heat energy interacts with the physical state and dimensions of matter. Key lessons include:

Temperature and Heat: Defining temperature as a measure of "hotness" and heat as the energy transferred due to a temperature difference.
Measurement of Temperature: Understanding the Kelvin, Celsius, and Fahrenheit scales and their inter-conversion:
$\frac{T _{C}}{100} = \frac{T _{F} - 32}{180}$ .
Thermal Expansion:
Linear ( $α$ ), Area ( $β$ ), and Volume ( $γ$ ) Expansion: Understanding how dimensions increase with temperature.
Anomalous Expansion of Water: Why water is densest at $4^{\circ} C$ , allowing aquatic life to survive in frozen lakes.
Specific Heat Capacity: The amount of heat required to raise the temperature of a unit mass by one degree.
Calorimetry: The principle of conservation of energy—Heat lost by a hot body equals heat gained by a cold body.
Change of State:
Latent Heat: Energy required for a phase change (Fusion or Vaporization) without a change in temperature (Q = mL).
Triple Point: The temperature and pressure where all three phases (solid, liquid, gas) are in equilibrium together
Heat Transfer Mechanisms:
Conduction: Heat flow through solids via molecular vibration ( $Q / t = K A Δ T / L$ ).
Convection: Heat transfer through fluid movement (Sea breezes and Land breezes).
Radiation: Heat transfer through electromagnetic waves (Black body radiation).
Newton’s Law of Cooling: The rate of loss of heat is proportional to the temperature difference between the body and the surroundings.

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 10

SOLVED EXAMPLES

Show that the coefficient of area expansion, $(Δ A / A) /Δ T$ , of a rectangular sheet of the solid is twice its linear expansivity, $α_{1}$ . Sol.

Consider a rectangular sheet of the solid material of length a and breadth b (figure). When the temperature increases by $Δ T$ , a increases by $Δ a = α_{1} a Δ T$ and b increases by $Δ b = α_{1} b Δ T$ . From figure, the increase in area

$Δ A Δ A = Δ A_{1} + Δ A_{2} + Δ A_{3} = a Δ b + b Δ a + (Δ a) (Δ b) = a_{1} b Δ T + b α_{1} a Δ T + (α_{1})^{2} ab (Δ T)^{2} = α_{1} ab Δ T (2 + α_{1} Δ T) = α_{1} A Δ T (2 + α_{1} Δ T)$

Since $α_{1} ≃ 1 0^{- 5} K^{- 1}$ , the produce $α_{1} Δ T$ for fractional temperature is small in comparison with 2 and may be neglected.

$Hence, (\frac{Δ A}{A}) \frac{1}{Δ T} ≃ 2 α_{1}$

A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m , respectively at $2 7^{\circ} C$ . To what temperature should the ring be heated so as to fit the rim of the wheel? Sol. Given, $T_{1} = 2 7^{\circ} C$

$L_{T_{1}} = 5.231 m$

$L_{T_{2}} = 5.243 m$

So, $L_{T_{2}} = L_{T_{1}} [1 + α_{1} (T_{2} - T_{1})]$

$5.243 = 5.231 [1 + 1.20 \times 1 0^{- 5} \times (T_{2} - 27)] 5.243 = 5.231 + 5.231 \times 1.2 \times 1 0^{- 5} (T_{1} - 27) \frac{5.243 - 5.231}{5.243 \times 1.2 \times 1 0 ^{- 5}} = T_{2} - 27 \Rightarrow T_{2} = \frac{0.012}{5.231 \times 1.2 \times 1 0 ^{- 5}} + 27 or T_{2} = 21 8^{\circ} C$

A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at $10 0^{\circ} C$ . It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg water at 20°C. The temperature of water rises and attains a steady state at $2 3^{\circ} C$ . Calculate the specific heat capacity of aluminium. Sol. In solving this example, we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter. Mass aluminium sphere $(m_{1}) = 0.047 kg$ Initial temperature of aluminium sphere

$= 10 0^{\circ} C$

Final temperature $= 2 3^{\circ} C$ Change in temperature ( $Δ T$ )

$= (10 0^{\circ} C - 2 3^{\circ} C) = 7 7^{\circ} C$

Let specific heat capacity of aluminium be $s_{Al}$ . The amount of heat lost by the aluminium sphere $= m_{1} S_{Al} Δ T = 0.047 kg \times S_{Al} \times 7 7^{\circ} C$ Mass of water $(m_{2}) = 0.25 kg$ Mass of calorimeter $(m_{3}) = 0.14 kg$ Initial temperature of water and calorimeter $= 2 0^{\circ} C$ Final temperature of the mixture $= 2 3^{\circ} C$ Change in temperature

$(Δ T_{2}) = 2 3^{\circ} C - 2 0^{\circ} C = 3^{\circ} C$

Specific heat capacity of water $(s_{w})$

$= 4.18 \times 1 0^{3} J kg^{- 1} K^{- 1}$

Specific heat capacity of copper calorimeter

$= 0.386 \times 1 0^{3} J kg^{- 1} K^{- 1}$

The amount of heat gained by water and calorimeter $= m_{2} s_{w} Δ T_{2} + m_{3} s_{cu} Δ T_{2}$

$= (m_{2} s_{w} + m_{3} s_{cu}) (Δ T_{2}) = (0.25 \times 4.18 \times 1 0^{3} + 0.14 \times 0.386 \times 1 0^{3}) (2 3^{\circ} C - 2 0^{\circ} C)$

In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.

$So, 0.047 kg^{\times} s_{Al} \times 7 7^{\circ} C = (0.25 \times 4.18 \times 1 0^{3} + 0.14 \times 0.386 \times 1 0^{3}) (3^{\circ} C) s_{Al} = 0.911 kJ kg^{- 1} K^{- 1}$

When 0.15 kg of ice at 0°C is mixed with 0.30 kg of water at $5 0^{\circ} C$ in a container, the resulting temperature is $6. 7^{\circ} C$ . Calculate the heat of fusion of ice. $(s_{water} = 4186 J kg^{- 1} K^{- 1})$ Sol. Heat lost by water $= ms_{w} (θ_{f} - θ_{i})_{w}$

$= (0.30) (4186) (50. 0^{\circ} C - 6. 7^{\circ} C) = 54376.14 J$

Heat required to melt ice

$= m_{2} L_{f} = (0.15) L_{f}$

Heat required to raise temperature of ice water to final temperature $= m_{1} s_{w} (Q_{f} - Q_{i})_{l}$

$= (0.15 kg) (4186) (6. 7^{\circ} C - 0^{\circ} C) = 4206.93 J$

Heat lost = heat gained

$54376.14 J = (0.15 kg_{e}) L_{f} + 4206.93 J L_{f} = 3.34 \times 1 0^{5} J kg^{- 1} .$

Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice $= 2100 J kg^{- 1} K^{- 1}$ , specific heat capacity of water $= 4186 J kg^{- 1} K^{- 1}$ , latent heat of fusion of ice $= 3.35 \times 1 0^{5} J kg^{- 1}$ and latent heat of steam $= 2.256 \times 1 0^{6} J kg^{- 1}$ .

Sol. We have Mass of the ice, $m = 3 kg$ specific heat capacity of ice,

$s_{icc} = 2100 J kg^{- 1} K^{- 1}$

specific heat capacity of water,

$s_{water} = 4186 J kg^{- 1} K^{- 1}$

latent heat of fusion of ice,

$L_{fice} = 3.35 \times 1 0^{5} J kg^{- 1}$

Latent heat of steam,

$L_{steam} = 2.256 \times 1 0^{6} J kg^{- 1}$

Now,

$Q = heat required to convert 3 kg of ice at - 1 2^{\circ} C to steam at 10 0^{\circ} C .$

$Q_{1} = heat required to convert ice at - to ice at 0^{\circ} C . = ms_{ice} Δ T_{1} = 3 \times 2100 [0 - (- 12)] = 75600 J$

$Q_{2} =$ heat required to melt ice at 0°C to water at 0°C

$= mL_{fice} = 3 \times 3.35 \times 1 0^{5} = 1005000 J$

$Q_{3} =$ heat required to convert water at 0°C to water at 100°C.

$= ms_{w} Δ T_{2} = 3 \times 4186 \times 100 = 1255800 J$

$Q_{4} =$ heat required to convert water at $10 0^{\circ} C$ to steam at $10 0^{\circ} C$ .

$= m L_{steam} = 3 \times 2.256 \times 1 0^{6} = 6768000 J$

So,

$Q = = = Q_{1} + Q_{2} + Q_{3} + Q_{4} 75600 J + 1005000 J + 1255800 J + 6768000 J 9.1 \times 1 0^{6} J$

What is the temperature of the steel-copper junction in the steady state of the system shown in figure. Length of the steel $rod = 15.0$ cm, length of the copper rod = 10.0 cm, temperature of the furnace $= 30 0^{\circ} C$ , temperature of the other end $= 0^{\circ} C$ . The area of cross-section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel $= 50.2 J s^{- 1} m^{- 1} K^{- 1}$ ; and of copper $= 385 J s^{- 1}$ $m^{- 1} K^{- 1}$ ).

Sol. The insulating material around the rods reduces heat loss from the sides of the rods. Therefore, heat flows only along the length of the rod. Consider any cross-section of the rod. In the steady state, heat flowing into the element must equal the heat flowing out of it; otherwise there would be a net gain or loss of heat by the element and its temperature would not be steady. Thus in the steady state, rate of heat flowing across a cross-section of the rod is the same at every point along the length of the combined steel-coper rod. Let T be the temperature of the steel-copper junction in the steady state. Then,

$\frac{K _{1} A _{1} ( 300 - T )}{L _{1}} = \frac{K _{2} A _{2} ( T - 0 )}{L _{2}}$

where 1 and 2 refer to the steel and copper rod respectively. For $A_{1} = 2 A_{2}, L_{1} = 15.0 cm$ . $L_{2} = 10.0 cm, K_{1} = 50.2 J s^{- 1} m^{- 1} K^{- 1}, K_{2} = 385 J$ $s^{- 1} m^{- 1} K^{- 1}$ , we have

$\frac{50.2 \times 2 ( 300 - T )}{15} = \frac{385 T}{10}$

which gives $T = 44. 4^{\circ} C$

An iron bar $(L_{1} = 0.1 m, A_{1} = 0.02 m^{2}$ , $K_{1} = 79 W m^{- 1} K^{- 1})$ and a brass bar $(L_{2} = 0.1 m$ , $A_{2} = 0.02 m^{2}, K_{2} = 109 W m^{- 1} K^{- 1}$ ) are soldered end to end as shown in figure. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar, and (iii) the heat current through the compound bar.

Sol. Given, $L_{1} = L_{2} = L = 0.1 m$ ,

$A_{1} = A_{2} = A = 0.02 m^{2} K_{1} = 79 W m^{- 1} K^{- 1} K_{2} = 109 Wm^{- 1} K^{- 1} T_{1} = 373 K^{'} and T_{2} = 273 K$

Under steady state condition, the heat current $(H_{1})$ through iron bar is equal to the heat current $(H_{2})$ through brass bar.

$So, H = H_{1} = H_{2} = \frac{K _{1} A _{1} ( T _{1} - T _{0} )}{L _{1}} = \frac{K _{2} A _{2} ( T _{0} - T _{2} )}{L _{2}}$

For $A_{1} = A_{2} = A$ and $L_{1} = L_{2} = L$ , this equation leads to

$K_{1} (T_{1} - T_{0}) = K_{2} (T_{0} - T_{2})$

Thus, the junction temperature $T_{0}$ of the two bars is

$T_{0} = \frac{( K _{1} T _{1} + K _{2} T _{2} )}{( K _{1} + K _{2} )}$

Using this equation, the heat current H through either bar is

$H = \frac{K _{1} A ( T _{1} - T _{0} )}{L} = \frac{K _{2} A ( T _{0} - T _{2} )}{L} = (\frac{K _{1} K _{2}}{K _{1} + K _{2}}) \frac{A ( T _{1} - T _{0} )}{L} = \frac{A ( T _{1} - T _{2} )}{L ( \frac{1}{K _{1}} + \frac{1}{K _{2}} )}$

Using these equations, the heat current H' through the compound bar of length $L_{1} + L_{2} = 2 L$ and the equivalent thermal conductivity K', of the compound bar are given by

$H^{'} = \frac{K ^{'} A ( T _{1} - T _{2} )}{2 L} = H K^{'} = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}}$

(i)

$T_{0} = \frac{K _{1} T _{1} + K _{2} T _{2}}{K _{1} + K _{2}} = \frac{79 \times 373 + 109 \times 273}{79 + 109} = 315 K$

(ii)

$K^{'} = \frac{2 K _{1} K _{2}}{K _{1} + K _{2}} = \frac{2 \times 79 \times 109}{79 + 109} = 91.6 Wm^{- 1} K^{- 1}$

(iii)

$H^{'} = H = \frac{K ^{'} A ( T _{1} - T _{2} )}{2 L} = \frac{91.6 \times 0.02 \times ( 373 - 273 )}{2 \times 0.1} = 916.1 W$

A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long with it take to cool from 71°C to 69°C? Sol. The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature. Under these conditions the pan cools 8°C in 2 minutes Using Eq. we have $\frac{Change in temperature}{Time} = K Δ T$

$\Rightarrow \frac{94 - 86}{2 min} = K (90 - 20) \frac{8}{2 min} = K (70)$

The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.

$\frac{71 - 69}{Time} = K (70 - 20) \frac{2}{Time} = K (50)$

When we divide above two equations, we have

$\frac{8/2 min}{2/ time} = \frac{K ( 70 )}{K ( 50 )} \Rightarrow 2 time = \frac{7}{5} \Rightarrow time = \frac{7}{10} Time = 0.7 min = 42 s$

EXERCISE QUESTIONS WITH SOLUTIONS

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales. Sol. Kelvin and Celsius scales are related as :

$T_{C} = T_{K} - 273.15$

Celsius and Fahrenheit scales are related as:

$T_{F} = \frac{9}{5} T_{C} + 32$

For neon :

$T_{k} = 24.57 K ∴ T_{c} = 24.57 - 273.15 = - 248.5 8^{\circ} C T_{F} = \frac{9}{5} T_{C} + 32 = \frac{9}{5} (- 248.58) + 32 = 415.4 4^{\circ} F$

For carbon dioxide :

$T_{K} = 216.55 K ∴ T_{C} = 216.55 - 273.15 = - 56.6 0^{\circ} C T_{F} = \frac{9}{5} T_{C} + 32 = \frac{9}{5} (- 56.60) + 32 = - 69.8 8^{\circ} C$

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between $T_{A}$ and $T_{B}$ ?

Sol. Triple point of water on absolute scale A, $T_{1} = 200 A$ . Triple point of water on absolute scale B, $T_{2} = 350 B$ . Triple point of water on Kelvin scale, $T_{K} = 279.15 K$ The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolutes scale A.

$T_{1} = T_{K} 200 A = 273.15 K ∴ A = \frac{273.15}{200}$

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.

$T_{2} = T_{K} 350 B = 273.15 ∴ B = \frac{273.15}{350}$

$T_{A}$ is triple point of water on scale A. $T_{B}$ is triple point of water on scale B.

$∴ \frac{273.15}{200} \times T_{A} = \frac{273.15}{350} \times T_{B} T_{A} = \frac{200}{350} T_{B}$

Therefore, the ratio $T_{A} : T_{B}$ is given as 4 : 7.

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law

$R = R_{0} [1 + α (T - T_{0})]$

The resistance is $101.6Ω$ at the triple-point of water 273.16 K, and $165.5Ω$ at the normal melting point of lead (600.5 K). What is the temperature when the resistance is $123.4Ω$ ? Sol. It is given that :

$R = R_{0} [1 + α (T - T_{0})]$

Where, $R_{0}$ and $T_{0}$ are the initial resistance and temperature respectively andR and T are the final resistance and temperature respectively $α$ is a constant. At the triple point of water, $T_{0} = 273.15 K$ Resistance of lead, $R_{0} = 101.6Ω$ At normal melting point of lead, $T = 600.5 K$ Resistance of lead, $R = 165.5Ω$ Substituting these values in equation (i), We get :

$R = R_{0} [1 + α (T - T_{0})] 165.5 = 101.6 [1 + α (600.5 - 273.15)] 1.629 = 1 + α (327.35) ∴ α = \frac{0.629}{327.35} = 1.92 \times 1 0^{- 3} K^{- 1}$

For resistance, $R_{1} = 123.4Ω$

$R_{1} = R_{0} [1 + α (T - T_{0})]$

Where, T is the temperature when the resistance of lead is $123.4Ω$

$123.4 = 101.6 [1 + 1.92 \times 1 0^{- 3} (T - 273.15)] 1.214 = 1 + 1.92 \times 1 0^{- 3} (T - 273.15) \frac{0.214}{1.92 \times 1 0 ^{- 3}} = T - 273.15 ∴ T = 384.61 K$

Answer the following : (a) The triple-point of water is a standard fixed point in modern thermometry. Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was original done in the Celsius scale)? (b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature $t_{c}$ on the Celsius scale by

$t_{c} = T - 273.15$

Why do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Sol. (a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular value because these points depend on pressure and temperature.

(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale. (c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K. Hence, absolute temperature (Kelvin scale)T, is related to temperature $t_{c}$ , on Celsius scale as :

$t_{c} = T - 273.15$

(d) Let $T_{F}$ be the temperature on Fahrenheit scale and $T_{K}$ be the temperature on absolute scale. Both the temperatures can be related as :

$\frac{T _{F} - 32}{180} = \frac{T _{K} - 273.15}{100}$

Let $T_{F_{1}}$ be the temperature on Fahrenheit scale and $T_{K_{1}}$ be the temperature on absolute scale. Both the temperatures can be related as :

$\frac{T _{F_{1}} - 32}{180} = \frac{T _{K_{1}} - 273.15}{100}$

It is given that :

$T_{K_{1}} - T_{k} - 1 K$

Subtracting equation (i) from equation (ii), we get :

$\frac{T _{F_{1}} - T _{F}}{180} = \frac{T _{K_{1}} - T _{K}}{100} = \frac{1}{100} T_{F_{1}} - T_{F} = \frac{1 \times 180}{100} = \frac{9}{5}$

Triple point of water = 273.16 K

$∴ Triple point of water on absolute scale = 273.16 \times \frac{9}{5} = 491.69$

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :

Temperature	Pressure Thermometer A	Pressure Thermometer B
Triple-point of water	$1.250 \times 1 0^{5} Pa$	$0.200 \times 1 0^{5} PA$
Normal melting point of sulphur	$1.797 \times 1 0^{5} Pa$	$0.287 \times 1 0^{5} Pa$

What is the absolute temperature of normal melting point of sulphur as ready by thermometers A and B? What do you think is the reason behind the slight difference in answers of thermometers A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Sol. Triple point of water, $T = 273.16 K$ At this temperature, pressure in thermometer A, $P_{A} = 1.250 \times 1 0^{5} Pa$ Let $T_{1}$ be the normal melting point of sulphur. At this temperature, pressure in thermometer $A, P_{1} = 1.797 \times 1 0^{5} Pa$ According to Charles' law, we have the relation :

$\frac{P _{A}}{T} = \frac{P _{1}}{T _{1}} ∴ T_{1} = \frac{P _{1} T}{P _{A}} = \frac{1.797 \times 1 0 ^{5} \times 273.16}{1.25 \times 1 0 ^{5}} = 392.69 K$

Therefore, the absolute temperature of the normal melting point of sulphur as ready by thermometer A is 392.69 K. At triple point 273.16 K, the pressure in thermometer $B, P_{B} = 0.200 \times 1 0^{5} Pa$ At temperature $T_{1}$ , the pressure in thermometer B, $P_{2} = 0.287 \times 1 0^{5} Pa$ According to Charles' law, we can write the relation :

$\frac{P _{B}}{T} = \frac{P _{2}}{T _{1}} \Rightarrow T_{1} = \frac{p _{2} \cdot T}{p _{B}} \frac{0.287 \times 1 0 ^{5}}{0.200 \times 1 0 ^{5}} \times 273.16 = 391.98 K$

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K. The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B. To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases.

A steel tape 1 m long is correctly calibrated for a temperature of $27. 0^{\circ} C$ . The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is $45. 0^{\circ} C$ . What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0°C? Coefficient of linear expansion of steel $= 1.20 \times 1 0^{- 5} K^{- 1}$ . Sol. Length of the steel tape at temperature

$T = 2 7^{\circ} C, l = 1 m = 100 cm$

At temperature $T_{1} = 4 5^{\circ} C$ , the length of the steel rod, $l_{1} = 63 cm$ Coefficient of linear expansion of steel,

$α = 1.20 \times 1 0^{- 5} K^{- 1}$

Let $l_{2}$ be the actual length of the steel rod and $l^{'}$ be the length of the steel tape at 45°C.

$l^{'} = l + α l (T_{1} - T) ∴ l^{'} = 100 + 1.20 \times 1 0^{- 5} \times 100 (45 - 27) = 100.0216 cm$

Hence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as :

$l_{2} = \frac{100.0216}{100} \times 63 = 63.0136 cm$

Therefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.

A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :

$α_{steel} = 1.20 \times 1 0^{- 5} K^{- 1} .$

Sol. The given temperature, $T = 2 7^{\circ} C$ can be written in Kelvin as :

$27 + 273 = 300 K$

Outer diameter of the steel shaft at T,

$d_{1} = 8.70 cm$

Diameter of the central hole in the wheel at T,

$d_{2} = 8.69 cm$

Coefficient of linear expansion of steel,

$α_{steel} = 1.20 \times 1 0^{- 5} K^{- 1}$

After the shaft is cooled using 'dry ice', its temperature becomes $T_{1}$ . The wheel will slip on the shaft, if the change in diameter,

$Δ d = 8.69 - 8.70 = - 0.01 cm$

Temperature $T_{1}$ , can be calculated from the relation :

$Δ d = d_{1} α_{steel} (T_{1} - T) - 0.01 = 8.70 \times 1.20 \times 1 0^{- 5} (T_{1} - 300) (T_{1} - 300) = - 95.78 ∴ T_{1} = 204.21 K = 204.21 - 273.16 = - 68.9 5^{\circ} C$

Therefore, the wheel will slip on the shaft when the temperature of the shaft is $- 6 9^{\circ} C$ .

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper $= 1.70 \times 1 0^{- 5} K^{- 1}$ . Sol. Initial temperature, $T_{1} = 27. 0^{\circ} C$ Diameter of the hole at $T_{1}, d_{1} = 4.24 cm$ Final temperature, $T_{2} = 22 7^{\circ} C$ Diameter of the hole at $T_{2} = d_{2}$ Co-efficient of linear expansion of copper,

$α_{Cu} = 1.70 \times 1 0^{- 5} K^{- 1}$

For co-efficient of superficial expansion $β$ , and change in temperature $Δ T$ , we have the relation : $\frac{Change in area ( Δ A )}{Original area ( A )} = β Δ T$

$\frac{( x \frac{d _{2}^{2}}{4} - π \frac{d _{1}^{2}}{4} )}{( π \frac{d _{1}^{2}}{4} )} = \frac{Δ A}{A} ∴ \frac{Δ A}{A} = \frac{d _{2}^{2} - d _{1}^{2}}{d _{1}^{2}}$

But $β = 2 α$

$\frac{d _{2}^{2}}{d _{1}^{2}} - 1 = 2 α (T_{2} - T_{1}) ∴ \frac{d _{2}^{2} - d _{1}^{2}}{d _{1}^{2}} = 2 α Δ T \frac{d _{2}^{2}}{( 4.24 ) ^{2}} = 2 \times 1.7 \times 1 0^{- 5} (227 - 27) + 1 d_{2}^{2} = 17.98 \times 1.0068 = 18.1 ∴ d_{2} = 4.2544 cm$

Change in diameter $= d_{2} - d_{1}$

$= 4.2544 - 4.24 = 0.0144 cm$

Hence, the diameter increases by

$1.44 \times 1 0^{- 2} cm .$

A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $- 3 9^{\circ} C$ , what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass $= 2.0 \times 1 0^{- 5} K^{- 1}$ ; Young's modulus of brass $= 0.91 \times 1 0^{11} Pa$ .

Sol. Initial temperature, $T_{1} = 2 7^{\circ} C$ Length of the brass wire at $T_{1}, l = 1.8 m$ Final temperature, $T_{2} = - 3 9^{\circ} C$ Diameter of the wire,

$d = 2.0 mm = 2 \times 1 0^{- 3} m$

Tension developed in the wire $= F$ Coefficient of linear expansion of brass,

$α = 2.0 \times 1 0^{- 5} K^{- 1}$

Young's modulus of brass,

$Y = 0.91 \times 1 0^{11} Pa$

Young's modulus is given by the relation :

$Y = \frac{Stress}{Strain} = \frac{F / A}{Δ L /Δ L} Δ L = \frac{F \times L}{A \times Δ L}$

Where, F = Tension developed in the wire A = Area of cross-section of the wire.

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a 'thermal stress' developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass $= 2.0 \times 1 0^{- 5} K^{- 1}$ , steel $= 1.2 \times 1 0^{- 5} K^{- 1}$ ). Sol. Initial Temperature, $T_{1} = 4 0^{\circ} C$ Final temperature, $T_{2} = 2 5^{\circ} C$ Change in temperature $Δ T = T_{2} - T_{1} = 21 0^{\circ} C$ Length of the brass rod at $T_{1}, l_{1} = 50 cm$ Diameter of the brass red at $T_{1}, d_{1} = 3.0 mm$ Length of the steel rod at $T_{2}, l_{2} = 50 cm$ Diameter of the steel rod $T_{2}, d_{2} = 3.0 mm$ Coefficient of linear expansion of brass,

$α_{1} = 2.0 \times 1 0^{- 5} K^{- 1}$

Coefficient of linear expansion of steel,

$α_{2} = 1.2 \times 1 0^{- 5} K^{- 1}$

For the expansion in the brass rod, we have:

$\frac{Change in length ( Δ l _{1} )}{Original length ( l _{1} )} = α_{1} Δ T ∴ Δ l_{1} = 50 \times (2.1 \times 1 0^{- 5}) \times 210 = 0.2205 cm$

For the expansion in the steel rod, we have: $\frac{Change in length ( Δ l _{2} )}{Original length ( l _{2} )} = α_{2} Δ T$ $∴ Δ l_{2} = 50 \times (1.2 \times 1 0^{- 5}) \times 210 = 0.126 cm$ Total change in the lengths of brass and steel,

$Δ l = Δ l_{1} + Δ l_{2} = 0.2205 + 0.126 = 0.346 cm$

Total change in the lengths of the combined rod $= 0.346 cm$ Since the rod expands freely from both ends, no thermal stress is developed at the junction.

The coefficient of volume expansion of glycerine is $49 \times 1 0^{- 5} K^{- 1}$ . What is the fractional change in its density for a 30°C rise in temperature? Sol. Coefficient of volume expansion of glycerine,

$α_{v} = 49 \times 1 0^{- 5} K^{- 1}$

Rise in temperature, $T = 3 0^{\circ} C$ Fractional change in its volume $= \frac{Δ V}{V}$ This change is related with the change in temperature as :

$\frac{Δ V}{V} = α_{1} Δ T V_{T_{2}} - V_{T_{1}} = V_{T_{1}} α_{V} Δ T \frac{m}{ρ _{T_{2}}} - \frac{m}{ρ _{T_{1}}} = \frac{m}{ρ _{T_{1}}} α_{V} Δ T$

Where,

$m = Mass of glycerine ρ_{T_{1}} = Initial density at T_{1} ρ_{T_{2}} = Final density at T_{2} \frac{ρ _{T_{1}} - ρ _{T_{2}}}{ρ _{T_{2}}} = α_{V} Δ T$

Where, $\frac{ρ _{T_{1}} - ρ _{T_{2}}}{ρ _{T_{2}}}$ = Fractional change in density ∴ Fractional change in the density of glycerine $= 49 \times 1 0^{- 5} \times 30 = 1.47 \times 1 0^{- 2}$

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

$= 10 kW = 10 \times 1 0^{3} W$

Mass of the aluminium block,

$m = 8.0 kg = 8 \times 1 0^{3} g$

Time for which the machine is used,

$t = 2.5 min = 2.5 \times 60 = 150 s$

Specific heat of aluminium,

$c = 0.91 J g^{- 1} K^{- 1}$

Rise in the temperature of the block after

$drilling = δ T$

Total energy of the drilling machine $= Pt$

$= 10 \times 1 0^{3} \times 150 = 1.5 \times 1 0^{6} J$

It is given that only 50% of the power is useful. Useful energy,

$Δ Q = \frac{50}{100} \times 1.5 \times 1 0^{6} = 7.5 \times 1 0^{5} J$

But $Δ Q = mc Δ T$

$∴ Δ T = \frac{Δ Q}{mc} = \frac{7.5 \times 1 0 ^{5}}{8 \times 1 0 ^{3} \times 0.91} = 10 3^{\circ} C$

Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is $10 3^{\circ} C$ .

A copper block of mass 2.5 kg is heated in a furnace to a temperature of $50 0^{\circ} C$ and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper $= 0.39 Jg^{- 1} K^{- 1}$ ; heat of fusion of water $= 335 J g^{- 1}$ ) Sol. Mass of the copper block,

$m = 2.5 kg = 2500 g$

Rise in the temperature of the copper block

$Δ θ = 50 0^{\circ} C$

Specific heat of copper, $C = 0.39 J g^{- 1} C^{- 1}$ Heat of fusion of water, $L = 335 Jg^{- 1}$ The maximum heat the copper block can lose,

$Q = mC Δ θ = 2500 \times 0.39 \times 500 = 487500 J$

Let $m_{1} g$ be the amount of ice that melts when the copper block is placed on the ice block. The heat gained by the melted ice, $Q = m_{1} L$

$∴ m_{1} = \frac{Q}{L} = \frac{487500}{335} = 1455.22 g$

Hence, the maximum amount of ice that can melt is 1.45 kg

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg ) containing $150 cm^{3}$ of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal? Sol. Mass of the metal $m, = 0.20 kg = 200 g$ Initial temperature of the metal, $T_{1} = 15 0^{\circ} C$ Final temperature of the metal, $T_{2} = 4 0^{\circ} C$ Calorimeter has water equivalent of mass,

$m^{'} = 0.025 kg = 25 g$

Volume of water, $V = 150 cm^{3}$ Mass (M) of water at temperature

$T = 2 7^{\circ} C : M = 150 \times 1 = 150 g$

Fall in the temperature of the metal :

$Δ T = T_{1} - T_{2} = 150 - 40 = 11 0^{\circ} C$

Specific heat of water,

$C_{w} = 4.186 J g^{- 1} K^{- 1}$

Specific heat of the metal = C Heat lost by the metal,

$θ = mC Δ T$

Rise in the temperature of the water and calorimeter system :

$Δ T^{'} = 40 - 27 = 1 3^{\circ} C$

Heat gained by the water and calorimeter system :

$Δ θ^{''} = m_{1} C_{w} Δ T^{1} = (M + m^{'}) C_{w} Δ T^{1}$

Heat lost by the metal = Heat gained by the water and calorimeter system

$mC Δ T = (M + m^{1}) C_{w} Δ T^{1} 200 \times C \times 110 = (150 + 25) \times 4.186 \times 13 ∴ C = \frac{175 \times 4.186 \times 13}{110 \times 200} = 0.43 Jg^{- 1} K^{- 1}$

Is some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat (CV) (cal mol $^{- 1} K^{- 1}$ )
Hydrogen	4.87
Nitrogen	4.97
Oxygen	5.02
Nitric oxide	4.99
Carbon monoxide	5.01
Chlorine	6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is $2.92 cal / mol K$ . Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Sol. The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion).

Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.

If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas $= \frac{5}{2} R$

$= \frac{5}{2} \times 1.98 = 4.95 cal mol^{- 1} K^{- 1}$

With the exception of chlorine, all the observations in the given table agree with $(\frac{5}{2} R)$ .

This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

A child running a temperature of $10 1^{\circ} F$ is given an antipyrin (i.e. medicine that lowers fever) which cause an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about $580 cal g^{- 1}$ .

Sol. Initial temperature of the body of the child,

$T_{1} = 10 1^{\circ} F$

Final temperature of the body of the child,

$T_{2} = 9 8^{\circ} F$

Change in temperature,

$Δ T = [(101 - 98) \times \frac{5}{9}]^{\circ} C = \frac{5}{3}^{\circ} C$

Time taken to reduce the temperature, $t = 20 min$

Mass of the child, $m = 30 kg = 30 \times 1 0^{3} g$ Specific heat of the human body = Specific heat of water $= c = 1000 cal kg^{- 1}^{\circ} C^{- 1}$ Latent heat of evaporation of water, $L = 580 cal g^{- 1}$ The heat lost by the child is given as:

$Δ θ = mc Δ T = 30 \times 1000 \times \frac{5}{3} = 50000 cal$

Let $m_{1}$ be the mass of the water evaporated from the child's body in 20 min. Loss of heat through water is given by :

$Δ θ = m_{1} L ∴ m_{1} = \frac{Δ θ}{L} = \frac{50000}{580} = 86.2 g$

∴ Average rate of extra evaporation caused by the $drug = \frac{m _{1}}{t} = \frac{86.2}{20} = 4.3 g / min$

A 'thermacole' icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box. estimate the amount of ice remaining after 6h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermacole is $0.01 Js^{- 1} m^{- 1} K^{- 1}$ . [Heat of fusion of water $= 335 \times 1 0^{3} J kg^{- 1}$ ]

Sol. Side of the given cubical ice box,

$s = 30 cm = 0.3 m$

Thickness of the ice box,

$l = 5.0 cm = 0.05 m$

Mass of ice kept in the ice box, $m = 4 kg$ Time gap, $t = 6 h = 6 \times 60 \times 60 s$ Outside temperature, $T = 4 5^{\circ} C$

Coefficient of thermal conductivity of thermacole, $K = 0.01 J s^{- 1} m^{- 1} K^{- 1}$ Heat of fusion of water, $L = 335 \times 1 0^{3} J kg^{- 1}$ Let m' be the total amount of ice that melts in 6 h . The amount of heat lost by the food :

$θ = \frac{KA ( T - 0 ) t}{l}$

Where,

$A θ = Surface area of the box = 6 s^{2} = 6 \times (0.3)^{2} = 0.54 m^{3} = \frac{0.01 \times 0.54 \times ( 45 ) \times 6 \times 60 \times 60}{0.05} = 104976 J$

But $θ =$ m'L $^{'}$

$∴ m^{'} = \frac{θ}{L} = \frac{104976}{335 \times 1 0 ^{3}} = 0.313 kg$

Mass of ice left $= 4 - 0.313 = 3.687 kg$ Hence, the amount of ice remaining after 6h is 3.687 kg.

A brass boiler has a base area of $0.15 m^{2}$ and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when place on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass $= 109 Js^{- 1} m^{- 1} K^{- 1}$ ; Heat of vaporisation of water $= 2256 \times 1 0^{3} Jkg^{- 1}$ . Sol. Base area of the boiler, $A = 0.15 m^{2}$ Thickness of the boiler, $l = 1.0 cm = 0.01 m$ Boiling rate of water, $R = 6.0 kg / min$ Mass, $m = 6 kg$ Time, $t = 1 min = 60 s$ Thermal conductivity of brass,

$K = 109 J s^{- 1} m^{- 1} K^{- 1}$

Heat of vaporisation, $L = 2256 \times 1 0^{3} J kg^{- 1}$ The amount of heat flowing into water through the brass base of the boiler is given by :

$θ = \frac{KA ( T _{1} - T _{2} ) t}{l}$

Where, $T_{1} =$ Temperature of the flame in contact with the boiler $T_{2} =$ Boiling point of water $= 10 0^{\circ} C$ Heat required for boiling the water.

$θ = mL$

Equating equation (i) and (ii), we get :

$∴ mL = \frac{KA ( T _{1} - T _{2} ) t}{l} T_{1} - T_{2} = \frac{mL l}{KAt} = \frac{6 \times 2256 \times 1 0 ^{3} \times 0.01}{109 \times 0.15 \times 60} = 137.9 8^{\circ} C$

Therefore, the temperature of the part of the flame in contact with the boiler is $237.9 8^{\circ} C$ .

Explain why:

(a) a body with large reflectivity is a poor emitter. (b) a brass tumbler feels much colder than a wooden tray on a chilly day. (c) an optical pyrometer (for measuring high temperature) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open but gives a correct value for the temperature when the same piece is in the furnace. (d) the earth without its atmosphere would be inhospitably cold. (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

Sol. A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter. Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.

Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and does not feel cool. Thus, a brass tumbler feels colder than a wooden tray on a chilly day. An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open. Black body radiation equation is given by : $E = σ (T^{4} - T_{0}^{4})$ Where, $E =$ Energy radiation T = Temperature of optical pyrometer $T_{0} =$ Temperature of open space $σ =$ Constant Hence, an increase in the temperature of open space reduces the radiation energy. When the same piece of iron is placed in furnace, the radiation energy, $E = σ T^{4}$ Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth's surface. A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C. Sol. According to Newton's law of cooling, we have :

$- \frac{dT}{dt} = K (T - T_{0}) \frac{dT}{( T - T _{0} )} = Kdt$

Where, Temperature of the body = T Temperature of the surroundings

$= T_{0} = 2 0^{\circ} C$

K is a constant Temperature of the body falls from 80°C to $5 0^{\circ} C$ in time, $t = 5 min = 300 s$

Integrating equation (i), we get :

$\int_{50}^{80} \frac{dT}{( T - T _{0} )} = - \int_{0}^{300} Kdt [lo g_{e} (T - T_{0})]_{50}^{80} = - K [t]_{0}^{300} \frac{2.3026}{K} lo g_{10} \frac{80 - 20}{50 - 20} = - 300 \frac{2.3026}{K} lo g_{10} \frac{80 - 20}{50 - 20} = - 300 \frac{- 2.3026}{300} lo g_{10} 2 = K \dots . (ii)$

The temperature of the body falls from 60°C to 30°C in time $t = t$ Hence, we get :

$\frac{2.3026}{K} lo g_{10} \frac{60 - 20}{30 - 20} = - t \frac{- 2.3026}{t} lo g_{10} 4 = K$

Equating equation (ii) and (iii), we get :

$\frac{- 2.3026}{t} lo g_{10} 4 = \frac{- 2.3026}{300} lo g_{10} 2 ∴ t = 300 \times 2 = 600 s = 10 min$

Therefore, the time taken to cool the body from $6 0^{\circ} C$ to $3 0^{\circ} C$ is 10 minutes.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Use NCERT Solutions for Class 11 Physics chapter-wise to understand textbook questions with clear and easy explanations. Learn key concepts, follow the correct steps to solve problems, and build a strong foundation in Physics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 2	Motion in a Straight Line
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Wave

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 10

Step-by-Step Calorimetry Solutions: Solve numerical problems involving heat transfer, calorimetry, and ice-water-steam mixtures with clear calculations and units.
Thermal Expansion Relations: Understand the relationship between linear, superficial, and volume expansion coefficients through clear derivations.
Thermal Conductivity Explained: Learn the role of thermal conductivity in heat transfer and its applications in insula ted containers and heat sinks.
Heating and Cooling Curves: Understand heating curves and cooling curves and why temperature remains constant during a change of state.
Prepared by ALLEN Subject Experts: Get accurate and detailed NCERT Solutions for Class 11 Physics Chapter 10, prepared according to the latest NCERT syllabus.
Simple and Concept- Based Explanation : Study about thermal expansion, heat transfer, calorimetry, specific heat capacity, latent heat and change of state with clear explanations and step-wise solutions.
Complete NCERT Exercise Solutions: Complete NCERT Solutions for Class 11 Physics Chapter 10 examples and exercise questions and important concepts for CBSE, JEE and NEET Preparation.

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