Q.A molecule of a substance has a permanent electric dipole moment of magnitude $10^{-29}\ \text{C m}$. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude $10^{6}\ \text{V m}^{-1}$. The direction of the field is suddenly changed by an angle of $60^\circ$. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.

Showing the 12 most recent of 34 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.An iron rod of mass 2 kg and length 1.5 m is supplied 18.4 kJ of heat. If the specific heat capacity of iron is $460 \, \mathrm{J\,kg^{-1}\,K^{-1}}$ and its coefficient of linear expansion is $1.2 \times 10^{-5} \, ^\circ \mathrm{C}^{-1}$, then the increase in the length of the rod is (A) 0.24 mm (B) 0.36 mm (C) 0.12 mm (D) 0.18 mm

   ›Reveal solutionSolution

   The rod’s temperature rise is found from the heat supplied, mass, and specific heat capacity; then the linear expansion formula gives the increase in length. The final increase is 0.36 mm, so option (B) is correct.

   Concept & Intuition

   When heat is added to a solid, its temperature rises according to $Q = m c \Delta T$. Once we know the temperature change, the rod expands linearly: $\Delta L = \alpha L_0 \Delta T$. The key is to connect the thermal energy input to the dimensional change via the temperature rise — a two‑step process that is straightforward if we keep units consistent.

   ---

   1. Find the temperature rise from the heat supplied The heat added is $Q = 18.4\ \text{kJ} = 18.4 \times 10^3\ \text{J}$. Mass $m = 2\ \text{kg}$, specific heat $c = 460\ \text{J kg}^{-1}\text{K}^{-1}$. Using $Q = m c \Delta T$:

   $\Delta T = \frac{Q}{m c} = \frac{18.4 \times 10^3}{2 \times 460} = \frac{18400}{920} = 20\ ^\circ\text{C (or K)}.$

   So the rod’s temperature increases by $20\ ^\circ\text{C}$.

   2. Apply the linear expansion formula Original length $L_0 = 1.5\ \text{m}$, coefficient $\alpha = 1.2 \times 10^{-5}\ {^\circ\text{C}}^{-1}$. The increase in length is:

   $\Delta L = \alpha L_0 \Delta T = (1.2 \times 10^{-5}) \times 1.5 \times 20.$

   Compute step by step:

   $1.2 \times 10^{-5} \times 1.5 = 1.8 \times 10^{-5},$

   then

   $$1.8 \times 10^{-5} \times 20 = 3.6 \times 10^{-4}\ \text{m}. …$$

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The rms value of electric field at a distance of 6 m from a 100 W bulb of efficiency 1.2% is (A) $4\, \mathrm{V\,m^{-1}}$ (B) $1\, \mathrm{V\,m^{-1}}$ (C) $2\, \mathrm{V\,m^{-1}}$ (D) $3\, \mathrm{V\,m^{-1}}$

   ›Reveal solutionSolution

   The rms electric field is found by converting the bulb’s electrical power into radiated intensity, then using the relation between intensity and rms electric field. The result is $1\ \mathrm{V\,m^{-1}}$, so the correct option is (B).

   Concept & Intuition

   A light bulb radiates electromagnetic waves. Only a fraction of its electrical power (the efficiency) becomes light. At a distance $r$, that power spreads uniformly over a sphere of area $4\pi r^2$, giving the intensity $I$ (power per unit area). For an electromagnetic wave in vacuum, the intensity is related to the rms electric field $E_{\text{rms}}$ by $I = \varepsilon_0 c\, E_{\text{rms}}^2$. So we can work backwards: from the bulb’s radiated power, find $I$, then solve for $E_{\text{rms}}$.

   Step-by-step solution

   1. Find the actual radiated power The bulb draws 100 W, but only 1.2% is converted to light.

   $P_{\text{rad}} = 100 \times \frac{1.2}{100} = 1.2\ \text{W}$

   2. Compute the intensity at 6 m The radiated power spreads uniformly over a sphere of radius $r = 6\ \text{m}$.

   $I = \frac{P_{\text{rad}}}{4\pi r^2} = \frac{1.2}{4\pi \times 36} = \frac{1.2}{144\pi} = \frac{1}{120\pi}\ \text{W/m}^2$

   3. Relate intensity to rms electric field For a plane electromagnetic wave in vacuum, the time‑averaged intensity is

   $I = \varepsilon_0 c\, E_{\text{rms}}^2$

   where $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ and $c = 3 \times 10^8\ \text{m/s}$.

   4. Solve for $E_{\text{rms}}$

   $E_{\text{rms}} = \sqrt{\frac{I}{\varepsilon_0 c}} = \sqrt{\frac{1/(120\pi)}{(8.85 \times 10^{-12})(3 \times 10^8)}}$

   First compute the denominator:

   $$\varepsilon_0 c = (8.85 \times 10^{-12})(3 \times 10^8) = 2.655 \times 10^{-3} …$$

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.At a distance of 20 cm from the centre of a charged conducting sphere of radius 10 cm, the electric field due to the sphere is $9000\, \mathrm{NC}^{-1}$. If an electric charge of $3\, \mu \mathrm{C}$ is placed at a distance of 30 cm from the centre of the sphere, then the electrostatic force acting on the charge is (A) 9 mN (B) 12 mN (C) 18 mN (D) 24 mN

   ›Reveal solutionSolution

   Outside a charged sphere the field is $E=kQ/r^2$; use the given field to find $kQ$, then the field (and force) at 30 cm. Force $=12~\text{mN}$ (B).

   Step 1 — Find $kQ$ from the given data.

   Outside the sphere it behaves like a point charge at the centre:

   $E=\frac{kQ}{r^2}\;\Rightarrow\; kQ = E\,r^2 = 9000\times(0.20)^2 = 360~\text{N m}^2\,\text{C}^{-1}$

   Step 2 — Field at 30 cm from the centre. …

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4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The amplitude of the magnetic field of a plane electromagnetic wave travelling along positive x-axis in vacuum is 6 mT. A particle of charge $5~\mu \mathrm{C}$ is travelling with a velocity of $6 \times 10^5~\mathrm{ms}^{-1}$ along the positive y-axis. If the magnetic field is oriented along positive z-axis, then the maximum force exerted on the particle due to electric field of the wave is (A) $15~\mathrm{N}$ (B) $18~\mathrm{N}$ (C) $9~\mathrm{N}$ (D) $12~\mathrm{N}$

   ›Reveal solutionSolution

   The maximum electric force on the particle is found by first relating the electric and magnetic field amplitudes in an EM wave via $E_0 = c B_0$, then using $F = qE_0$. The result is $9\ \mathrm{N}$, so the correct option is (C).

   Concept & Intuition

   In a plane electromagnetic wave in vacuum, the electric and magnetic fields are perpendicular to each other and to the direction of propagation. Their amplitudes are related by $E_0 = c B_0$, where $c$ is the speed of light. The force from the magnetic field depends on velocity, but the question asks specifically for the maximum force due to the electric field — that is simply $qE_0$, independent of the particle’s motion. The particle’s velocity along y is irrelevant for the electric force; it only matters if we were asked about the magnetic force.

   Step-by-step solution

   1. Identify the given data

      • Magnetic field amplitude: $B_0 = 6\ \mathrm{mT} = 6 \times 10^{-3}\ \mathrm{T}$
      • Charge: $q = 5\ \mu\mathrm{C} = 5 \times 10^{-6}\ \mathrm{C}$
      • Speed of light: $c = 3 \times 10^8\ \mathrm{m/s}$
      • The wave travels along $+x$, magnetic field along $+z$, so electric field is along $+y$ (since $\mathbf{E} \times \mathbf{B}$ gives propagation direction). The particle moves along $+y$, but that does not affect the electric force magnitude.

   2. Relate electric and magnetic field amplitudes

      For any electromagnetic wave in vacuum,

   $E_0 = c B_0$

   This follows from Maxwell’s equations: the ratio of field strengths is fixed by the speed of light.

   3. Compute the electric field amplitude $E_0 = (3 \times 10^8) \times (6 \times 10^{-3}) = 1.8 \times 10^6\ \mathrm{V/m}$ …

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the average kinetic energy of the molecules of a gas at a temperature of $30\,^\circ\mathrm{C}$ is $U$, then the temperature at which the average kinetic energy of the molecules of the gas becomes $2U$ is (A) $939\,^\circ\mathrm{C}$ (B) $303\,^\circ\mathrm{C}$ (C) $333\,^\circ\mathrm{C}$ (D) $60\,^\circ\mathrm{C}$

   ›Reveal solutionSolution

   The average kinetic energy of gas molecules is directly proportional to the absolute temperature (in Kelvin). Doubling the energy means doubling the Kelvin temperature, then converting back to Celsius gives $333\,^\circ\mathrm{C}$.

   The key idea here is that the average kinetic energy of molecules in an ideal gas depends only on the absolute temperature — not on the type of gas, pressure, or volume. This is a direct consequence of the kinetic theory of gases.

   The average kinetic energy per molecule is given by $\frac{3}{2} k_B T$, where $k_B$ is Boltzmann's constant and $T$ is the absolute temperature in Kelvin. So if the energy doubles, the absolute temperature must double as well. The trap most students fall into is forgetting to convert Celsius to Kelvin before doing the ratio — Celsius is a shifted scale, not an absolute one.

   Let’s work through it step by step.

   1. Convert the given Celsius temperature to Kelvin.

      The formula is $T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273$.

      So $T_1 = 30 + 273 = 303\ \mathrm{K}$.

   2. Relate kinetic energy to temperature.

      The average kinetic energy $U \propto T$ (in Kelvin). So if $U$ becomes $2U$, the new absolute temperature $T_2$ must satisfy

   $\frac{T_2}{T_1} = \frac{2U}{U} = 2.$

   Hence $T_2 = 2 \times 303 = 606\ \mathrm{K}$.

   3. Convert the new temperature back to Celsius. …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.At a constant pressure of $2 \times 10^5$ Nm$^{-2}$, if the volume of 4 moles of a monoatomic gas changes from 1000 cc to 1500 cc, then the change in internal energy of the gas is (A) 150 J (B) 250 J (C) 200 J (D) 600 J

   ›Reveal solutionSolution

   At constant $P$: $W=P\Delta V=100\ \text{J}$ and for a monoatomic gas $\Delta U=\tfrac32 P\Delta V=150\ \text{J}$.

   Given $P=2\times10^5\ \mathrm{N\,m^{-2}}$, $\Delta V=1500-1000=500\ \text{cc}=5\times10^{-4}\ \mathrm{m^3}$.

   Work done by the gas at constant pressure:

   $W=P\,\Delta V=(2\times10^5)(5\times10^{-4})=100\ \text{J}.$

   For a monoatomic ideal gas, $C_v=\tfrac32 R$ and $P\Delta V=nR\Delta T$, so …

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7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the molar specific heat of a rigid diatomic gas at constant pressure is C, then the molar specific heat of a monoatomic gas at constant volume is (A) $\dfrac{2C}{5}$ (B) $\dfrac{3C}{5}$ (C) $\dfrac{5C}{7}$ (D) $\dfrac{3C}{7}$

   ›Reveal solutionSolution

   The key is to relate the given $C$ (rigid diatomic, constant pressure) to the monatomic constant‑volume specific heat using the equipartition of energy. The answer is $\frac{3C}{7}$, which corresponds to option (D).

   ---

   Concept & Intuition

   Molar specific heats depend on how many degrees of freedom a gas molecule has.

   • A rigid diatomic molecule has 5 degrees of freedom (3 translational + 2 rotational).
   • A monatomic molecule has only 3 translational degrees of freedom.

   For an ideal gas, the molar specific heat at constant volume is $C_V = \frac{f}{2}R$, where $f$ is the number of degrees of freedom. The constant‑pressure specific heat is $C_P = C_V + R = \frac{f+2}{2}R$.

   We are told that for the rigid diatomic gas at constant pressure, $C = C_P^{\text{diatomic}}$. We want $C_V^{\text{monatomic}}$ in terms of $C$.

   ---

   Step‑by‑step reasoning

   1. Write $C$ for the rigid diatomic gas at constant pressure. For a rigid diatomic gas, $f = 5$.

   $C_V^{\text{diatomic}} = \frac{5}{2}R, \quad C_P^{\text{diatomic}} = C_V^{\text{diatomic}} + R = \frac{7}{2}R.$

   Hence

   $C = \frac{7}{2}R.$

   2. Express $R$ in terms of $C$. From the above,

   $R = \frac{2C}{7}.$

   3. Find $C_V$ for a monatomic gas. For a monatomic gas, $f = 3$, so

   $C_V^{\text{monatomic}} = \frac{3}{2}R.$

   4. Substitute $R$ from step 2. $$$$

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8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.A bar magnet of magnetic moment $1.8\, \mathrm{Am}^2$ is free to rotate about a vertical axis passing through its centre at a place where the vertical component of earth’s magnetic field is $0.3 \times 10^{-4}\ \mathrm{T}$ and the dip angle is $45^\circ$. If the magnet at rest in east-west direction is released, then the kinetic energy (in $\mu$J) of the magnet when it reaches north-south direction is (A) 72 (B) 54 (C) 27 (D) 36

   ›Reveal solutionSolution

   Rotation about a vertical axis is governed by the horizontal field $B_H$; with dip $45^\circ$, $B_H=B_V=0.3\times10^{-4}\,\text{T}$, so $KE = mB_H = 54\,\mu\text{J}$.

   The magnet rotates about a vertical axis, so only the horizontal component of the earth's field exerts a torque. With dip angle $\delta = 45^\circ$:

   $\tan\delta = \frac{B_V}{B_H} \Rightarrow B_H = \frac{B_V}{\tan 45^\circ} = B_V = 0.3\times10^{-4}\,\text{T}$ …

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9. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.A bar magnet of magnetic moment $1.5\, \mathrm{A\, m^2}$ is initially placed in the direction of a uniform magnetic field of $18 \times 10^{-2}\ \mathrm{T}$. The work to be done to rotate the magnet so that its magnetic moment becomes perpendicular to the direction of the magnetic field is (A) $540\ \mathrm{mJ}$ (B) $135\ \mathrm{mJ}$ (C) $270\ \mathrm{mJ}$ (D) $90\ \mathrm{mJ}$

   ›Reveal solutionSolution

   The work done to rotate a magnetic dipole from alignment to perpendicular orientation equals the change in potential energy, given by $W = mB$, which here is $0.27\ \mathrm{J}$ or $270\ \mathrm{mJ}$.

   The key concept is the potential energy of a magnetic dipole in a uniform magnetic field. A bar magnet behaves like a magnetic dipole, and its potential energy depends on the angle between its magnetic moment $\vec{m}$ and the field $\vec{B}$.

   When the magnet is aligned with the field ($\theta = 0^\circ$), the system is in its lowest energy state. To rotate it to a perpendicular orientation ($\theta = 90^\circ$), you must do work against the torque exerted by the field. This work is stored as the increase in potential energy.

   The formula for potential energy is $U = - mB \cos\theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$. The work done by an external agent to change the orientation from $\theta_1$ to $\theta_2$ is simply the difference in potential energy: $W = U(\theta_2) - U(\theta_1)$.

   Let's work through the calculation step by step.

   1. Identify the given quantities.

      Magnetic moment, $m = 1.5\ \mathrm{A\,m^2}$.

      Magnetic field strength, $B = 18 \times 10^{-2}\ \mathrm{T} = 0.18\ \mathrm{T}$.

      Initial angle, $\theta_1 = 0^\circ$ (magnet aligned with field).

      Final angle, $\theta_2 = 90^\circ$ (magnet perpendicular to field).

   2. Write the potential energy at each orientation.

      At $\theta_1 = 0^\circ$: $U_1 = - mB \cos 0^\circ = - mB$.

      At $\theta_2 = 90^\circ$: $U_2 = - mB \cos 90^\circ = 0$.

   3. Compute the work done.

      The work required is the change in potential energy:

   $W = U_2 - U_1 = 0 - (- mB) = mB.$

   Substitute the values:

   $W = (1.5) \times (0.18) = 0.27\ \mathrm{J}.$

   4. Convert to millijoules. Since $1\ \mathrm{J} = 1000\ \mathrm{mJ}$, …

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10. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.Two moles of a gas at a temperature of $327~^\circ\mathrm{C}$ expands adiabatically such that its volume increases by $700\%$. If the ratio of the specific heat capacities of the gas is $\dfrac{4}{3}$, then the work done by the gas is (Universal gas constant $= 8.3~\mathrm{J~mol^{-1}~K^{-1}}$) (A) 14.94 kJ (B) 29.88 kJ (C) 44.82 kJ (D) 59.76 kJ

    ›Reveal solutionSolution

    For an adiabatic process, work is done at the expense of internal energy. Using the adiabatic relation $TV^{\gamma-1} = \text{constant}$ and the formula $W = \frac{nR(T_i - T_f)}{\gamma - 1}$, the work done is found to be 14.94 kJ, corresponding to option (A).

    Concept & Intuition

    In an adiabatic expansion, no heat enters or leaves the system ($Q = 0$). The first law of thermodynamics then says $\Delta U = -W$, so the work done by the gas comes entirely from a drop in its internal energy. For an ideal gas, internal energy depends only on temperature, so if we can find the final temperature after the expansion, we can compute the work directly. The key is the adiabatic relation between temperature and volume: $TV^{\gamma-1} = \text{constant}$, where $\gamma = C_p/C_v$ is given as $4/3$.

    Step-by-step solution

    1. Convert initial temperature to Kelvin The initial temperature is $327^\circ\mathrm{C}$.

    $T_i = 327 + 273 = 600~\mathrm{K}$

    2. Interpret the volume increase “Volume increases by 700%” means the final volume is the initial volume plus 700% of it:

    $V_f = V_i + 7V_i = 8V_i$

    So the volume ratio is $V_f/V_i = 8$.

    3. Apply the adiabatic condition For an adiabatic process, $T V^{\gamma-1} = \text{constant}$. Thus:

    $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$

    Rearranging:

    $T_f = T_i \left( \frac{V_i}{V_f} \right)^{\gamma-1}$

    Here $\gamma = 4/3$, so $\gamma - 1 = 1/3$.

    $T_f = 600 \times \left( \frac{1}{8} \right)^{1/3}$

    Since $8^{1/3} = 2$, we have:

    $T_f = 600 \times \frac{1}{2} = 300~\mathrm{K}$

    4. Compute the work done For an adiabatic process, the work done by the gas is: $W = \frac{nR(T_i - T_f)}{\gamma - 1}$ …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.A thin film of water is formed between two straight parallel wires each of length 8 cm separated by distance of 0.6 cm. The work done to increase the distance between the wires to 0.8 cm is (Surface tension of water = $0.07~\mathrm{Nm}^{-1}$) (A) $33.6~\mu\mathrm{J}$ (B) $22.4~\mu\mathrm{J}$ (C) $11.2~\mu\mathrm{J}$ (D) $44.8~\mu\mathrm{J}$

    ›Reveal solutionSolution

    The work done equals the change in surface energy of the water film, which has two surfaces. The increase in area times surface tension times 2 gives the work: $W = 2 \times 0.07 \times (0.08 \times 0.002) = 2.24 \times 10^{-5}~\mathrm{J} = 22.4~\mu\mathrm{J}$, so option (B) is correct.

    Concept & Intuition

    A thin film of water has two free surfaces (top and bottom), each contributing to the surface energy. When you pull the wires apart, you increase the area of both surfaces. The work required is simply the increase in surface energy, because the film is in equilibrium and the only resistance comes from surface tension. The key is remembering that a film has two surfaces — a common oversight.

    Step-by-step solution

    1. Identify the geometry

       The wires are parallel, each of length $L = 8~\text{cm} = 0.08~\text{m}$. The initial separation is $d_1 = 0.6~\text{cm} = 0.006~\text{m}$, and the final separation is $d_2 = 0.8~\text{cm} = 0.008~\text{m}$. The film is rectangular between the wires.

    2. Calculate the change in area of one surface

       The area of one surface of the film is $L \times \text{separation}$.

       Initial area: $A_1 = L \times d_1 = 0.08 \times 0.006 = 4.8 \times 10^{-4}~\text{m}^2$

       Final area: $A_2 = L \times d_2 = 0.08 \times 0.008 = 6.4 \times 10^{-4}~\text{m}^2$

       Increase in area for one surface: $\Delta A = A_2 - A_1 = 1.6 \times 10^{-4}~\text{m}^2$

    3. Account for both surfaces

       The water film has two surfaces (top and bottom). So the total increase in surface area is $2 \times \Delta A = 3.2 \times 10^{-4}~\text{m}^2$. …

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.A thin film of water is formed between two straight parallel wires each of length 8 cm separated by distance of 0.6 cm. The work done to increase the distance between the wires to 0.8 cm is (Surface tension of water = $0.07~\mathrm{Nm}^{-1}$) (A) $22.4~\mu \mathrm{J}$ (B) $33.6~\mu \mathrm{J}$ (C) $44.8~\mu \mathrm{J}$ (D) $11.2~\mu \mathrm{J}$

    ›Reveal solutionSolution

    The work done equals the change in surface energy of the film, which has two surfaces. The area increases as the wires are pulled apart, so work = surface tension × total area change = $0.07 \times 2 \times (0.08 \times 0.002) = 2.24 \times 10^{-5}~\mathrm{J} = 22.4~\mu\mathrm{J}$. The correct option is (A).

    Concept & Intuition

    A thin film of water between two wires has two free surfaces (top and bottom), each contributing to surface energy. Work done to stretch the film increases its surface area, and that work is stored as additional surface energy. The key formula: Work = Surface tension × Total change in surface area. Since the film has two surfaces, the total area change is twice the change in the area of one face.

    Step-by-step solution

    1. Identify the geometry

       The wires are parallel, each of length $L = 8~\text{cm} = 0.08~\text{m}$. Initial separation $d_1 = 0.6~\text{cm} = 0.006~\text{m}$, final separation $d_2 = 0.8~\text{cm} = 0.008~\text{m}$. The film is rectangular between the wires.

    2. Area of one surface

       Initial area of one surface: $A_1 = L \times d_1 = 0.08 \times 0.006 = 4.8 \times 10^{-4}~\text{m}^2$

       Final area of one surface: $A_2 = L \times d_2 = 0.08 \times 0.008 = 6.4 \times 10^{-4}~\text{m}^2$

       Change in area of one surface: $\Delta A_{\text{one}} = A_2 - A_1 = 1.6 \times 10^{-4}~\text{m}^2$

    3. Total area change (two surfaces)

       The film has two surfaces (top and bottom), so total area change: …

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