Topics

A device in which heat measurement can be made is called a calorimeter. It consists a metallic vessel and stirrer of the same material like copper or aluminium. The vessel is kept inside a wooden jacket which contains heat insulating materials like glass wool etc. The outer jacket acts as a heat shield and reduces the heat loss from the inner vessel. There is an opening in the outer jacket through which a mercury thermometer can be inserted into the calorimeter.

From this experiment he came up with new physical quantities.
Heat capacity \(\left( C^{'} \right):\ Q = \int_{}^{}\ C^{'}dT\)
specific heat capacity: \(\ Q = m\int_{T_{j}}^{T_{j}}\mspace{2mu} sdT = ms_{avg}\Delta T\)
Molar heat Capacity: \(\ \ C = \frac{C^{'}}{n}\ \) (n - no. of moles)
The branch of thermodynamics which deals with the measurement of Heat is called colorimetry. When two bodies at different temperature are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature. Principle of colorimetry represents the law of conservation of Heat Energy.

Heat lost \(=\) Heat gained

Specific Heat capacity

The amount of heat needed to raise the temperature of unit mass of a material by unit degree of measurement is known as the specific heat capacity of that material. If \(Q\) amount of heat raises the temperature of mass \(m\) of a material by \(\Delta T\), then its specific heat capacity is given as :

\[s = \frac{Q}{\text{ }m\Delta\text{ }T}\ \Rightarrow \ Q = ms\Delta\text{ }T\]

Also the amount of heat supplied per unit increase in temperature for any body is known as its heat capacity, \(c = \frac{Q}{\Delta T} = ms\).

Latent Heat

Heat required for the change of phase or state. No chage in temperature is involved when substance changes its state or phase. ( \(Q = mL,L =\) Latent Heat)

Latent Heat of Fusion : The Heat supplied to a substance which changes it from solid to liquid state at its melting point and 1 atm . pressure is called latent Heat of fusion. \(\left( Q = {mL}_{f} \right)\)
Latent heat of fusion of Ice \(\left( L_{f} \right) = 80cal/gm\).
Latent Heat of Vapourization : The Heat supplied to a substance which changes it from liquid to vapour state at its boiling point and 1 atm pressure is called latent heat of vapourization. \(\left( Q = {mL}_{v} \right)\) Latent heat of vapouriztion of water \(\left( L_{v} \right) = 540cal/g\).

If to a given mass ( m ) of a solid (Ice), Heat is supplied at constant rate P and a graph is poltted between temperature and time

(1) In the region OA

Temperature of solid is changing with time

\[\begin{matrix} Q = ms\Delta\text{ }T & \\ P(\Delta t) = ms\Delta\text{ }T & \left( \Delta t = t_{1} - 0,\Delta\text{ }T = 0 - ( - T) \right) \\ \frac{P}{\text{ }ms} = \frac{\Delta T}{\Delta t} & \frac{\Delta\text{ }T}{\Delta t} = \text{~}\text{slope of line}\text{~}OA \end{matrix}\]

(2) In the region AB

Temperature is constant, here substance changes its phase solid to liquid, between \(A\) and \(B\).

\[\begin{matrix} & Q = {mL}_{f} \\ & P\Delta t = {mL}_{f} \\ & {\text{ }L}_{f} = \frac{P\left( t_{2} - t_{1} \right)}{m} \\ & {\text{ }L}_{f} = \text{~}\text{length of line}\text{~}AB \end{matrix}\]

Latent Heat of fusion is proportional to length of line.
(3) In the Region BC

Temp. of liquid is increasing with time

\[\begin{matrix} Q = ms\Delta\text{ }T & \left( \Delta t = t_{3} - t_{2},\Delta\text{ }T = 100 - 0 \right) \\ P\Delta t = ms\Delta\text{ }T & \end{matrix}\]

(4) In the regin \(CD\), temperature is constant, so it represents change of state.

\[\begin{matrix} & Q = {mL}_{v} \\ & \frac{P\left( t_{4} - t_{3} \right)}{m} = L_{v} \\ & {\text{ }L}_{v} = \text{~}\text{length of line}\text{~}CD \end{matrix}\]

(5) The line DE represents gaseous state of substance with its temperature increasing linearly with time. The reciprocal of slope of line will be proportional to specific heat of substance in vapour state.

It is a equivalent mass of water (w) that has same heat capacity as that of the given body (b). In other words,

\[C = m_{w}{\text{ }s}_{w} = m_{b}{\text{ }s}_{b}\]

It is a convenient way to represent the heat capacity of the calorimeter

Practice Exercise

Q. 11 kg of ice at \(- 10^{\circ}C\) is mixed with 1 kg water at \(100^{\circ}C\), then find the equilibrium temperature and mixture content.
Q. 25 gm of ice at \(0^{\circ}C\) is mixed with 10 gm of steam at \(100^{\circ}C\). Find the final temperature and composition of the mixture if the mixing is done in a calorimeter of water equivalent 13 gm , initially at \(0^{\circ}C\).
Q. 3 A lumb of ice of 0.1 kg at \(- 10^{\circ}C\) is put in 0.15 kg of water at \(20^{\circ}C\). How much water and ice will be found in the mixtre when it has reached thermal equilibrium.

Answers

Q. \(1\text{ }T = {7.5}^{\circ}C\ \) Q. \(2\ 245/27gm\) of water, \(160/27gm\) of steam
Q. 3 Amount of water amd Ice are 181.25 gm and 68.75 gm respetively.

When matter is heated without change in state, it usually expands. According to atomic theory of matter, asymmetry in potential energy curve is responsible for thermal expansion as with rise in temperature say from \(T_{1}\) to \(T_{2}\) the amplitude of vibration and hence energy of atoms increases from \(E_{1}\) to \(E_{2}\) and hence the average distance between atoms increases from \(r_{1}\) to \(r_{2}\).

Due to this increase in distance between atoms, the matter as a whole expands. Had the potential energy curve been symmetrical, no thermal expansion would have taken place in spite of heating.

Linear Expansion of solids

To varying extents, most materials expand when heated and contract when cooled. The increase in any one dimension of a solid is called linear expansion, linear in the sense that the expansion occurs along a line. A rod whose length is \(L_{0}\) when the temperature is \(T_{0}\) when the temperature increases to \(T_{0} + \Delta T\), the length becomes \(L_{0} + \Delta L\), where \(\Delta T\) and \(\Delta L\) are the magnitudes of the changes in temperature and length, respectively.
Conversely, when the temperature decreases to \(T_{0} - \Delta T\), the length decreases to \(L_{0} - \Delta L\). For small temperature changes, experiments show that the change in length is directly proportional to the change in temperature ( \(\Delta L \propto \Delta T\) ). In addition, the change in length is proportional to the initial length of the rod,

Equation \(\Delta L = \alpha L_{0}\Delta\text{ }T\) expresses the fact that \(\Delta L\) is proportional to both \(L_{0}\) and \(\Delta T\left( \Delta L \propto L_{0}\Delta\text{ }T \right)\) by using a proportionality constant \(\alpha\), which is called the coefficient of linear expansion. Common unit for the coefficient of linear expansion \(\left( C^{\circ} \right)^{- 1}\).

A bimetallic strip is made from two thin strips of metal that have different coefficients of linear expansion, as fig. (a) A bimetallic strip and how it behaves when (b) heated and (c) cooled

(a)
(b) Heated
(c) Cooled

Often brass \(\left\lbrack \alpha = 19 \times 10^{- 6}\left( C^{\circ} \right)^{- 1} \right\rbrack\) and steel \(\left\lbrack \alpha = 12 \times 10^{- 6}\left( C^{\circ} \right)^{- 1} \right\rbrack\) are selected. The two pieces are welded or riveted together. When the bimetallic strip is heated, the brass, having the larger value of \(\alpha\), expands more than the steel. Since the two metals are bonded together, the bimetallic strip bends into an arc as in fig. (b), with the longer brass piece having a larger radius than the steel piece. When the strip is cooled, the bimetallic strip bends in the opposite direction, as in fig. (c).

Mathematical analysis

\[\begin{matrix} & \left( R + \frac{d}{2} \right)\phi = L_{0}\left( 1 + \alpha_{1}\Delta\theta \right)\ (\Delta\theta\text{~}\text{increase in temp.}\text{~}) \\ & \left( R - \frac{d}{2} \right)\phi = L_{0}\left( 1 + \alpha_{2}\Delta\theta \right) \end{matrix}\]

On dividing above equations we get

\[\frac{R + \frac{d}{2}}{R - \frac{d}{2}} = \frac{1 + \alpha_{1}\Delta\theta}{1 + \alpha_{2}\Delta\theta}\]

by above euqation we can find mean radius \(R\) of bimetallic strip.

\[R = \frac{d}{\left( \alpha_{1} - \alpha_{2} \right)\Delta\theta}\]

Area and Volume Expansion

Area Expansion

Volume Expansion

If the temperature of a two-dimensional object (lamina) is changed, its area changes. If the coefficient of linear expansion of the material of lamina is small and constant, then its final area is given by \(A = A_{0}(1 + \beta\Delta T)\), where \(A_{0}\) is the initial area. \(\Delta T\) is the change in temperature and \(\beta\) is the area coefficient of thermal expansion. For isotropic bodies it can be shown the \(\beta = 2\alpha\).

The volume \(V_{0}\) of an object change by an amount \(\Delta V\) when its temperature changes by an amount \(\Delta T\). \(\Delta V = \gamma V_{0}\Delta\text{ }T\) where \(\gamma\) is the coefficient of volume expansion. Common Unit for the coefficient of volume Expansion : \(\left( C^{\circ} \right)^{- 1}\). The unit for \(\gamma\), like that for \(\alpha\), is \(\left( C^{\circ} \right)^{- 1}\). Values for \(\gamma\) depend on the nature of the material. The values of \(\gamma\) for liquids are substantially larger than those for solids, because liquids typically expand more than solids, given the same initial volumes and temperature expansion is three times greater than the coefficient of linear expansion : \(\gamma = 3\alpha\).
If a cavity exists within a solid object, the volume of the cavity increases when the object expands, just as if the cavity were filled with the surrounding material. The expansion of the cavity is analogous to the expansion of a hole in a sheet of material. Accordingly, the change in volume of a cavity can be found using the relation \(\Delta V = \gamma V_{0}\Delta\text{ }T\), where \(\gamma\) is the coefficient of volume expansion of the material that surrounds the cavity.
Similar ( \(Here\gamma \approx 3\alpha\) ) is known as the coefficient of volume expansion

\[\alpha:\beta:\gamma::1:2:3\]

Illustration : (The expansion of holes)

Do holes expand or contract when the temperature increases?
Figure (a) shows eight square tiles that are arranged to form a square pattern with a hole in the centre. If the tiles are heated, what happens to the size of the hole?

(a) unheated

(b) Heated

(c)

Sol. We can analyze this problem by disassembling the pattern into separate tiles, heating, it is evident from figure (b) that the heated pattern expands and so does the hole in the centre. In fact, if we had a ninth tile that was identical to and also heated like the others, it would fit exactly into the centre hole, as figure (c) indicates. Thus, not only does the hole in the pattern expand, but it expands exactly as much as one of the tiles. Since the ninth tile is made of the same material as the others, we see that the hole expands just as if it were made of the material of the surrounding tiles.
The thermal expansion of the hole and the surrounding material is analogous to a photographic enlargement ; in both situations everything is enlarged, including holes.
Thus, it follows that a hole in a piece of solid material expands when heated and contracts when cooled, just as if it were filled with the material that surrounds it. If the hole is circular, the equation \(\Delta L = \alpha L_{0}\Delta T\) can be used to find the change in any linear dimension of the hole, such as its radius or diameter. Example illustrates this type of linear expansion.

A change in shape/size i.e., dimensions need not necessarily imply a strain. For example, if a body is heated to expand, its volume change, as it acquires a new size, due to expansion. However, the strain remains zero. Unless and until, internal elastic forces operate, to bring the body to the original state, no strain exists. When a body is heated, the total energy of molecule increase, owing to an increase in the kinetic energy of the molecules. This results in a shift (increment) of the "equilibrium distance" of molecules and the body acquires a new shape and size, in the expanded form, whereby the molecules are in "zero force" state. Hence, there is no strain. However, if the body is resistricted to expand, during the process of heating, then the molecules become "strained", and even if there is no apparent change in dimensions of the body, there is strain. In such cases, strain is measured as the ratio. In dimension that would have occured, and the change in dimension that would have occured, had the body been free to expand or contract, to the original dimension.

When a metal rod is heated or cooled it tends to expand or contract. If it is left free to expand or contract, no temperature stresses will be induced. However, if the rod be restricted to change its length, then temperature stresses are generated within it. Stress induced due to temperature change can be understood as follows:

Consider a uniform rod AB fixed rigidly between two supports. (fig.) If L be its length, \(\alpha\) the coefficient of linear expansion, then a change in temperature of \(\Delta\theta\), would tend to bring a change in its length by \(l = L\alpha\Delta\theta\). Had the rod been free (say one of its ends) its length would have changed by \(l\). Now, let a force be gradually applied so as to restore the natural length. Since the rod, tends to remain in the new state, due to a change in temperature, so when a force F is applied, thermal stress is induced. In equilibrium.

\[\frac{F}{\text{ }A} = \frac{l}{(\text{ }L \pm l)}Y\ \lbrack\because\text{~}\text{stress}\text{~} = strain \times Y\rbrack\]

Neglecting \(l\) in comparison to L ,

\[F = \frac{l\text{ }A}{\text{ }L}Y = AY\alpha\Delta\theta\]

Now, if the two ends remain fixed, then this external force is provided from the support.
Clearly strain \(= \frac{\mathcal{l}}{L} = \alpha\Delta\theta\)

Like solids, liquids also, in general, expand on heating; however, their expansion is much large compared to solids for the same temperature rise. A noteworthy point to be taken into account during the expansion of liquid is that they are always contained in a vessel or a container and hence the expansion of the vessel also comes into picture. Further, linear or superficial expansion in case of a liquid does not carry any sense. Consider a liquid contained in a round bottomed flask fitted with a long narrow stem as shown in fig. Let the initial level of the liquid be X . When it is heated the level falls initially to Y .

However, after sometime, the liquid level eventually rises to \(Z\). The entire phenomenon can be understood as follows: Upon being heated, the container gets heated first and hence expands. As a result, the capacity of the flask increases and hence the liquid level falls.
After sometime, the heat gets conducted from the vessel to the liquid and hence liquid also expands thereby rising its level eventually to Z . Since, the volume expansivity of liquids, in general, are far more than that of solids, so the level Z will be above the level X .

Effect of temperature on density

When a solid or liquid is heated, it expands, with mass remaining constant. Density being the ratio of mass to volume, it decreases. Thus, if \(V_{0}\) and \(V_{t}\) be the respective volumes of a substance at \(0^{\circ}C\) and \(t^{\circ}C\) and if the corresponding values of densities be \(\rho_{0}\) and \(\rho_{t}\), then the mass \(m\) of the substance is given by

\[\begin{matrix} m = V_{0}\rho_{0} = V_{t}\rho_{t} \\ \text{~}\text{But}\text{~}\ V_{t} = V_{0}(1 + \gamma t),\text{~}\text{so}\text{~}\rho_{t} = \rho_{0}(1 + \gamma t)^{- 1} \end{matrix}\]

Practice Exercise

Q. 1 The apparatus shown in fiugre consists of four glass colums connected by horizontal sections. The heights of two central columns \(B\) and \(C\) are 49 cm each. The two outer column \(A\) and \(D\) are open to atmosphere. A and C are maintained at a temperature of \(95^{\circ}C\), while the column B and D are maintained at \(5^{\circ}C\). The heights of liquid in A and D measured from base line are 52.8 cm and 51 cm respectively. Determine the coefficient of thermal expansion of the liquid.

Answers

Q. \(1\ \gamma = 1.96 \times 10^{- 4}/\ ^{\circ}C\)

While most substances expand when heated, a few do not. For instant, if water at \(0^{\circ}C\) is heated, its volume decreases until the temperature reaches \(4^{\circ}C\). Above \(4^{\circ}C\) water behaves normally, and its volume increases as the temperature increases.
Because a given mass of water has a minimum volume at \(4^{\circ}C\), the density (mass per unit volume) of water is greatest at \(4^{\circ}C\), as figure shows.

The density of water in the temperature range from 0 to \(10^{\circ}C\). Water has a maximum density of \(999.973\text{ }kg/m^{3}\) at \(4^{\circ}C\). (This value for the density is equivalent to the often quoted density of 1.000 grams per milliliter)
When the air temperature drops, the surface layer of water is chilled. As the temperature of the surface layer drops toward \(4^{\circ}C\), this layer becomes more dense than the warmer water below. The denser water sinks and pushes up the deeper and warmer water, which in turn is chilled at the surface. This process continues until the temperature of the entire lake reaches \(4^{\circ}C\). Further cooling of the surface water below \(4^{\circ}C\) makes it less dense than the deeper layers; consequently, the surface layer does not \(sink\) but stays on top. Continued cooling of the top layer to \(0^{\circ}C\) leads to the formation of ice that floats on the water, because ice has a smaller density than water at any temperature. Below the ice, however, the water temperature remains above \(0^{\circ}C\). The sheet of ice acts as an insulator that reduces the loss of heat from the lake, especially if the ice is covered with a blanket of snow, which is also an insulator. As a result, lakes usually do not freeze solid, even during prolonged cold spells, so fish and other aquatic life can survive.

Q. 1 Each of the three blocks \(P,Q\) and R shown in figure has a mass of 3 kg . Each of th wires A and B has cross-sectional area \(0.005{\text{ }cm}^{2}\) and Young's modulus \(2 \times 10^{11}\text{ }N/m^{2}\). Neglect friction. Find the longitudinal strain developed in each of the wires. Take \(g = 10\text{ }m/s^{2}\).

Sol. The block R will descend vertically and the blocks P and Q will move on the frictionless horizontal table. Let the common magnitude of the acceleration be a . Let the tensions in the wires A and B be \(T_{A}\) and \(T_{n}\) respectively.

Writing the equation of motion of the blocks \(P,Q\) and R , we get,

\[\begin{array}{r} T_{A} = (3)a\#(i) \\ {\text{ }T}_{B} - T_{A} = (3)a\#(ii) \end{array}\]

and (3) \(g - T_{B} = (3)a\)

By (i) and (ii)

\[T_{B} = 2{\text{ }T}_{A}\]

By (i) and (iii)

\[T_{A} + T_{B} = (3)g = 30\text{ }N\]

or \(\ 3{\text{ }T}_{A} = 30\text{ }N\)
or \(\ T_{A} = 10\text{ }N\) and \(T_{B} = 20\text{ }N\).

Longitudinal strain \(= \frac{\text{~}\text{Longitudinal stress}\text{~}}{\text{~}\text{You}\text{ng's modules}\text{~}}\)

Strain in wire \(A = \frac{10\text{ }N/0.005{\text{ }cm}^{2}}{2 \times 10^{11}\text{ }N/m^{2}} = 10^{- 4}\)

Strain in wire \(B = \frac{20\text{ }N/0.005{\text{ }cm}^{2}}{2 \times 10^{11}\text{ }N/m^{2}} = 2 \times 10^{- 4}\)
Q. 2 A light rod of length 200 cm is suspended from the ceiling horizontally by means of two vertical wires of equal length, tied to its ends. One of the wiresis made of steel and is of cross-section \(0.1{\text{ }cm}^{2}\) and the other of brass of cross-section \(0.2{\text{ }cm}^{2}\). Along the rod at which distance may a weight be hung to produce (a) equal stresses in both the wires (b) equal strains in both the wires? Y for brass and steel are \(10 \times 10^{11}\) and \(20 \times 10^{11}\) dyne \(/{cm}^{2}\) respectively.

Sol. (a) According to the problem stresses are equal, so we have

\[\frac{T_{1}}{{\text{ }A}_{1}} = \frac{T_{2}}{{\text{ }A}_{2}},\]

i.e. \(\ \frac{T_{1}}{{\text{ }T}_{2}} = \frac{A_{1}}{{\text{ }A}_{2}} = \frac{0.1}{0.2}\)

or

\[\begin{array}{r} T_{2} = 2{\text{ }T}_{1}\#(i) \end{array}\]

As the rod is in equilibrium,

\[\Sigma F_{y} = T_{1} + T_{2} - W = 0\]

or

\[\begin{array}{r} T_{1} + T_{2} = W\#(ii) \end{array}\]

From equation (i) and (2), we get

\[\begin{array}{r} T_{1} = (W/3)\text{~}\text{and}\text{~}T_{2} = (2\text{ }W/3)\#(iii) \end{array}\]

Let x be the distance of weight W from steel wire, Torque balance for rotational equilibrium of rod.

\[\Sigma\tau = T_{1}x - T_{2}(2 - x) = 0\]

or \(\ (W/3)x = (2\text{ }W/3)(2 - x)\),
i.e. \(\ x = (4/3)m\)
(b) According to the problem, strains are equal.

\[\frac{T_{1}}{{\text{ }A}_{1}Y_{1}} = \frac{T_{2}}{{\text{ }A}_{2}Y_{2}}\ \left\lbrack \text{~}\text{as strain}\text{~} = \frac{\text{~}\text{stress}\text{~}}{Y} \right\rbrack\]

So, \(\ \frac{T_{1}}{{\text{ }T}_{2}} = \frac{A_{1}Y_{1}}{{\text{ }A}_{2}Y_{2}} = \frac{0.1 \times 20 \times 10^{11}}{0.2 \times 10 \times 10^{11}} = 1\)

\[\begin{array}{r} T_{1} = T_{2}\#(iv) \end{array}\]

i.e. from equation (ii) and (iv), we get

\[T_{1} = T_{2} = (W/2)\]

and for rotational equilibrium of rod

\[\Sigma\tau = T_{1}x - T_{2}(2 - x) = 0\]

or

\[(W/2)x = (W/2)(2 - x),\ \text{~}\text{i.e.}\text{~}x = 1m\]

Q. 3 A steel wire of diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart and in the same plane. A body is hung from the middle point of the wire such that the middle point sages 1 cm lower from the original position. Calculate the mass of the body. Given that Young's modulus of the material of wire is \(2 \times 10^{11}\text{ }N/m^{2}\).

Sol. As shown in figure, mass M is in equilibrium

\[\begin{array}{r} Mg = 2\text{ }Tsin\theta\#(i) \end{array}\]

But from the geometry of figure, for small angle \(\theta\), we have

\[\begin{array}{r} sin\theta \approx tan\theta = (x/L)\#(ii) \end{array}\]

and by definition of Young's modulus, we have

\[\begin{array}{r} T = \frac{YA}{\text{ }L}\Delta\text{ }L = \frac{YA}{\text{ }L}\left\lbrack \left( {\text{ }L}^{2} + x^{2} \right)^{1/2} - L \right\rbrack \approx \frac{{YAx}^{2}}{2{\text{ }L}^{2}}\#(iii) \end{array}\]

So substituting the values of \(sin\theta\) and T from equation and (iii) in (i), we get

\[\begin{matrix} & & Mg = 2 \times \frac{YAx^{2}}{2L^{2}} \times \frac{x}{L} \\ & M = \frac{YAx^{3}}{gL^{3}} & \text{(iv)} \end{matrix}\]

Now as here \(2\text{ }L = 1\text{ }m,x = 1\text{ }cm = 10^{- 2}\text{ }m\)
and

\[\begin{matrix} & A = \pi r^{2} = p\left\lbrack (0.8/2) \times 10^{- 3} \right\rbrack^{2} \\ & \ = \pi \times \left( 4 \times 10^{- 4} \right)^{2}{\text{ }m}^{2} \end{matrix}\]

so \(M = \frac{2 \times 10^{11} \times \pi\left( 4 \times 10^{- 4} \right)^{2} \times \left( 10^{- 2} \right)^{3}}{9.8 \times (1/2)^{3}}\text{ }kg = 82\text{ }g\)
from equation (iv), we have

\[x = L\left( \frac{Mg}{YA} \right)^{1/3}\]

So the angle \(\theta\) can be determined from

\[\theta = \frac{x}{\text{ }L} = \left( \frac{Mg}{YA} \right)^{1/3}\]

Q. 4 A thin unifrom metallic rod of length 0.5 m and radius 0.1 m rotates with an angular velocicty \(400rad/s\) in a horizontal plane about a vertical axis passing through one of its ends. Calculate (a) the tension in the rod and (b) the elongation of the rod. The density of the meterial of the rod is \(10^{4}\text{ }kg/m^{3}\) and the Young's modulus is \(2 \times 10^{11}\text{ }N/m^{2}\).

Sol. (a) We take a differential element of the length dr at a distance r form the axis of rotation as shown in figure. The centripetal force acting on this element is

\[dT = dmr\omega^{2} = (\rho Adr)r\omega^{2}\]

The tension in the rod at a distance \(r\) from the axis of rotation will be due to the centripetal force due to all elements between \(x = r\) to \(x = L\),
i.e., \(\ T = \int_{r}^{L}\mspace{2mu}\rho A\omega^{2}rdr = \frac{1}{2}\rho\text{ }A\omega^{2}\left\lbrack {\text{ }L}^{2} - r^{2} \right\rbrack\)

Thus, tension as function of r ,

\[\begin{matrix} & T = \frac{1}{2} \times 10^{4} \times \pi \times 10^{- 2} \times (400)^{2}\left\lbrack \left( \frac{1}{2} \right)^{2} - r^{2} \right\rbrack \\ & \ = 8\pi \times 10^{6}\left\lbrack \frac{1}{4} - r^{2} \right\rbrack N \end{matrix}\]

Note that tension in the rod is minimum at \(r = L\) and maximum at \(r = 0\)
(b) Let dy be the eleongation in the element of length dr at position \(r\) due to tension, T. From definition of Young's modulus,

\[\begin{matrix} & & \text{~}\text{Strain}\text{~} = \frac{\text{~}\text{stress}\text{~}}{Y} \\ & \frac{dy}{dr} = \frac{T}{AY} & \text{(so)} \end{matrix}\]

From equation (i), we have \(dy = \frac{1}{2}\frac{\rho\omega^{2}}{Y}\left\lbrack L^{2} - r^{2} \right\rbrack dr\)
So the elongation of the entire rod,

\[\begin{matrix} & \Delta L = \frac{\rho\omega^{2}}{2Y}\int_{0}^{L}\mspace{2mu}\mspace{2mu}\left\lbrack L^{2} - r^{2} \right\rbrack dr \\ & \ = \frac{1}{3}\frac{\rho\omega^{2}L^{3}}{Y} \end{matrix}\]

Here \(\ \Delta L = \frac{1}{3} \times \frac{10^{4} \times (400)^{2}(0.5)^{3}}{2 \times 10^{11}}\)

\[= \frac{1}{3} \times 10^{- 3}\text{ }m\]

Q. 5 Estimate the pressure deep inside the sea at a depth \(h\) below the surface. Assume that the density of water is \(\rho_{0}\) at sea level and its bulk modulus is B.
Sol. In a static fluid the pressure variation is given by

\[\begin{array}{r} \frac{dP}{dh} = - \rho g\#(i) \end{array}\]

The bulk modulus is defined as

\[\begin{array}{r} B = - \frac{dP}{dV/V}\#(ii) \end{array}\]

Where \(dV/V\) is fractional change in volume of a element subjected to isotropic pressure increase dP .We consider a sample of the fluid having mass M , its volume \(V = M/\rho\), so that

\[\begin{array}{r} dv = \frac{- M}{\rho^{2}}\text{ }d\rho\#(iii) \end{array}\]

Hence \(\ \frac{dV}{V} = - \frac{d\rho}{\rho}\)
combining equations (ii) and (iii), we get

\[\frac{Bd\rho}{\rho} = \rho gdh\]

or

\[\int_{\rho_{0}}^{\rho}\mspace{2mu}\frac{d\rho}{\rho^{2}} = \int_{0}^{h}\mspace{2mu}\frac{gdh}{B}\]

or

\[\begin{array}{r} \frac{1}{\rho^{0}} - \frac{1}{\rho} = \frac{gh}{\text{ }B}\#(iv) \end{array}\]

as

\[dP = - \frac{BdV}{\text{ }V} = B\frac{\text{ }d\rho}{\rho}\]

Hence

\[\int_{P_{0}}^{P}\mspace{2mu} dP = \int_{\rho_{0}}^{\rho}\mspace{2mu} B\frac{\text{ }d\rho}{\rho}\]

or

\[\begin{array}{r} P - P_{0} = B\mathcal{l}n\frac{\rho}{\rho_{0}}\#(v) \end{array}\]

On multiplying equation (iv) by \(\rho_{0}\), we get

\[1 - \frac{\rho_{0}}{\rho} = \frac{\rho_{0}gh}{\text{ }B}\]

so that

\[In\frac{\rho}{\rho_{0}} = - \mathcal{l}n\left( 1 - \frac{\rho_{0}gh}{B} \right)\]

Substituting this in equation(v), we get

\[P = P_{0} - B\mathcal{l}n\left( 1 - \frac{\rho_{0}gh}{B} \right)\]

This is the required expression for P .
Q. 6 A 5 g piece of ice at \(- 20^{\circ}C\) is put into 10 g of water at \(30^{\circ}C\). Assuming that heat is exchanged only between the ice and the water, find the final the final temperature of the mixture. Specific heat capacity of ice \(= 2100\text{ }J/kg - \ ^{\circ}C\) specific heat capacity of water \(= 4200\text{ }J/kg - \ ^{\circ}C\) and latent heat of fusion of ice \(= 3.36 \times 10^{5}\text{ }J/kg\).
Sol. The heat given by the water when it cools down from \(30^{\circ}C\) to \(0^{\circ}C\) is

\[(0.01\text{ }kg)\left( 4200\text{ }J/k - \ ^{\circ}C \right)\left( 30^{\circ}C \right) = 1260\text{ }J\]

The heat required to bring the ice to \(0^{\circ}C\) is \((0.005\text{ }kg)\left( 2100\text{ }J/kg - \ ^{\circ}C \right)\left( 20^{\circ}C \right) = 210\text{ }J\).
The heat required to melt 5 g of ice is \(\ (0.005\text{ }kg)\left( 3.36 \times 10^{5}\text{ }J/kg \right) = 1680\).
We see that whole of the ice cannot be melted as the required amount of heat is not proved by the water. Also, the heat is enough to bring the ice to \(0^{\circ}C\). Thus the final temperature of the mixture is \(0^{\circ}C\) with some of the ice melted.
Q. 7 Steam at \(100^{\circ}C\) is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at \(15^{\circ}C\) till the temperature of the calorimeter and its contents riseds to \(80^{\circ}C\). The mass of the steam condensed in kilogram is
(A) 0.130
(B) 0.065
(C) 0.260
(D) 0.135

Sol. Heat required to bring water and calorimeter from \(15^{\circ}C\) to \(80^{\circ}C\)

\[\begin{matrix} & Q = mC\Delta\text{ }T \\ & Q = (1.1 + 0.02) \times 1 \times (80 - 15) \\ & Q = 72.8kcal \end{matrix}\]

Amount of steam condensed to provide 72.8 kcal heat \(= m_{1}\text{ }L\)

\[\begin{matrix} \Rightarrow \ & 72.8 \times 1000 = m_{1}\lbrack 540\rbrack \\ & m_{1} = 134.8gm \\ & {\text{ }m}_{1} = 0.1348\text{ }kg \end{matrix}\]

Q. 8 A bullet of mass of 10 gm moving with a speed of \(20\text{ }m/s\) hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. How much ice will melt if \(50\%\) of the lost kinetic energy goes to ice? (Temperature of ice block \(= 0^{\circ}C\) )

Sol. Velocity of bullet + ice block, \(V = \frac{(10gm) \times (20\text{ }m/s)}{1000gm} = 0.2\text{ }m/s\)

\[\begin{matrix} \text{~}\text{Loss of K.E.}\text{~} & = \frac{1}{2}{mv}^{2} - \frac{1}{2}(\text{ }m + M)V^{2} \\ & = \frac{1}{2}\left\lbrack 0.01 \times (20)^{2} - 1 \times (0.2)^{2} \right\rbrack = \frac{1}{2}\lbrack 4 - 0.04\rbrack = 1.98\text{ }J \\ \therefore & \text{~}\text{Heat generated}\text{~} = \frac{1.98}{4.2}Cal \\ \therefore & \text{~}\text{Heat }\text{recived}\text{ by ice block}\text{~} = \frac{1.98}{4.2 \times 2}Cal = 0.24cal \\ \therefore & \text{~}\text{Mass of ice malted}\text{~} = \frac{(0.24Cal.)}{(80Cal/gm)} = 0.003gm \end{matrix}\]

Q. 9 A calorimeter of water equivalent 1 kg contains 10 kg of ice & 10 kg of water in thermal equilibrium. The atmospheric temperature is \(15^{\circ}\) below freezing point due to which the calorimeter loses heat. -As a result ice is formed inside the calorimeter at a rate of 10.8 gm per second. To try to compensate for this heat loss, steam at \(100^{\circ}C\) is supplied to the calorimeter at a rate of r . ( \(L_{V} = 540cal/gm,L_{f} = 80cal/gm\), sp heat of water \(1cal/gm\ ^{\circ}C\).) Column-I gives the value of r and column-II gives the situation just after the introduction of steam.

Column-I

(A) \(r = 1.6gm/sec\)
(B) \(r = 1.35gm/sec\)
(C) \(r = 1.2gm/sec\)
(D) \(r = 1gm/sec\)

Column-II

(P) Amount of ice in calorimeter increases.
(Q) Amount of water in calorimeter increases.
(R) Amount of ice remains constant at 10 kg
(S) Amount of water remains constant at 10 kg
(T) Amount of ice in calorimeter decreases.
[Ans. (A) Q, T; (B) Q, R; (C) P, S; (D) P ]

Sol. Rate of heat loss \(= 80 \times 10.8 = 54 \times 16cal/sec\).
(A) \(r = 1.6\)
\(\Rightarrow \ \) rate of heat supplies by forming steam to water at \(0^{\circ}\ = 1.6 \times 640 > 54 \times 16\)
\(\therefore\ \) additional ice will melt. Correct options are \(Q\) and \(T\)
(B) Rate of heat loss \(= 54 \times 16 = 64 \times 13.5cal/sec\).
\[r = 1.35 =\] rate of heat supplied for converting steam to water at \(0^{\circ}C = 1.35 \times 640 = 13.5 \times 64\).
no additional ice will melt or water will fuse. Correct options are \(Q\) and \(R\)
(C) Rate of heat loss \(= 54 \times 16 = 72 \times 12cal/sec\).

Rate of heat supplied by converting steam to ice at \(0^{\circ}C = 1.20 \times 720 = 12 \times 72cal/sec\) no additional ice will melt or water will fuse. Correct options are P and S
(D) Additional water will fuse to ice. Correct option is P.
Q. 10 An ice cube of mass 0.1 kg at \(0^{\circ}C\) is placed in an isolated container which is at \(227^{\circ}C\). The specific Heat S of the container varies with temperature T according to relation \(S = A + BT\), where \(A = 100cal/kg - k\) and \(B = 2 \times 10^{- 2}cal/kg - k^{2}\). If the final temperature of the container is \(27^{\circ}C\), Determine the mass of the container.
Sol. Specific heat of container is temperature dependent so we have to calculate heat lost for a small temperature change dT and then integrate it from initial temperature to final temperature.

Heat lost by container \(= - \int_{500}^{300}\mspace{2mu}{\text{ }m}_{c}(A + B + T)dT\)

\[= - m_{c}\left\lbrack AT + \frac{{BT}^{2}}{2} \right\rbrack_{500}^{300}\ = 21600{\text{ }m}_{c}\]

Heat gained by Ice \(= mL + mC\Delta T\)

\[\begin{matrix} & \ = 0.1 \times 1000 \times 80 + 0.1 \times 1000 \times 27 \\ & \ = 10700Cal \end{matrix}\]

from principle of colorimetry
Heat lost by container \(=\) Heat gained by Ice

\[\begin{matrix} & 21600{\text{ }m}_{c} = 10700 \\ & {\text{ }m}_{c} = 0.495\text{ }kg \end{matrix}\]

Q. 11 A certain clock with an iron pendulum is made so as to keep correct time at \(10^{\circ}C\).

Given \(\alpha_{\text{iron}\text{~}} = 12 \times 10^{- 6}\) per \(\ ^{\circ}C\). How will the rate alter if the temperature rises to \(25^{\circ}C\) ?
Sol. When the pendulum keeps correct time, its period of vibration is 2 sec and so it makes

\[\frac{20 \times 60 \times 60}{2} = 43200\text{~}\text{Vibrations}\text{~}/\text{~}\text{day}\text{~}\]

If length of pendulum at \(10^{\circ}C\) is \(\mathcal{l}_{10}\) and at \(25^{\circ}C\) is \(\mathcal{l}_{25}\)
\[\therefore\ \mathcal{l}_{25} = \mathcal{l}_{10}\lbrack 1 + \alpha(25 - 10)\rbrack = \mathcal{l}_{10}\lbrack 1 + 15\alpha\rbrack \]as \(\ T = 2\pi\sqrt{\frac{\mathcal{l}}{g}}\)
i.e. \(\ T \propto \sqrt{\mathcal{l}}\)
i.e. \(\ n \propto \frac{1}{\sqrt{\mathcal{l}}},n\) is no. of vibrations per sec.

\[\begin{matrix} \therefore & \frac{n_{25}}{n_{lo}} = \sqrt{\frac{\mathcal{l}_{lo}}{\mathcal{l}_{25}}} = \lbrack 1 + 15\alpha\rbrack^{- 1/2} \approx 1 - \frac{15}{2}\alpha \\ \therefore & n_{25} = n_{lo}\left( 1 - \frac{15}{2} \times 12 \times 10^{- 6} \right) \\ & = 43200\lbrack 1 - 0.00009\rbrack = 43196.12 \end{matrix}\]

That is the clock makes \((43200 - 43196.12) = 3.88\) vibration loss per day. That is clock losses \(3.88 \times 2 = 7.76sec\) per day
Q. 12 A sphere of silver is floating in a mercury bath. If temperature is increased will the sphere sink deeper or rise ? It is given \(\gamma_{\text{silver}\text{~}} > \gamma_{\text{mercury}\text{~}}\)
Sol. Rise
Q. 13 A glass vessel of volume \(V_{0}\) is completely filled with a liquid and its temperature is raised by \(\Delta T\). What volume of the liquid will overflow? Coefficient of linear expansion of glass \(= \alpha_{g}\) and coefficient of volume expansion of the liquid \(= \gamma\).
Sol. Volume of the liquid over flown

\[\begin{matrix} & \ = \text{~}\text{increase in the volume of the liquid}\text{~} - \text{~}\text{increase in the volume of the container}\text{~} \\ & \ = \left\lbrack V_{0}\left( 1 + \gamma_{\mathcal{l}}\Delta T \right) - V_{0} \right\rbrack - \left\lbrack V_{0}\left( 1 + \gamma_{g}\Delta\text{ }T \right) - V_{0} \right\rbrack \\ & \ = V_{0}\Delta\text{ }T\left( \gamma_{\mathcal{l}} - \gamma_{g} \right) = V_{0}\Delta\text{ }T\left( \gamma_{\mathcal{l}} - 3\alpha_{g} \right)\ (\therefore\gamma \approx 3\alpha) \end{matrix}\]

Q. 14 A liquid having coefficient of volume expansion \(\gamma_{0}\) is filled in a glass vessel. The coefficient of linear expansion of glass is \(\alpha\). When the arrangement is heated to raise the temperature of the liquid and the glass container by \(\Delta T\), exapnsion takes place in both. The expansion may be different or equal. Depending on the values of \(\gamma_{0}\) and \(\alpha\) you may find that level of the liquid rises with respect to ground or it may fall with repect to ground.

(i) What is the relation between \(\gamma_{0}\) and \(\alpha\) so that the fraction of volume of container occupied by the liquid does not change with rise in temperature?
(ii) What is the relation between \(\gamma_{0}\) and \(\alpha\) so the liquid does not change with respect to ground?
(iii) What is the relation between \(\gamma_{0}\) and \(\alpha\) so that the level of the liquid does not change with respect to the container itself?

Sol. (i) Let \(V_{C}\) be the volume of the container and \(V_{\mathcal{l}}\), be the volume of the liquid. According to the condition

\[\frac{V_{\mathcal{l}}}{V_{C}} = \text{~}\text{constant (i.e. independent of }\text{temperture}\text{)}\text{~}\]

\(\Rightarrow \ \frac{V_{\mathcal{l}}^{'}}{V_{C}^{'}} = \frac{V_{\mathcal{l}}}{V_{C}}\ \) or \(\ \frac{V_{\mathcal{l}}^{'}}{V_{\mathcal{l}}} = \frac{V_{C}^{'}}{V_{C}}\ \) (Here \(V_{\mathcal{l}}^{'}\) and \(V_{C}^{'}\) are volume on heating)
\[\Rightarrow \ \frac{V_{\mathcal{l}}^{'} - V_{\mathcal{l}}}{V_{\mathcal{l}}} = \frac{V_{c}^{'} - V_{c}}{V_{c}}\]

\[\begin{matrix} \Rightarrow \ \frac{\Delta V_{\mathcal{l}}}{V_{\mathcal{l}}} & \ = \frac{\Delta V_{C}}{{\text{ }V}_{C}} \\ \gamma_{0}\Delta\text{ }T & \ = (3\alpha)\Delta T \end{matrix}\]

[For container, coefficient of volume expansion will be \(3\alpha\) ]
(ii) If on heating, H does not change, than the increase in volume of the liquid is accomodated by the increase in base area of the vessel. Let the area be A.

\[\begin{matrix} & \text{~}\text{Initial volume of liquid}\text{~} = V_{\mathcal{l}} = A \times H \\ & \text{~}\text{Final volume of liquid}\text{~} = V_{\mathcal{l}}^{'} = A^{'} \times H \end{matrix}\]

\[\Rightarrow \ \frac{V_{\mathcal{l}}^{'}}{V_{\mathcal{l}}} = \frac{A^{'}}{A} \Rightarrow \frac{\Delta V_{\mathcal{l}}^{'}}{V_{\mathcal{l}}} = \frac{\Delta A^{'}}{A} \Rightarrow \ \gamma_{0}\Delta\text{ }T = \beta\Delta T\] or \(\gamma_{0} = 2\alpha\ (\therefore\beta = 2\alpha)\)
(iii) This is exactly same as part (i)

\[\gamma = 3\alpha.\]