Q.A 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side $30\ \text{cm}$ has a thickness of $5.0\ \text{cm}$. If $4.0\ \text{kg}$ of ice is put in the box, estimate the amount of ice remaining after $6\ \text{h}$. The outside temperature is $45\ ^\circ\text{C}$, and coefficient of thermal conductivity of thermacole is $0.01\ \text{J s}^{-1}\ \text{m}^{-1}\ \text{K}^{-1}$. [Heat of fusion of water $= 335 \times 10^{3}\ \text{J kg}^{-1}$]

The problem is about steady-state heat conduction through rods. The rate of heat flow (heat current) through a rod in steady state is given by Fourier's law:

where $k$ is thermal conductivity, $A$ is cross-sectional area, $L$ is length, and $\Delta T$ is the temperature difference across the ends.

We are given ratios for $k$, $L$, and volume $V$, and we need to find $\Delta\theta$ such that the ratio of heat currents $H_A : H_B = 1 : 16$.

Let’s work through it step by step.

1. Write the given ratios clearly.

   Let $k_A : k_B = 2 : 3$, so $k_A = 2x$, $k_B = 3x$ for some $x$.

   Let $L_A : L_B = 2 : 1$, so $L_A = 2y$, $L_B = y$.

   Let $V_A : V_B = 1 : 2$, so $V_A = V$, $V_B = 2V$.

2. Find the area ratio from volume and length.

   Volume = area $\times$ length, so $A = V/L$.

   Therefore:

The ratio of areas is:

3. Write the heat current ratio using Fourier’s law.

Given $\Delta T_A = 60\ ^\circ\mathrm{C}$ and $\Delta T_B = \Delta\theta$, and $H_A : H_B = 1 : 16$, so:

4. Substitute the ratios.

Plug in:

$\frac{1}{16} = \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{60}{\Delta\theta}$ …

The key idea is that when two rods are joined in series and the system is in steady state, the rate of heat flow (heat current) through each rod must be identical — there is no accumulation of heat anywhere, so what flows into rod A per second must flow out of rod B per second.

For a rod of length $L$, cross-sectional area $A$, and thermal conductivity $K$, the heat current $H$ is given by Fourier's law:

where $\Delta T$ is the temperature difference between its ends.

Here both rods have equal dimensions — same $L$ and same $A$. Rod A has conductivity $K$, rod B has conductivity $2K$. The temperature difference across rod A alone is given as $\Delta T_A = 12\,^\circ\mathrm{C}$.

1. Heat current through rod A:

2. Heat current through rod B: let $\Delta T_B$ be the temperature difference across rod B:

3. Equate the currents (steady state, series connection):

Cancelling $K$, $A$, and $L$ (all non-zero):

4. Total temperature difference between the open ends: since the rods are in series, the total drop from the free end of A to the free end of B is the sum of the drops across each rod: $\Delta T_{\text{total}} = \Delta T_A + \Delta T_B = 12 + 6 = 18\,^\circ\mathrm{C}$ …

Concept & Intuition

When two rods are welded end-to-end, the total expansion of the composite rod for a given temperature rise is the sum of the expansions of each part. Since expansion depends on original length and coefficient of linear expansion, the effective coefficient is not simply the average of the two coefficients — it must be weighted by the lengths. This is because a longer rod contributes more to the total length change.

1. Define the variables

   Let the cross-sectional area be the same for both rods (given).

   Rod 1: length $L_1 = L$, coefficient $\alpha_1 = 2\alpha$

   Rod 2: length $L_2 = 2L$, coefficient $\alpha_2 = \alpha$

   Composite rod: total length $L_{\text{total}} = L_1 + L_2 = 3L$, unknown effective coefficient $\alpha_{\text{eff}}$.

2. Write the expansion for each rod

   For a temperature increase $\Delta T$, the change in length of each rod is:

3. Total expansion of the composite The total expansion is the sum:

4. Express the same total expansion using the effective coefficient By definition, for the composite rod:

5. Equate and solve $$$$

Concept & Intuition

Strain measures how much a material deforms relative to its original size. For a wire under tension, it’s simply the ratio of the change in length to the original length. No units — it’s a pure number. The trick is to keep the units consistent (both in cm here) and then express the result in the form given in the options.

Step-by-step solution

1. Identify the given quantities

   Original length of wire, $L = 40 \, \text{cm}$

   Extension (change in length), $\Delta L = 0.1 \, \text{cm}$

2. Recall the definition of strain

3. Plug in the numbers

4. Convert to the form given in the options Options are all in multiples of $10^{-4}$. $0.0025 = 25 \times 10^{-4}$ …

The key idea is that the wire’s tensile strength gives its maximum tension per unit area — that tells us the cross‑sectional area. Then, using Young’s modulus, we relate a small stretch to the tension produced, assuming the wire stays within its elastic limit (which the problem states it does).

A common pitfall is to forget that the area isn’t given directly; you must extract it from the breaking condition. Once you have the area, the rest is a straightforward application of $F = \frac{Y A}{L} \Delta L$.

1. Find the cross‑sectional area $A$ from the breaking data. Tensile strength is the maximum stress the material can bear:

So

This is a tiny area — about the cross‑section of a thin wire, which makes sense.

2. Apply Hooke’s law for a stretched wire. Young’s modulus $Y$ relates stress and strain:

Rearranging for the tension $F$:

This is the same as $F = k \Delta L$ with $k = YA/L$, the effective spring constant of the wire.

3. Plug in the numbers.
   • $Y = 2 \times 10^{11} \ \mathrm{N/m^2}$
   • $A = 3 \times 10^{-6} \ \mathrm{m^2}$
   • $L = 1.5 \ \mathrm{m}$
   • $\Delta L = 0.20 \ \mathrm{cm} = 0.0020 \ \mathrm{m}$

First compute the factor $\frac{YA}{L}$:

The key idea: when a solid is immersed in a liquid, its apparent weight is its true weight minus the buoyant force. The buoyant force equals the weight of the displaced liquid, which depends on the volume of the solid and the density of the liquid. When temperature changes, both the solid’s volume (due to its own expansion) and the liquid’s density (given) change. By writing the apparent weight at two temperatures, we can solve for the solid’s volume expansion coefficient, and then get its linear expansion coefficient.

Let’s work through it step by step.

1. Set up the first condition (at $32^\circ\text{C}$).

   True weight in air: $W = 49\ \text{gf}$ (gram-force; we can work in grams since $g$ cancels).

   Apparent weight in liquid: $W_{\text{app},1} = 39\ \text{gf}$.

   Buoyant force $B_1 = W - W_{\text{app},1} = 49 - 39 = 10\ \text{gf}$.

   Buoyant force also equals the weight of displaced liquid: $B_1 = V_1 \rho_1 g$, where $V_1$ is the metal’s volume at $32^\circ\text{C}$ and $\rho_1 = 1.2 \times 10^3\ \text{kg/m}^3$.

   In gram-force units, $g$ cancels if we treat densities in $\text{g/cm}^3$:

   $\rho_1 = 1.2\ \text{g/cm}^3$ (since $1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$).

   So $10 = V_1 \times 1.2$, giving $V_1 = \frac{10}{1.2} = \frac{25}{3}\ \text{cm}^3$.

2. Set up the second condition (at $42^\circ\text{C}$).

   Apparent weight: $W_{\text{app},2} = 40\ \text{gf}$.

   Buoyant force $B_2 = 49 - 40 = 9\ \text{gf}$.

   Liquid density at $42^\circ\text{C}$: $\rho_2 = 1.0 \times 10^3\ \text{kg/m}^3 = 1.0\ \text{g/cm}^3$.

   Let $V_2$ be the metal’s volume at $42^\circ\text{C}$. Then $9 = V_2 \times 1.0$, so $V_2 = 9\ \text{cm}^3$.

3. Relate the volume change to thermal expansion.

   The volume expansion coefficient $\gamma$ of the metal satisfies:

   $V_2 = V_1 (1 + \gamma \Delta T)$, where $\Delta T = 42 - 32 = 10^\circ\text{C}$. …

The copper hole ($9.98\,\mathrm{cm}^2$) starts smaller than the steel rod ($10.00\,\mathrm{cm}^2$). Because copper has the larger expansion coefficient, heating both together closes the gap. The ring just slips on when the two areas become equal.

The coefficient of area expansion is $\beta = 2\alpha$:

1. Equal-area condition using $A = A_0(1 + \beta\,\Delta T)$:

2. Expand: …

The problem asks for the total change in length of a combined rod made of two different metals when its temperature increases. This involves the concept of linear thermal expansion.

Concept and Intuition

Most materials expand when heated and contract when cooled. This phenomenon is called thermal expansion. For a one-dimensional object like a rod, the change in length is primarily governed by its initial length, the change in temperature, and a material property called the coefficient of linear expansion.

Imagine a rod made of a single material. When its temperature increases, the atoms within the material vibrate more vigorously and move slightly further apart on average. This results in an overall increase in the rod's length. The amount of expansion is directly proportional to the original length of the rod (a longer rod will expand more for the same temperature change), and also directly proportional to the change in temperature. The proportionality constant is the coefficient of linear expansion, which is unique to each material.

When two different rods are joined end-to-end, they behave as independent segments in terms of their thermal expansion. Each rod will expand or contract based on its own material properties and initial length, for the same temperature change. The total change in length of the combined rod is simply the sum of the individual changes in length of each segment.

Let's apply this concept to the given problem.

Step-by-step Derivation

1. Identify Given Parameters and Calculate Temperature Change:

   We are given the following information for rods A and B:

   • Initial length of rod A, $L_{A0} = 50 \text{ cm}$
   • Initial length of rod B, $L_{B0} = 50 \text{ cm}$
   • Initial temperature, $T_0 = 30^\circ \mathrm{C}$
   • Final temperature, $T_f = 230^\circ \mathrm{C}$
   • Coefficient of linear expansion for rod A, $\alpha_A = 2.0 \times 10^{-5}/^\circ \mathrm{C}$
   • Coefficient of linear expansion for rod B, $\alpha_B = 1.0 \times 10^{-5}/^\circ \mathrm{C}$

   The diameter of the rods (4.0 mm) is provided but is not relevant for calculating the linear expansion. Linear expansion depends only on the length, coefficient of expansion, and temperature change.

   First, calculate the change in temperature ($\Delta T$):

2. Calculate Change in Length for Rod A: Using the formula $\Delta L = L_0 \alpha \Delta T$ for rod A: …

When a material is subjected to an external force, it deforms. For solids, this deformation can be an elongation (stretching) or compression. The relationship between the applied force and the resulting deformation, within the elastic limit, is described by Hooke's Law. For a solid rod, this relationship is quantified using Young's modulus.

Young's modulus ($Y$) is a measure of a material's stiffness or its resistance to elastic deformation under tensile or compressive stress. It is defined as the ratio of stress to strain:

Combining these definitions, we get:

From this, we can find the elongation $\Delta L$:

This formula allows us to calculate how much a rod will stretch under a given load, provided we know its dimensions, the applied force, and the material's Young's modulus.

Let's apply this concept to the given problem.

1. Identify Given Values and Convert Units:

   First, we list the given parameters and ensure they are in consistent SI units.

   • Length of the rod, $L = 1.0\,\text{m}$
   • Cross-sectional area, $A = 0.50\,\text{cm}^2$. We convert this to square meters: $A = 0.50 \times (10^{-2}\,\text{m})^2 = 0.50 \times 10^{-4}\,\text{m}^2 = 5.0 \times 10^{-5}\,\text{m}^2$
   • Mass of the platform, $m = 500\,\text{kg}$
   • Young's modulus, $Y = 10^{11}\,\text{Pa}$ (Pascals, which is $\text{N/m}^2$)
   • Acceleration due to gravity, $g = 10\,\text{m/s}^2$

2. Calculate the Force ($F$):

   The force acting on the rod is the weight of the platform hanging from its lower end.

   $F = mg = 500\,\text{kg} \times 10\,\text{m/s}^2 = 5000\,\text{N}$

3. Calculate the Stress ($\sigma$):

   Stress is the force per unit area. …

When a solid material is heated, its dimensions increase. This phenomenon is called thermal expansion. For a one-dimensional object like a rod, we talk about linear expansion. For a two-dimensional object like a sheet, we consider superficial or area expansion. The problem asks for the ratio of the coefficient of linear expansion to the coefficient of superficial expansion.

The core idea is to express the change in area of the sheet in terms of the change in its linear dimensions, and then relate the respective coefficients of expansion.

1. Define Linear Expansivity: The coefficient of linear expansion, denoted by $\alpha$, describes how the length of a material changes with temperature. If an object has an initial length $L_0$ at a certain temperature and its temperature increases by $\Delta T$, its new length $L$ is given by:

2. Consider a Rectangular Sheet:

   Let's take a rectangular sheet with initial length $L_0$ and initial width $W_0$ at some initial temperature.

   The initial area of the sheet is $A_0 = L_0 W_0$.

3. Calculate New Dimensions after Temperature Change:

   When the temperature of the sheet increases by $\Delta T$, both its length and width will expand according to the linear expansivity $\alpha$.

   The new length will be $L = L_0 (1 + \alpha \Delta T)$.

   The new width will be $W = W_0 (1 + \alpha \Delta T)$.

4. Calculate the New Area:

   The new area $A$ of the sheet will be the product of its new length and new width:

Since $A_0 = L_0 W_0$, we can write:

5. Expand the Expression for New Area: Expand the term $(1 + \alpha \Delta T)^2$:

The term $\alpha \Delta T$ is typically very small (on the order of $10^{-3}$ or less for common materials and temperature changes). Therefore, $(\alpha \Delta T)^2$ will be an even smaller term (on the order of $10^{-6}$ or less) and can be neglected without significant error for practical purposes.
So, we approximate the new area as:

6. Define Superficial Expansivity: …

The key idea here is thermal resistance. In heat conduction, a rod resists the flow of heat just like a resistor resists electric current. For a rod of length $l$, cross-sectional area $A$, and thermal conductivity $k$, the thermal resistance is $R = \frac{l}{kA}$. When two rods are placed end to end (in series), the total resistance is simply the sum of their individual resistances. The question asks for the single conductivity $k_{\text{eq}}$ that would give the same resistance for a rod of the combined length $l_1 + l_2$ and the same area $A$.

Let’s work through it.

1. Write the thermal resistance of each rod. For rod 1: $R_1 = \dfrac{l_1}{k_1 A}$ For rod 2: $R_2 = \dfrac{l_2}{k_2 A}$ Since they are in series, the total resistance of the composite rod is

2. Now imagine a single uniform rod of length $l_1 + l_2$, same area $A$, and unknown conductivity $k_{\text{eq}}$. Its resistance would be

3. Set the two resistances equal (because the system’s overall conductivity is defined by this equivalence):

The area $A$ cancels on both sides, leaving

4. Solve for $k_{\text{eq}}$. Combine the right-hand side over a common denominator:

So …