Q.Underline the correct alternative:

Showing the 12 most recent of 39 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A block is at rest at the top of an inclined plane of angle of inclination $30^\circ$. The coefficient of kinetic friction between the upper half surface of the plane and the block is $\dfrac{1}{2\sqrt{3}}$, and the coefficient of kinetic friction between the lower half surface of the plane and the block is $\dfrac{1}{4\sqrt{3}}$. When the block slides down, ratio of the velocities of the block at the midpoint of the plane and at the bottom of the plane is (A) $2:3$ (B) $\sqrt{2}:\sqrt{5}$ (C) $2:5$ (D) $\sqrt{2}:\sqrt{3}$

   ›Reveal solutionSolution

   The block slides down two halves of the incline with different friction. Using work–energy separately for each half gives the velocity ratio $\sqrt{2}:\sqrt{5}$, which is option (B).

   The problem gives a single incline split into two halves, each with a different coefficient of kinetic friction. The block starts from rest at the top. We want the ratio of its speed at the midpoint (end of the first half) to its speed at the bottom (end of the second half).

   Why use work–energy? Because the forces are constant along each half, and we care about speeds, not time. Work–energy directly connects net work done to change in kinetic energy, avoiding the need to solve for acceleration and time separately.

   Let the total length of the incline be $L$. Then the upper half has length $L/2$ with $\mu_1 = \frac{1}{2\sqrt{3}}$, and the lower half has length $L/2$ with $\mu_2 = \frac{1}{4\sqrt{3}}$. The angle of incline is $\theta = 30^\circ$.

   1. Forces and work on the upper half The block slides down, so the component of gravity along the incline is $mg\sin\theta$. Friction opposes motion, so kinetic friction force is $\mu_1 mg\cos\theta$. Net force along the incline:

   $F_{\text{net},1} = mg\sin\theta - \mu_1 mg\cos\theta$

   Work done by this net force over distance $L/2$ equals the gain in kinetic energy from rest to speed $v_1$ at the midpoint:

   $\frac{1}{2} m v_1^2 = \left( mg\sin\theta - \mu_1 mg\cos\theta \right) \frac{L}{2}$

   Cancel $m$ and multiply both sides by 2:

   $v_1^2 = gL \left( \sin\theta - \mu_1 \cos\theta \right)$

   2. Plug in numbers for the upper half $\sin 30^\circ = \frac12$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\mu_1 = \frac{1}{2\sqrt{3}}$.

   $\sin\theta - \mu_1 \cos\theta = \frac12 - \left( \frac{1}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \right) = \frac12 - \frac{1}{4} = \frac14$

   So

   $v_1^2 = gL \cdot \frac14$

   3. Forces and work on the lower half At the start of the lower half, the block already has speed $v_1$. The net force on the lower half uses $\mu_2 = \frac{1}{4\sqrt{3}}$:

   $F_{\text{net},2} = mg\sin\theta - \mu_2 mg\cos\theta$

   Work done over the lower half ($L/2$) adds to the existing kinetic energy:

   $\frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = \left( mg\sin\theta - \mu_2 mg\cos\theta \right) \frac{L}{2}$ …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The time taken by a body projected vertically upwards from the ground to reach 75% of the maximum height it can reach is 2 s. The ratio of the kinetic energy of the body at a time $t = 2\,\text{s}$ and the potential energy of the body at a time $t = 3\,\text{s}$ is (Acceleration due to gravity $= 10\,\text{ms}^{-2}$) (A) $2 : 15$ (B) $4 : 15$ (C) $2 : 5$ (D) $4 : 5$

   ›Reveal solutionSolution

   The key is to first find the initial velocity using the given time to reach 75% of max height, then compute kinetic energy at t=2 s and potential energy at t=3 s, and finally take their ratio. The ratio is 4:15, so option (B) is correct.

   Concept and Intuition

   When a body is projected vertically upward, its motion is symmetric under constant gravity. The maximum height is reached when velocity becomes zero. The time to reach a certain fraction of that height is not simply proportional to the time to reach the top — we must use the kinematic equations. Here, knowing the time to reach 75% of max height lets us solve for the initial speed. Then we can find the speed at t=2 s (for kinetic energy) and the height at t=3 s (for potential energy). The ratio of these energies simplifies nicely.

   Step-by-step solution

   1. Relate maximum height to initial velocity For vertical projection with initial speed $u$, the maximum height $H$ is given by

   $H = \frac{u^2}{2g}$

   where $g = 10 \, \text{m/s}^2$.

   2. Height at time $t$ The height at any time $t$ is

   $h(t) = u t - \frac{1}{2} g t^2$

   We are told that at $t = 2$ s, the body is at 75% of $H$:

   $h(2) = \frac{3}{4} H$

   Substitute:

   $u \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2 = \frac{3}{4} \cdot \frac{u^2}{2 \cdot 10}$

   Simplify:

   $2u - 20 = \frac{3}{4} \cdot \frac{u^2}{20} = \frac{3u^2}{80}$

   Multiply through by 80:

   $160u - 1600 = 3u^2$

   Rearrange:

   $3u^2 - 160u + 1600 = 0$

   3. Solve for $u$ Divide by 1 (or solve directly):

   $u = \frac{160 \pm \sqrt{160^2 - 4 \cdot 3 \cdot 1600}}{2 \cdot 3} = \frac{160 \pm \sqrt{25600 - 19200}}{6} = \frac{160 \pm \sqrt{6400}}{6}$

   $u = \frac{160 \pm 80}{6}$

   So $u = \frac{240}{6} = 40$ or $u = \frac{80}{6} \approx 13.33$.

   Which one is correct? At $t=2$ s, the body is still rising (since it hasn't reached the top yet). The time to reach maximum height is $T = u/g$. For $u=40$, $T=4$ s, so at 2 s it is indeed rising. For $u=13.33$, $T=1.33$ s, so at 2 s it would already be falling — but the problem says it reaches 75% of max height at 2 s, which can happen on the way up or down. However, the phrase "time taken ... to reach 75% of the maximum height" usually implies the first time it gets there (on the way up). So we take $u = 40 \, \text{m/s}$. …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The correct statement regarding fundamental forces in nature is (A) Electromagnetic force acts over large distances (B) Electromagnetic force needs an intervening medium (C) Strong nuclear force is a repulsive force (D) Strong nuclear force is a long-range force

   ›Reveal solutionSolution

   The key idea is to recall the range and nature of each fundamental force. Electromagnetic force obeys an inverse-square law and does not require a medium, so it acts over large distances — making option (A) correct.

   The four fundamental forces in nature are gravitational, electromagnetic, strong nuclear, and weak nuclear. Each has a distinct range, relative strength, and whether it requires a medium. The question tests your grasp of these properties, especially for electromagnetic and strong nuclear forces.

   Let’s examine each option carefully.

   1. Option (A): Electromagnetic force acts over large distances

      The electromagnetic force between charged particles follows an inverse-square law ($F \propto 1/r^2$), just like gravity. This means its influence extends to infinity, though it weakens with distance. It does not need a medium — it can act through vacuum (e.g., light from the Sun reaches Earth). So this statement is true.

   2. Option (B): Electromagnetic force needs an intervening medium

      This is false. Electromagnetic interactions are mediated by photons, which can travel through empty space. No material medium is required. This was a historical misconception (the "aether" idea) that has been disproven.

   3. Option (C): Strong nuclear force is a repulsive force …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A ball of mass 5 kg moving with a kinetic energy of 90 J collides head-on with another ball of mass 4 kg at rest. If the relative velocity of separation between the two balls after collision is $3\,\mathrm{ms}^{-1}$, then the loss of kinetic energy due to the collision is (A) 30 J (B) 60 J (C) 90 J (D) 45 J

   ›Reveal solutionSolution

   Using conservation of momentum and the given relative velocity of separation, we find the final velocities and compute the loss in kinetic energy. The loss is 30 J.

   The problem involves a head-on collision between two balls. You are given the masses, the initial kinetic energy of the moving ball, and the relative velocity of separation after the collision. The key is to combine conservation of momentum with the definition of relative velocity to find the final velocities, then compare the total kinetic energy before and after.

   Let’s break it down.

   1. Find the initial velocity of the 5 kg ball. Kinetic energy is $K = \frac{1}{2} m v^2$. For the 5 kg ball, $K = 90$ J.

   $\frac{1}{2} \times 5 \times v_1^2 = 90 \quad \Rightarrow \quad v_1^2 = \frac{180}{5} = 36 \quad \Rightarrow \quad v_1 = 6\ \mathrm{m/s}$

   The 4 kg ball is at rest, so its initial velocity $v_2 = 0$.

   2. Apply conservation of linear momentum. Let the velocities after collision be $v_1'$ (5 kg ball) and $v_2'$ (4 kg ball), both in the same direction as the initial motion (say positive). Initial momentum: $5 \times 6 + 4 \times 0 = 30\ \mathrm{kg\,m/s}$. Final momentum: $5 v_1' + 4 v_2'$.

   $5 v_1' + 4 v_2' = 30 \qquad(1)$

   3. Use the relative velocity of separation. The relative velocity of separation is given as $3\ \mathrm{m/s}$. For a head-on collision, separation velocity = $v_2' - v_1'$ (since both move forward, the faster one ahead).

   $v_2' - v_1' = 3 \qquad(2)$

   4. Solve for $v_1'$ and $v_2'$. From (2): $v_2' = v_1' + 3$. Substitute into (1):

   $5 v_1' + 4(v_1' + 3) = 30 \quad \Rightarrow \quad 5 v_1' + 4 v_1' + 12 = 30 \quad \Rightarrow \quad 9 v_1' = 18 \quad \Rightarrow \quad v_1' = 2\ \mathrm{m/s}$

   Then $v_2' = 2 + 3 = 5\ \mathrm{m/s}$.

   5. Compute the kinetic energy after collision. …

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.When an electron and a positron at rest annihilate each other, the momentum of each of the photons emitted is (Rest mass of electron $= 9 \times 10^{-31}$ kg) (A) $36 \times 10^{-23}~\mathrm{kg\,ms^{-1}}$ (B) $9 \times 10^{-23}~\mathrm{kg\,ms^{-1}}$ (C) $27 \times 10^{-23}~\mathrm{kg\,ms^{-1}}$ (D) $18 \times 10^{-23}~\mathrm{kg\,ms^{-1}}$

   ›Reveal solutionSolution

   Each photon carries energy $m_ec^2$, so its momentum is $p=m_ec = 27\times10^{-23}\ \text{kg m s}^{-1}$ — option (C).

   Setup

   An electron and positron at rest annihilate into two photons. The total initial momentum is zero, so by momentum conservation the two photons fly off in opposite directions with equal and opposite momenta. By symmetry they share the total energy equally.

   Step 1 — Energy of each photon

   Total rest energy released:

   $E_{\text{tot}} = 2m_ec^2.$

   Shared equally, each photon has

   $E_\gamma = m_ec^2.$

   Step 2 — Photon momentum

   For a photon $E = pc$, so …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If a capacitor is charged by connecting it to a battery, then the ratio of work done by the battery and the energy stored in the capacitor is (A) $3:2$ (B) $1:2$ (C) $1:1$ (D) $2:1$

   ›Reveal solutionSolution

   When a battery charges a capacitor, the battery does twice the work that ends up stored as electrostatic energy; the missing half is lost as heat in the circuit resistance. The ratio of work done by the battery to energy stored is therefore 2:1, which corresponds to option (D).

   ---

   Why this ratio is not 1:1

   A common first guess is that all the work done by the battery goes into the capacitor’s stored energy — after all, the capacitor ends up with charge $Q$ and voltage $V$. But that would violate energy conservation if we think carefully: the battery must also push charge through any resistance in the circuit, and that dissipates energy as heat. Even if the resistance is tiny, the process is not reversible; the capacitor’s final state is the same regardless of resistance, but the battery’s total work depends on the path.

   The key insight: The battery supplies a constant voltage $V$ and delivers a total charge $Q$. The work done by the battery is simply $W_{\text{battery}} = QV$. Meanwhile, the energy stored in the capacitor is $\frac{1}{2} QV$. The factor of $\frac{1}{2}$ appears because the capacitor’s voltage builds from 0 to $V$ as charge accumulates, so the average voltage during charging is $V/2$.

   ---

   Step-by-step reasoning

   1. Set up the charging process

      A battery of emf $V$ is connected to an initially uncharged capacitor of capacitance $C$ through some resistance (even if small). The battery maintains a constant potential difference $V$ across its terminals.

   2. Work done by the battery

      The battery pushes a total charge $Q = CV$ from its negative to its positive terminal. Since the potential difference is constant $V$, the work done is

   $W_{\text{battery}} = Q \cdot V = CV^2.$

   3. Energy stored in the capacitor The energy stored in a capacitor is given by

   $U_{\text{cap}} = \frac{1}{2} C V^2 = \frac{1}{2} Q V.$

   This is the area under the $Q$-$V$ graph for the capacitor (a triangle), whereas the battery’s work is the area of the rectangle $Q \times V$.

   4. Find the ratio

   $\frac{W_{\text{battery}}}{U_{\text{cap}}} = \frac{CV^2}{\frac{1}{2} CV^2} = \frac{2}{1}.$

   So the ratio is $2:1$.

   5. Where does the extra energy go? …

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7. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A body of mass $4\,\text{kg}$ moving with a velocity of $12\,\text{ms}^{-1}$ collides head-on with a stationary body of mass $2\,\text{kg}$. If the relative velocity of separation of the two bodies after collision is $6\,\text{ms}^{-1}$, then the percentage loss of kinetic energy of the body of mass $4\,\text{kg}$ is (A) 75 (B) 25 (C) 15 (D) 50

   ›Reveal solutionSolution

   Using momentum conservation with the given separation speed, the $4\,\text{kg}$ body slows from $12$ to $6\ \text{m/s}$, losing $75\%$ of its kinetic energy — option (A).

   Given: $m_1=4\,\text{kg}$, $u_1=12\ \text{m/s}$; $m_2=2\,\text{kg}$, $u_2=0$; relative velocity of separation $v_2-v_1=6\ \text{m/s}$.

   Conservation of momentum:

   $m_1u_1+m_2u_2=m_1v_1+m_2v_2 \implies 48 = 4v_1+2v_2 \implies 2v_1+v_2=24.$

   Separation condition: $v_2-v_1=6$, so $v_2=v_1+6$. Substituting: …

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The work done in blowing a soap bubble of diameter 3 cm is (Surface tension of soap solution = $0.035~\mathrm{Nm}^{-1}$) (A) $792~\mu \mathrm{J}$ (B) $99~\mu \mathrm{J}$ (C) $396~\mu \mathrm{J}$ (D) $198~\mu \mathrm{J}$

   ›Reveal solutionSolution

   The work done equals the increase in surface energy, which is surface tension times the total increase in surface area. A soap bubble has two surfaces (inner and outer), so the total area is $2 \times 4\pi r^2 = 8\pi r^2$. With diameter 3 cm (radius 1.5 cm) and surface tension $0.035~\mathrm{N/m}$, the work is $198~\mu\mathrm{J}$, corresponding to option (D).

   ---

   The key idea is that work done in forming a bubble is stored as surface energy. A soap bubble has two surfaces — an inner and an outer film — each contributing to the total surface area. The work required is the surface tension multiplied by the total increase in surface area (from essentially zero to the final bubble).

   Why this works: Surface tension $\sigma$ is the energy per unit area. So if you create a new surface of area $A$, the energy cost is $\sigma A$. For a bubble blown from a tiny film (negligible initial area), the work done is $\sigma \times (\text{final total surface area})$.

   ---

   Step-by-step solution

   1. Find the radius from the diameter

      Diameter = 3 cm = $0.03~\mathrm{m}$, so radius $r = \dfrac{0.03}{2} = 0.015~\mathrm{m}$.

   2. Surface area of one spherical surface

      Area of a sphere = $4\pi r^2$.

   3. Total surface area of the bubble

      A soap bubble has two surfaces (inner and outer), so the total area is

   $A_{\text{total}} = 2 \times 4\pi r^2 = 8\pi r^2.$

   4. Work done = surface tension × total area

   $W = \sigma \times 8\pi r^2 = 0.035 \times 8\pi (0.015)^2.$

   5. Evaluate $$$$

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Due to global warming, if the ice in the polar region melts and some of this water flows to the equatorial region, then (A) Angular momentum of the earth increases and duration of day increases (B) Angular momentum of the earth decreases and duration of day decreases (C) Angular momentum of the earth is constant and duration of day decreases (D) Angular momentum of the earth is constant and duration of day increases

   ›Reveal solutionSolution

   The Earth’s angular momentum is conserved (no external torque), but redistributing mass toward the equator increases the moment of inertia, which decreases the angular speed, lengthening the day. The correct option is (D).

   ---

   Concept & Intuition

   The key idea is conservation of angular momentum. For the Earth–water system, no external torque acts (ignoring tidal friction for this problem). When polar ice melts and the water moves toward the equator, mass is shifted farther from the Earth’s rotation axis. This increases the Earth’s moment of inertia $I$. Since angular momentum $L = I\omega$ is constant, an increase in $I$ forces a decrease in angular speed $\omega$. A slower spin means a longer day.

   ---

   Step-by-step reasoning

   1. Identify the system and external torques The Earth plus the melted water is an isolated system for rotation (neglecting gravitational interactions with the Moon and Sun). No external torque acts about the Earth’s spin axis, so angular momentum is conserved:

   $L_{\text{initial}} = L_{\text{final}}.$

   2. Write angular momentum in terms of moment of inertia and angular speed

   $L = I \omega,$

   where $\omega = \frac{2\pi}{T}$ and $T$ is the duration of a day. Conservation gives:

   $I_i \omega_i = I_f \omega_f.$

   3. Analyze the change in moment of inertia
      • Initially, ice is near the poles (close to the rotation axis), contributing little to $I$.
      • After melting, the water spreads toward the equator (far from the axis).
      • Moment of inertia depends on $r^2$ (distance from axis). Moving mass outward increases $I$:

   $I_f > I_i.$

   4. Determine the effect on angular speed and day length From conservation: $\omega_f = \frac{I_i}{I_f} \omega_i.$ …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.A solid sphere of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal surface. The ratio of the rotational and translational kinetic energies of the sphere is (A) $3:5$ (B) $2:5$ (C) $4:5$ (D) $7:5$

    ›Reveal solutionSolution

    For a solid sphere rolling without slipping, the rotational kinetic energy is $\frac{2}{5}$ of the translational kinetic energy, so the ratio is $2:5$, which corresponds to option (B).

    The key here is that rolling without slipping couples the rotational and translational motion through the condition $v = \omega R$. The kinetic energy splits into two parts: translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$). For a solid sphere, the moment of inertia is $I = \frac{2}{5}mR^2$. Once you substitute the rolling condition, the ratio emerges directly from these constants — no extra physics needed.

    1. Write the two kinetic energy expressions.

       Translational: $K_{\text{trans}} = \frac{1}{2} m v^2$

       Rotational: $K_{\text{rot}} = \frac{1}{2} I \omega^2$

    2. Insert the moment of inertia for a solid sphere.

       $I = \frac{2}{5} m R^2$, so

       $K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{5} m R^2 \right) \omega^2 = \frac{1}{5} m R^2 \omega^2$

    3. Use the rolling condition $v = \omega R$ to eliminate $\omega$.

       Replace $\omega$ with $v/R$:

       $K_{\text{rot}} = \frac{1}{5} m R^2 \left( \frac{v}{R} \right)^2 = \frac{1}{5} m v^2$

    4. Form the ratio. …

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.During the disintegration of a radioactive nucleus of mass number 208 at rest, two alpha particles each with kinetic energy E are emitted. The total kinetic energy of the emitted alpha particles and the daughter nucleus after the disintegration is (A) $\dfrac{51E}{50}$ (B) $\dfrac{26E}{25}$ (C) $\dfrac{52E}{25}$ (D) $\dfrac{51E}{25}$

    ›Reveal solutionSolution

    The parent (mass number $208$) leaves a daughter of mass number $200$. The two $\alpha$-particles ($4$ u each) carry momentum $p$ in the same direction, so the daughter recoils with momentum $2p$ and kinetic energy $\frac{2E}{25}$; total $= 2E + \frac{2E}{25} = \frac{52E}{25}$.

    Setting up

    Mass numbers: parent $=208$, each $\alpha =4$, so the daughter $=208-2(4)=200$. Taking mass $\propto$ mass number, $m_\alpha : m_d = 4:200 = 1:50$, i.e. $m_d = 50\,m_\alpha$.

    Each $\alpha$ has kinetic energy $E$, so its momentum satisfies

    $p^2 = 2\,m_\alpha E.$

    Momentum conservation

    The parent is at rest, so the total momentum after disintegration is zero. The two identical $\alpha$-particles leave with equal momentum $p$ in the same direction, together carrying momentum $2p$; the daughter must recoil with equal and opposite momentum: …

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12. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.In the $\beta^+$ decay, the particle emitted along with neutron and positron is (A) Electron (B) Proton (C) Neutrino (D) Anti-neutrino

    ›Reveal solutionSolution

    In $\beta^+$ decay, a proton transforms into a neutron, a positron, and a neutrino — the missing particle is the neutrino, option (C).

    The key to this question is understanding what conservation laws demand in a nuclear decay. In any radioactive process, charge, mass-energy, and lepton number must all be conserved. Lepton number is a quantum number: leptons (like electrons, positrons, neutrinos) have lepton number $+1$, antileptons have $-1$, and all other particles have $0$. This is the hidden rule that tells us exactly which particle is emitted alongside the positron.

    In $\beta^+$ decay, a proton-rich nucleus converts a proton into a neutron. The reaction inside the nucleus is:

    $p \rightarrow n + e^+ + \nu$

    Here $e^+$ is the positron (the antiparticle of the electron), and $\nu$ is a neutrino. The positron carries away positive charge, and the neutrino carries away energy and lepton number — without it, lepton number would not balance.

    Let’s walk through the reasoning step by step.

    1. Identify the decay process. In $\beta^+$ decay, a proton in the nucleus changes into a neutron. The atomic number $Z$ decreases by 1, while the mass number $A$ stays the same. The emitted particles are a positron ($e^+$) and a neutral, nearly massless particle.

    2. Check charge conservation. The proton has charge $+1$. The neutron is neutral ($0$), and the positron has charge $+1$. So charge is already balanced: $+1 \rightarrow 0 + (+1)$. No other charged particle is needed.

    3. Check lepton number conservation. This is the crucial step. The proton and neutron are not leptons — they have lepton number $0$. The positron is an antilepton, so it has lepton number $-1$. To conserve lepton number, the other emitted particle must have lepton number $+1$ (so that $0 \rightarrow -1 + (+1) = 0$). The only particle that fits is a neutrino ($\nu$), which is a lepton with lepton number $+1$.

    4. Eliminate the wrong options.

       • (A) Electron: An electron has lepton number $+1$, but it is negatively charged. Adding an electron would give total charge $+1 + (-1) = 0$, which doesn’t match the proton’s $+1$ charge. Also, $\beta^+$ decay emits a positron, not an electron.
       • (B) Proton: Emitting a proton would change the nucleus in a completely different way (like proton emission), and it would not conserve lepton number or match the known decay mode. …

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