Q.One mole of an ideal gas at 300 K is contained in a cubical vessel of volume $V$ whose eight corners are labelled A, B, C, D, E, F, G, H, so that ABCD and EFGH are two opposite (parallel) faces of the cube. The face EFGH is made of a material that totally absorbs every gas molecule that strikes it (the molecules are not reflected back), while the opposite face ABCD reflects molecules in the usual way. At any given instant, which statement about the pressure on the faces is correct?

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Two brass spheres approaching each other with the same speed collide head-on elastically. After collision, if one of the spheres of radius $R$ comes to rest, then the radius of the other sphere is (A) $\dfrac{R}{\sqrt[3]{2}}$ (B) $\sqrt[3]{3}R$ (C) $\sqrt[3]{2}R$ (D) $\dfrac{R}{\sqrt[3]{3}}$

   ›Reveal solutionSolution

   In a head-on elastic collision, both momentum and kinetic energy are conserved. Using the fact that one sphere comes to rest after the collision, the mass ratio is determined, and since mass is proportional to the cube of the radius, the other sphere's radius is found to be $\dfrac{R}{\sqrt[3]{3}}$.

   Concept and Intuition

   When two objects collide elastically, both momentum and kinetic energy are perfectly conserved. Here, the two brass spheres have the same material density, so their masses are proportional to the cube of their radii. They approach each other with equal speed, and after the collision one sphere (radius $R$) stops completely. This special outcome forces a specific mass ratio. By solving the conservation equations, we can find the unknown radius.

   ---

   1. Set up variables and conservation of momentum Let the sphere of radius $R$ have mass $m_1 = \rho \cdot \frac{4}{3}\pi R^3$, and the other sphere of unknown radius $r$ have mass $m_2 = \rho \cdot \frac{4}{3}\pi r^3$. Both approach with speed $v$ in opposite directions. Take the direction of $m_1$ as positive. Then initial velocities: $u_1 = +v$, $u_2 = -v$. After collision, $m_1$ comes to rest: $v_1 = 0$. Let $v_2$ be the velocity of $m_2$. Conservation of momentum:

   $m_1 v + m_2(-v) = m_1(0) + m_2 v_2$

   $(m_1 - m_2)v = m_2 v_2 \quad \Rightarrow \quad v_2 = \frac{m_1 - m_2}{m_2} v.$

   2. Apply conservation of kinetic energy Initial kinetic energy:

   $\frac{1}{2}m_1 v^2 + \frac{1}{2}m_2 v^2 = \frac{1}{2}(m_1 + m_2)v^2.$

   Final kinetic energy:

   $\frac{1}{2}m_1(0)^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_2 v_2^2.$

   Equate:

   $\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2}m_2 v_2^2.$

   Substitute $v_2$ from step 1:

   $(m_1 + m_2)v^2 = m_2 \left( \frac{m_1 - m_2}{m_2} v \right)^2 = \frac{(m_1 - m_2)^2}{m_2} v^2.$

   Cancel $v^2$ (nonzero):

   $m_1 + m_2 = \frac{(m_1 - m_2)^2}{m_2}.$

   3. Solve for the mass ratio Multiply both sides by $m_2$:

   $m_2(m_1 + m_2) = (m_1 - m_2)^2.$

   Expand:

   $m_1 m_2 + m_2^2 = m_1^2 - 2m_1 m_2 + m_2^2.$

   Cancel $m_2^2$:

   $m_1 m_2 = m_1^2 - 2m_1 m_2 \quad \Rightarrow \quad 3m_1 m_2 = m_1^2.$

   If $m_1 \neq 0$, divide by $m_1$:

   $3m_2 = m_1 \quad \Rightarrow \quad \frac{m_1}{m_2} = 3.$

   Tip

   Notice that the mass ratio is exactly 3:1. This is a classic result: in a head-on elastic collision where a heavier moving object is brought to rest by a lighter one, the mass ratio is 3.

   4. Relate mass to radius Since both spheres are brass (same density $\rho$),

   $\frac{m_1}{m_2} = \frac{\rho \cdot \frac{4}{3}\pi R^3}{\rho \cdot \frac{4}{3}\pi r^3} = \frac{R^3}{r^3} = 3.$

   Thus,

   $r^3 = \frac{R^3}{3} \quad \Rightarrow \quad r = \frac{R}{\sqrt[3]{3}}.$

   Watch out

   A common point of confusion is which sphere stops. Here the sphere of radius $R$ is the one that comes to rest, and the algebra gives $m_1 = 3m_2$ — so the stopping sphere is the heavier one. This is physically consistent: in a head-on elastic collision a heavier sphere can be brought to rest by a lighter one moving toward it. The unknown sphere is therefore the lighter one, with the smaller radius.

   Thus the radius of the other sphere is $R / \sqrt[3]{3}$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.During adiabatic compression of an ideal gas at an initial pressure P, the final density of the gas becomes n times its initial value. The final pressure of the gas is ($\gamma$ is the ratio of the specific heats of the gas at constant pressure and constant volume) (A) $n^{(1-\gamma)}P$ (B) $n^{(\gamma-1)}P$ (C) $n^{-\gamma}P$ (D) $n^{\gamma}P$

   ›Reveal solutionSolution

   During an adiabatic compression, the relationship between pressure and density is $P \propto \rho^\gamma$. Given that the final density is $n$ times the initial density, the final pressure becomes $n^\gamma$ times the initial pressure. The final pressure is $\boxed{n^\gamma P}$.

   When an ideal gas undergoes an adiabatic process, it means there is no heat exchange with the surroundings. This specific condition leads to a particular relationship between the pressure ($P$) and volume ($V$) of the gas, which is $PV^\gamma = \text{constant}$. To solve this problem, we need to connect this relationship with the given information about the change in density.

   The key insight is to express the volume of the gas in terms of its density. For a fixed mass of gas ($m$), volume ($V$) and density ($\rho$) are inversely related: $V = m/\rho$. Since the mass of the gas does not change during compression, we can substitute this expression for volume into the adiabatic equation to find a relationship between pressure and density.

   1. State the adiabatic process equation: For an adiabatic process, the product of pressure and volume raised to the power of $\gamma$ (the ratio of specific heats) remains constant. If $P_1$ and $V_1$ are the initial pressure and volume, and $P_2$ and $V_2$ are the final pressure and volume, then:

   $P_1 V_1^\gamma = P_2 V_2^\gamma$

   2. Relate volume to density: The density $\rho$ of a gas is defined as its mass $m$ per unit volume $V$, i.e., $\rho = m/V$. Therefore, the volume can be expressed as $V = m/\rho$. Since the mass of the gas remains constant throughout the compression, we can write:

   $V_1 = \frac{m}{\rho_1} \quad \text{and} \quad V_2 = \frac{m}{\rho_2}$

   where $\rho_1$ and $\rho_2$ are the initial and final densities, respectively.
   

   3. Substitute volume in terms of density into the adiabatic equation:

   Substitute the expressions for $V_1$ and $V_2$ from Step 2 into the adiabatic equation from Step 1:

   $P_1 \left(\frac{m}{\rho_1}\right)^\gamma = P_2 \left(\frac{m}{\rho_2}\right)^\gamma$

   $P_1 \frac{m^\gamma}{\rho_1^\gamma} = P_2 \frac{m^\gamma}{\rho_2^\gamma}$

   Since $m^\gamma$ is a non-zero constant on both sides, it cancels out:
   

   $P_1 \frac{1}{\rho_1^\gamma} = P_2 \frac{1}{\rho_2^\gamma}$

   Rearranging this equation to solve for $P_2$:
   

   $P_2 = P_1 \left(\frac{\rho_2}{\rho_1}\right)^\gamma$

   4. Use the given information about density change: The problem states that the final density of the gas becomes $n$ times its initial value. This means:

   $\rho_2 = n \rho_1$

   From this, we can write the ratio of final to initial density as:
   

   $\frac{\rho_2}{\rho_1} = n$

   5. Calculate the final pressure: Substitute the ratio $\rho_2/\rho_1 = n$ into the equation for $P_2$ from Step 3:

   $P_2 = P_1 (n)^\gamma$

   Given that the initial pressure is $P$, we have $P_1 = P$. So, the final pressure $P_2$ is:
   

   $P_2 = n^\gamma P$

   Important

   For an adiabatic process, the relationship between pressure $P$ and density $\rho$ is $P \rho^{-\gamma} = \text{constant}$, or equivalently, $P \propto \rho^\gamma$.

   ✓Final answer

   The final pressure of the gas is $\boxed{n^{\gamma}P}$.

3. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If the temperature of a gas is increased from $27\,^\circ\mathrm{C}$ to $159\,^\circ\mathrm{C}$, the increase in the rms speed of the gas molecules is (A) $142\%$ (B) $71\%$ (C) $80\%$ (D) $20\%$

   ›Reveal solutionSolution

   The rms speed scales with the square root of absolute temperature; converting to Kelvin and taking the ratio gives a 20% increase, so the correct option is (D).

   The key idea is that the root-mean-square speed of gas molecules is given by $v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$, where $T$ is the absolute temperature (in Kelvin). This means $v_{\text{rms}} \propto \sqrt{T}$. A common pitfall is to use Celsius temperatures directly, but the relationship only holds for Kelvin. The percentage increase in speed is not the same as the percentage increase in Celsius temperature.

   1. Convert temperatures to Kelvin. Absolute zero is $-273\,^\circ\mathrm{C}$, so:

   $T_1 = 27 + 273 = 300\,\mathrm{K}, \quad T_2 = 159 + 273 = 432\,\mathrm{K}.$

   2. Write the ratio of rms speeds. Since $v_{\text{rms}} \propto \sqrt{T}$,

   $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{432}{300}} = \sqrt{1.44} = 1.2.$

   3. Find the percentage increase. The increase factor is $1.2$, meaning the new speed is 120% of the original. The increase is $1.2 - 1 = 0.2 = 20\%$.

   Watch out

   If you mistakenly used Celsius temperatures, you’d get $\frac{159}{27} \approx 5.89$, leading to a nonsensical 143% increase — that’s option (A), a classic trap.

   Tip

   Notice that $432/300 = 1.44$ is a perfect square (1.2²), making the calculation clean. Always check if the ratio simplifies nicely.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If the temperature of a gas is increased from $27^\circ \mathrm{C}$ to $159^\circ \mathrm{C}$, then the percentage increase in the rms speed of the gas molecules is (A) 5 (B) 10 (C) 15 (D) 20

   ›Reveal solutionSolution

   rms speed $\propto\sqrt{T}$ (absolute temperature). Raising $T$ from $300\,$K to $432\,$K multiplies the speed by $\sqrt{1.44}=1.2$, a 20% increase.

   Solution

   The rms speed of gas molecules is

   $v_{rms}=\sqrt{\frac{3RT}{M}}\ \propto\ \sqrt{T},$

   with $T$ in kelvin.

   Convert the temperatures:

   $T_1=27^\circ\text{C}=300\,\text{K},\qquad T_2=159^\circ\text{C}=432\,\text{K}.$

   Ratio of speeds:

   $\frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{432}{300}}=\sqrt{1.44}=1.2.$

   So $v_2=1.2\,v_1$, i.e. an increase of $0.2\,v_1$:

   $\text{percentage increase}=\left(\frac{v_2-v_1}{v_1}\right)\times100\%=(1.2-1)\times100\%=20\%.$

   ✓Final answer

   Percentage increase in rms speed $=20\%$, option (D).

5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.Heavy water is used as moderator in nuclear reactor because (A) It controls the energy released in the reactor (B) It absorbs neutrons and stops chain reaction (C) It cools the reactor faster (D) It slows down the fast moving neutrons

   ›Reveal solutionSolution

   Heavy water slows down fast neutrons via elastic collisions without absorbing them, making it an ideal moderator. The correct option is (D).

   Concept and Intuition

   In a nuclear reactor, the fuel (e.g., uranium-235) fissions most efficiently when struck by slow (thermal) neutrons. The neutrons released from fission are very fast — about 2 MeV of kinetic energy. A moderator is a material placed in the reactor to reduce the speed of these fast neutrons to thermal energies (around 0.025 eV) through repeated collisions, without absorbing them. Heavy water (D₂O) is excellent at this because its deuterium nucleus has nearly the same mass as a neutron, allowing maximum energy transfer per collision, and it has a very low neutron absorption cross-section.

   Step-by-Step Reasoning

   1. Identify the role of a moderator

      A moderator’s job is to slow down fast neutrons so they can cause further fission. It does not control the overall energy release (that’s the job of control rods), nor does it absorb neutrons to stop the chain reaction (that would be a poison or control rod), nor is its primary purpose cooling (that’s the coolant). So options (A), (B), and (C) are incorrect by definition.

   2. Why heavy water slows neutrons effectively

      When a neutron collides elastically with a nucleus, the energy lost per collision is greatest when the target nucleus has a mass close to the neutron’s. Deuterium (²H) has a mass of about 2 u, while a neutron is about 1 u. This near-match means each collision can transfer a large fraction of the neutron’s kinetic energy. Ordinary water (H₂O) has hydrogen (mass 1 u), which is even better for slowing, but hydrogen absorbs neutrons too readily (forming deuterium), which would reduce the neutron population. Heavy water’s deuterium absorbs neutrons far less.

   3. Eliminate the distractors

      • (A) “Controls the energy released” — This is done by control rods (e.g., boron or cadmium) that absorb excess neutrons. The moderator merely sustains the chain reaction by slowing neutrons.
      • (B) “Absorbs neutrons and stops chain reaction” — Heavy water is chosen precisely because it does not absorb many neutrons. Absorbing would stop the reaction, which is undesirable.
      • (C) “Cools the reactor faster” — Cooling is the function of a coolant (e.g., water, liquid sodium). While heavy water can also act as a coolant in some designs, its primary moderator role is slowing neutrons, not cooling.

   4. Confirm the correct choice

      The only option that matches the fundamental purpose of a moderator — and specifically why heavy water is used — is (D): “It slows down the fast moving neutrons.”

   Watch out

   A common mistake is to confuse a moderator with a control rod. Control rods absorb neutrons to regulate the reaction; moderators slow neutrons to sustain it. Heavy water does the latter, not the former.

   Tip

   The “slowing power” of a moderator depends on two factors: the average energy loss per collision (higher for light nuclei) and the scattering cross-section. Heavy water’s low absorption cross-section makes it superior to ordinary water despite slightly less efficient slowing per collision.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
   Q.Match the columns for the first law of thermodynamics Column - I A) Adiabatic processes B) Constant - volume processes C) Isothermal processes D) Constant - pressure processes Column - II I) $W \neq 0, \ \Delta E_{int} \neq Q$ II) $Q = W; \ \Delta E_{int} = 0$ III) $Q = 0; \ \Delta E_{int} = -W$ IV) $W = 0; \ \Delta E_{int} = Q$ ($W$ – work done, $Q$ - heat absorbed, $\Delta E_{int}$ - change of internal energy) The correct match is (A) A B C D III II IV I (B) A B C D III IV II I (C) A B C D IV I III II (D) A B C D I III IV II

   ›Reveal solutionSolution

   The first law of thermodynamics, $\Delta E_{\text{int}} = Q - W$, is applied to four special processes. Matching each process to its characteristic conditions gives the sequence A→III, B→IV, C→II, D→I, which corresponds to option (B).

   The first law of thermodynamics is the energy‑conservation statement for thermodynamic systems:

   $\Delta E_{\text{int}} = Q - W,$

   where $Q$ is heat added to the system, $W$ is work done by the system, and $\Delta E_{\text{int}}$ is the change in internal energy.

   Each special process imposes a constraint (e.g., constant volume, no heat exchange), which simplifies the first law in a characteristic way. The trick is to remember the definition of each process and then substitute into the first law.

   1. Adiabatic processes (A)

      Definition: No heat exchange with the surroundings → $Q = 0$.

      First law becomes $\Delta E_{\text{int}} = -W$.

      Work is generally nonzero (e.g., compression or expansion).

      So A matches III: $Q = 0;\ \Delta E_{\text{int}} = -W$.

   2. Constant‑volume processes (B)

      Definition: Volume does not change → $W = \int P\,dV = 0$.

      First law becomes $\Delta E_{\text{int}} = Q$.

      Heat can flow in or out, changing internal energy directly.

      So B matches IV: $W = 0;\ \Delta E_{\text{int}} = Q$.

   3. Isothermal processes (C)

      Definition: Temperature is constant → for an ideal gas, $\Delta E_{\text{int}} = 0$ (internal energy depends only on temperature).

      First law gives $0 = Q - W$ → $Q = W$.

      Work is generally nonzero (e.g., isothermal expansion).

      So C matches II: $Q = W;\ \Delta E_{\text{int}} = 0$.

   4. Constant‑pressure processes (D)

      Definition: Pressure is constant; work done is $W = P\Delta V$, which is generally nonzero.

      Heat $Q$ is also nonzero, and internal energy changes because temperature changes.

      The only remaining description is I: $W \neq 0,\ \Delta E_{\text{int}} \neq Q$ (since $Q$ and $W$ are both present and not equal).

      So D matches I.

   Thus the sequence A→III, B→IV, C→II, D→I is exactly option (B).

   Watch out

   A common mistake is to think that constant‑pressure processes have $Q = \Delta H$ and forget that $\Delta E_{\text{int}} \neq Q$ because work is done. Also, do not confuse “adiabatic” with “isothermal” — one has $Q=0$, the other has $\Delta E_{\text{int}}=0$.

   Tip

   For quick recall:

   • Adiabatic → $Q=0$
   • Isochoric (constant volume) → $W=0$
   • Isothermal → $\Delta E_{\text{int}}=0$
   • Isobaric (constant pressure) → none of the above simplifications, so it’s the leftover.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.An ideal gas at pressure $P_0$ undergoes an isothermal expansion until its volume is $8.0$ times its initial volume. The gas is slowly and adiabatically compressed back to its original volume. If the adiabatic constant of the gas is $\gamma = 4/3$, then the ratio of the average kinetic energy per molecule in this final state to that in the initial state is (A) 1.44 (B) 1.68 (C) 2.0 (D) 1.2

   ›Reveal solutionSolution

   An ideal gas undergoes isothermal expansion followed by adiabatic compression. The average kinetic energy per molecule is directly proportional to the absolute temperature. The isothermal process keeps the temperature constant, while the adiabatic compression increases the temperature. The final temperature is twice the initial temperature, so the ratio of average kinetic energy per molecule in the final state to the initial state is $\boxed{2.0}$.

   The average kinetic energy per molecule of an ideal gas is directly proportional to its absolute temperature. This is a fundamental concept in the kinetic theory of gases. Therefore, to find the ratio of the average kinetic energy per molecule in the final state to that in the initial state, we need to determine the ratio of the final absolute temperature to the initial absolute temperature.

   We will analyze the process in two stages:

   1. Isothermal Expansion: The temperature remains constant.
   2. Adiabatic Compression: The temperature changes according to the adiabatic relation.

   Let's denote the initial state as state 1, the intermediate state (after isothermal expansion) as state 2, and the final state (after adiabatic compression) as state 3.

   1. Initial State (State 1):

      Let the initial pressure, volume, and temperature be $P_1 = P_0$, $V_1 = V_0$, and $T_1 = T_0$, respectively.

      The average kinetic energy per molecule in this state is $KE_1 \propto T_0$.

   2. Isothermal Expansion (State 1 to State 2):

      In an isothermal process, the temperature of the gas remains constant.

      Important

      For an isothermal process, $T = \text{constant}$.

      The gas expands until its volume is $8.0$ times its initial volume.

      So, $V_2 = 8.0 V_0$.

      Since the process is isothermal, the temperature at state 2 is the same as at state 1:

      $T_2 = T_1 = T_0$.

      The average kinetic energy per molecule in this state is $KE_2 \propto T_0$.

   3. Adiabatic Compression (State 2 to State 3):

      The gas is slowly and adiabatically compressed back to its original volume.

      Important

      For an adiabatic process, $PV^\gamma = \text{constant}$ or $TV^{\gamma-1} = \text{constant}$.

      The compression starts from state 2 ($V_2 = 8V_0$, $T_2 = T_0$) and ends at state 3, where the volume is the original volume, $V_3 = V_0$.

      We use the adiabatic relation involving temperature and volume:

      [!FORMULA]

      $T V^{\gamma-1} = \text{constant}$

      Applying this between state 2 and state 3:

      $T_2 V_2^{\gamma-1} = T_3 V_3^{\gamma-1}$

      Substitute the known values:

      $T_0 (8V_0)^{\gamma-1} = T_3 (V_0)^{\gamma-1}$

      To find $T_3$, rearrange the equation:

      $T_3 = T_0 \left(\frac{8V_0}{V_0}\right)^{\gamma-1}$

      $T_3 = T_0 (8)^{\gamma-1}$

      We are given the adiabatic constant $\gamma = 4/3$.

      Calculate $\gamma-1$:

      $\gamma-1 = \frac{4}{3} - 1 = \frac{1}{3}$

      Now substitute this value into the expression for $T_3$:

      $T_3 = T_0 (8)^{1/3}$

      Since $8^{1/3} = 2$ (because $2 \times 2 \times 2 = 8$), we have:

      $T_3 = T_0 (2)$

      So, the final temperature $T_3 = 2T_0$.

   4. Ratio of Average Kinetic Energy per Molecule:

      The average kinetic energy per molecule ($KE_{avg}$) of an ideal gas is given by:

      $KE_{avg} = \frac{f}{2} k_B T$

      where $f$ is the degrees of freedom, $k_B$ is the Boltzmann constant, and $T$ is the absolute temperature.

      Therefore, the average kinetic energy per molecule is directly proportional to the absolute temperature.

      The ratio of the average kinetic energy per molecule in the final state to that in the initial state is:

      $\frac{KE_3}{KE_1} = \frac{\frac{f}{2} k_B T_3}{\frac{f}{2} k_B T_1}$

      $\frac{KE_3}{KE_1} = \frac{T_3}{T_1}$

      Substitute $T_3 = 2T_0$ and $T_1 = T_0$:

      $\frac{KE_3}{KE_1} = \frac{2T_0}{T_0} = 2$

   The ratio of the average kinetic energy per molecule in the final state to that in the initial state is $2.0$.

   The correct option is (C).

   ✓Final answer

   The ratio of the average kinetic energy per molecule in the final state to that in the initial state is $\boxed{2.0}$.

8. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.Statement I : A device in which heat measurement can be made is called calorimeter. Statement II : Skating is possible on snow due to the formation of water below the skates. Water is formed due to the increase of temperature and ice melts. Statement III: Two bodies at different temperature are mixed in a calorimeter. Total internal energy of the two bodies remains conserved. Which of the following is correct? (A) Statements I, II and III are true (B) Statement I is true, but statements II and III are false (C) Both statements I and II are true, but statement III is false (D) Both statements I, III are true, but statement II is false

   ›Reveal solutionSolution

   We need to evaluate three statements about calorimetry and phase transitions. Statement I (calorimeter definition) is true, Statement II (skating mechanism) is false because it misidentifies the cause of ice melting, and Statement III (energy conservation in calorimetry) is true. The correct option is (D).

   Let me analyze each statement carefully to understand the physics involved.

   Statement I: Definition of a Calorimeter

   A calorimeter is indeed a device designed to measure heat transfer. The name comes from "calor" (heat) and "meter" (to measure). This is a straightforward definition and is TRUE.

   Statement II: The Physics of Ice Skating

   This statement claims that skating is possible because increased temperature melts the ice. Let's examine this critically:

   1. The traditional explanation (which this statement reflects) suggested that pressure from the skate blade lowers the melting point of ice, causing it to melt and form a lubricating water layer.

   2. Modern understanding shows this is incorrect for typical skating conditions. The pressure-melting effect is negligible at normal skating temperatures (around -5°C to 0°C). The pressure from a skate can only lower the melting point by about 0.5°C, which is insufficient.

   3. The actual mechanism involves frictional heating and a pre-existing quasi-liquid layer that exists on ice surfaces even below 0°C due to surface molecular mobility.

   The statement says water forms "due to the increase of temperature" without mentioning the actual mechanism (friction), and implies this is primarily a pressure effect. This is FALSE or at best misleading.

   Watch out

   The pressure-melting explanation for ice skating is a common misconception found in many textbooks, but modern research has shown it's not the primary mechanism at typical skating temperatures.

   Statement III: Energy Conservation in Calorimetry

   When two bodies at different temperatures are mixed in an isolated calorimeter:

   1. The system is thermally isolated from the surroundings (that's the purpose of a calorimeter).

   2. No work is done on or by the system during simple mixing.

   3. By the First Law of Thermodynamics: $\Delta U = Q - W$

   4. Since the calorimeter prevents heat exchange with surroundings ($Q_{external} = 0$) and no work is done ($W = 0$), the total internal energy of the system remains constant.

   5. Heat flows from the hotter body to the cooler body until thermal equilibrium is reached, but the total internal energy is conserved.

   This statement is TRUE.

   Principle of Calorimetry: For an isolated system, $m_1 c_1 (T_f - T_1) + m_2 c_2 (T_f - T_2) = 0$, which is a consequence of internal energy conservation.

   Evaluation Summary

   • Statement I: TRUE ✓
   • Statement II: FALSE ✗
   • Statement III: TRUE ✓

   Looking at the options, we need: I true, II false, III true.

   ✓Final answer

   The correct option is (D): Both statements I and III are true, but statement II is false.

   ANSWER: D

9. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
   Q.A hole is drilled at the center of a metallic plate. If the metal plate is heated, then the size of the hole will be (A) Expanding (B) Contracting (C) Remains same (D) Not predicted

   ›Reveal solutionSolution

   When a metal plate is heated uniformly, every linear dimension — including the diameter of a hole — expands according to the same coefficient of linear expansion. The hole expands.

   The key idea is that thermal expansion is a property of the material itself, not of the shape. When you heat a uniform metal plate, the atoms vibrate more and the average distance between them increases. This happens everywhere in the plate — in the solid region and along the boundary of the hole. The hole is not a separate object; it is simply the absence of metal. The metal around the hole expands outward, making the hole larger.

   A common mistake is to think the hole "fills in" as the metal expands inward. That would only happen if the plate were heated from the outside inward, but here the heating is uniform. Every point on the rim of the hole moves away from the center because the material between that point and the center expands.

   1. Think of the hole as being defined by a circle of metal atoms. If you mark a circle on the plate and then heat it, the circumference of that circle increases. Since circumference = $\pi d$, the diameter $d$ must also increase.

   2. Use the linear expansion formula. For a small temperature change $\Delta T$, the change in any linear dimension $L$ is $\Delta L = \alpha L \Delta T$, where $\alpha$ is the coefficient of linear expansion. This applies to the diameter of the hole just as it applies to the outer diameter of the plate.

   3. Consider a thought experiment. Imagine the plate before the hole is drilled. Mark a circle where the hole will be. Heat the plate — the circle expands. Now drill the hole along that expanded circle. The hole you get is larger than the original intended size. The same logic holds if you drill first and heat afterward: the hole expands.

   Watch out

   Do not confuse this with a bimetallic strip or non-uniform heating. In a uniformly heated isotropic material, there is no "inward expansion" — the material expands equally in all directions, so cavities expand.

   Tip

   A quick way to remember: if you heat a metal ring, the inner diameter increases. A hole in a plate is just a very short ring.

   4. The only exception would be if the plate were not free to expand (e.g., constrained by rigid boundaries), but the problem states no such constraint. Under normal conditions, the hole expands.

   ✓Final answer

   The correct option is (A) Expanding.

10. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.A gaseous mixture consists of molecules of type A, B and C with masses $M_A > M_B > M_C$. The correct relation between their average kinetic energy is (A) $K_A > K_B > K_C$ (B) $K_A < K_B < K_C$ (C) $K_A = K_B = K_C$ (D) $K_A = K_B < K_C$

    ›Reveal solutionSolution

    The average kinetic energy of gas molecules in a mixture depends only on the absolute temperature of the mixture, not on the individual masses of the molecules. Since all components of a gaseous mixture are at the same temperature, their average kinetic energies are equal. The correct option is (C).

    The core concept here comes from the kinetic theory of gases, which describes the macroscopic properties of gases in terms of the microscopic behavior of their molecules. A fundamental postulate of this theory is that the absolute temperature of an ideal gas is a direct measure of the average translational kinetic energy of its molecules.

    When different gases are mixed together, they quickly reach thermal equilibrium. This means that all components of the mixture will be at the same temperature. If the temperature is the same for all types of molecules in the mixture, then their average kinetic energies must also be the same, regardless of their individual masses.

    1. Understanding Average Kinetic Energy: For an ideal gas, the average translational kinetic energy per molecule is directly proportional to the absolute temperature of the gas. This is a key result from the kinetic theory.

    2. The Formula: The average translational kinetic energy ($K_{avg}$) of a single molecule in an ideal gas is given by:

       $K_{avg} = \frac{3}{2} k_B T$

       where $k_B$ is the Boltzmann constant (a universal constant) and $T$ is the absolute temperature of the gas in Kelvin.

    3. Applying to the Mixture: In the given problem, we have a gaseous mixture consisting of molecules of type A, B, and C. Since they are all part of the same mixture, they are in thermal equilibrium with each other. This implies that all three types of molecules (A, B, and C) are at the same absolute temperature, $T$.

    4. Comparing Kinetic Energies:

       • For molecule type A, the average kinetic energy is $K_A = \frac{3}{2} k_B T$.
       • For molecule type B, the average kinetic energy is $K_B = \frac{3}{2} k_B T$.
       • For molecule type C, the average kinetic energy is $K_C = \frac{3}{2} k_B T$.

       Since $k_B$ is a constant and $T$ is the same for all three types of molecules, it follows that their average kinetic energies must be equal.

    $K_A = K_B = K_C$

    Watch out

    Do not confuse average kinetic energy with average speed (or root-mean-square speed). While the average kinetic energy is the same for all molecules at a given temperature, their average speeds will differ if their masses are different. Specifically, molecules with smaller masses will have higher average speeds to maintain the same average kinetic energy. The root-mean-square speed is given by $v_{rms} = \sqrt{\frac{3k_B T}{M}}$, where $M$ is the mass of the molecule. Since $M_A > M_B > M_C$, it would follow that $v_{rms,A} < v_{rms,B} < v_{rms,C}$.

    ✓Final answer

    The correct relation between their average kinetic energy is $\boxed{K_A = K_B = K_C}$.

11. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
    Q.An ideal gas at temperature T & pressure P fills the chamber A which is separated from chamber B which has vacuum. The two chambers are thermally insulated. When plug is removed the gas fills both the chambers. Both A & B have same volume. What will be the pressure & temperature of the gas after it comes to equilibrium.? (A) $P,\,T$ (B) $P/2,\,T$ (C) $P/2,\,T/2$ (D) $P,\,T/2$

    ›Reveal solutionSolution

    The gas undergoes free expansion into a vacuum, which is an irreversible adiabatic process. For an ideal gas, internal energy depends only on temperature, so temperature remains constant; pressure halves because volume doubles. The correct option is (B).

    The key concept here is free expansion (also called Joule expansion). When a gas expands into a vacuum, it does no work (no opposing pressure) and, because the container is thermally insulated, no heat is exchanged. The first law of thermodynamics says $\Delta U = Q - W$. Here $Q = 0$ and $W = 0$, so $\Delta U = 0$. For an ideal gas, internal energy $U$ depends only on temperature, so constant $U$ means constant $T$.

    Now, let’s walk through the reasoning step by step.

    1. Identify the process.

       The gas is initially in chamber A at pressure $P$ and temperature $T$. Chamber B is evacuated. When the plug is removed, the gas rushes into B. This is free expansion — the gas expands against zero external pressure, so no work is done ($W = 0$). The walls are thermally insulated, so no heat flows ($Q = 0$).

    2. Apply the first law.

       $\Delta U = Q - W = 0 - 0 = 0$. The internal energy of the ideal gas does not change.

    3. Relate internal energy to temperature.

       For an ideal gas, $U = n C_V T$, where $n$ is the number of moles and $C_V$ is the molar heat capacity at constant volume. Since $\Delta U = 0$, we have $n C_V \Delta T = 0$, so $\Delta T = 0$. Thus the final temperature remains $T$.

    4. Find the final pressure.

       The gas now occupies both chambers, so the volume doubles: $V_{\text{final}} = 2V_{\text{initial}}$. The number of moles $n$ is unchanged, and temperature $T$ is unchanged. Using the ideal gas law $PV = nRT$, initially $P V = nRT$. Finally, let $P_f$ be the final pressure: $P_f (2V) = nRT$. Dividing the final equation by the initial gives $\frac{P_f \cdot 2V}{P V} = 1$, so $2P_f / P = 1$, hence $P_f = P/2$.

    5. Check the options.

       The final state is pressure $P/2$ and temperature $T$, which matches option (B).

    Watch out

    A common mistake is to think that because the volume doubles, the temperature must also change (e.g., using the adiabatic relation $TV^{\gamma-1} = \text{constant}$). That relation applies only to reversible adiabatic processes, not to free expansion, which is irreversible and involves no work.

    Tip

    Free expansion is a classic “trick” problem: the gas does no work, so internal energy (and thus temperature) stays constant. Always check whether work is done — if the expansion is into a vacuum, the answer is almost always constant temperature.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
    Q.The rms speed of H$_2$ molecules is C at 27$^\circ$C. The molecules break into atoms. What should be the new temperature such that atoms have same speed as molecules (Assume the mass of H$_2$ molecules is twice the mass of H- atom) (A) 600 K (B) 150 K (C) 100 K (D) 300 K

    ›Reveal solutionSolution

    The key idea is that rms speed depends on temperature and mass: $v_{\text{rms}} = \sqrt{3kT/m}$. For the same speed, the product $mT$ must be constant. Since an H atom has half the mass of H₂, the temperature must be halved from 300 K to 150 K. The correct option is (B).

    Concept and Intuition

    The root-mean-square speed of gas molecules is given by $v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$, where $k$ is Boltzmann’s constant, $T$ is absolute temperature, and $m$ is the mass of one molecule. If we want the atoms (each of mass $m_{\text{atom}}$) to have the same rms speed as the original molecules (each of mass $m_{\text{molecule}} = 2m_{\text{atom}}$), then the temperature must adjust to compensate for the mass change. Since $v_{\text{rms}}$ is proportional to $\sqrt{T/m}$, keeping $v_{\text{rms}}$ constant means $T/m$ must stay constant — so $T$ must scale with $m$.

    Step-by-step solution

    1. Convert given temperature to Kelvin

       The problem states $27^\circ\text{C}$.

       $T_{\text{initial}} = 27 + 273 = 300\ \text{K}$.

    2. Write the rms speed for H₂ molecules

       Let $m$ be the mass of an H atom. Then the mass of an H₂ molecule is $2m$.

       The rms speed of H₂ at 300 K is

    $v_{\text{rms, H₂}} = \sqrt{\frac{3k \cdot 300}{2m}}.$

    3. Write the rms speed for H atoms at unknown temperature $T$ Each H atom has mass $m$.

    $v_{\text{rms, H}} = \sqrt{\frac{3k T}{m}}.$

    4. Set the speeds equal We require $v_{\text{rms, H}} = v_{\text{rms, H₂}}$, so

    $\sqrt{\frac{3k T}{m}} = \sqrt{\frac{3k \cdot 300}{2m}}.$

    5. Cancel common factors and solve for $T$ Square both sides:

    $\frac{3k T}{m} = \frac{3k \cdot 300}{2m}.$

    Cancel $3k/m$ (nonzero):

    $T = \frac{300}{2} = 150\ \text{K}.$

    Tip

    Notice the elegant shortcut: since $v_{\text{rms}} \propto \sqrt{T/m}$, for constant speed we have $T \propto m$. Halving the mass means halving the temperature — no need to plug in numbers until the very end.

    Watch out

    A common mistake is to forget that temperature must be in Kelvin. Using Celsius (27°C) directly would give 13.5°C, which is not among the options and is physically wrong because rms speed is zero at 0 K, not at 0°C.

    ✓Final answer

    The correct option is (B).

    ANSWER: B