Q.Consider a coin of Question 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8\text{ kC}$. Suppose that these equal charges were concentrated in two point charges separated by

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1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.An electron is moving in a stable circular orbit of radius 0.1 m around a thin infinitely long positively charged straight wire. If orbital velocity of the electron around the wire is $4 \times 10^7~\mathrm{ms}^{-1}$, then linear charge density of the wire is nearly (A) $4.5 \times 10^{-7}~\mathrm{Cm}^{-1}$ (B) $9 \times 10^{-7}~\mathrm{Cm}^{-1}$ (C) $5 \times 10^{-7}~\mathrm{Cm}^{-1}$ (D) $2.5 \times 10^{-7}~\mathrm{Cm}^{-1}$

   ›Reveal solutionSolution

   The wire's field supplies the centripetal force: $\lambda=\dfrac{2\pi\varepsilon_0 m v^2}{e}\approx5\times10^{-7}\,\mathrm{Cm}^{-1}$.

   The radial field of an infinite line charge is $E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$. This provides the centripetal force on the electron:

   $eE=\frac{mv^2}{r}\;\Rightarrow\;\frac{e\lambda}{2\pi\varepsilon_0 r}=\frac{mv^2}{r}$

   The radius cancels:

   $\lambda=\frac{2\pi\varepsilon_0\, m v^2}{e}$ …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Two metallic spherical shells A and B of radii 3 cm and 4 cm are given electric charges 20 $\mu$C and 40 $\mu$C respectively. If the shells are arranged concentrically, then the ratio of the surface charge densities on the outer surfaces of the shells A and B is (A) $4 : 27$ (B) $4 : 3$ (C) $16 : 9$ (D) $16 : 27$

   ›Reveal solutionSolution

   When two charged spherical shells are arranged concentrically, the charge on the outer surface of the inner shell is its own charge, while the charge on the outer surface of the outer shell is the sum of the charges of both shells. The ratio of their surface charge densities is $\boxed{16:27}$.

   Concept and Intuition

   Surface charge density, denoted by $\sigma$ (sigma), is defined as the amount of electric charge per unit surface area. For a spherical conductor with charge $Q$ and radius $R$, its surface area is $4\pi R^2$, so the surface charge density is $\sigma = \frac{Q}{4\pi R^2}$.

   When we have two concentric conducting spherical shells, the distribution of charges follows specific rules due to electrostatic induction:

   1. Inner Shell (A): Since shell A is a conductor, any charge $Q_A$ given to it will reside entirely on its outer surface. This is because charges on a conductor repel each other and try to maximize their separation, pushing to the outermost boundary available to them.
   2. Outer Shell (B):
      • The charge $Q_A$ on the outer surface of shell A will induce an equal and opposite charge, $-Q_A$, on the inner surface of shell B. This happens to maintain zero electric field inside the material of the outer conductor (shell B).
      • Since shell B was initially given a total charge $Q_B$, and $-Q_A$ has appeared on its inner surface, the remaining charge must reside on its outer surface. This remaining charge is $Q_B - (-Q_A) = Q_A + Q_B$. This total charge $Q_A + Q_B$ will spread uniformly over the outer surface of shell B.

   Therefore, to find the ratio of surface charge densities on the outer surfaces of shells A and B, we need to use $Q_A$ for shell A and $Q_A + Q_B$ for shell B.

   Step-by-step Derivation

   1. Identify the given parameters:

      • Radius of shell A, $r_A = 3 \text{ cm}$
      • Charge on shell A, $Q_A = 20 \text{ } \mu\text{C}$
      • Radius of shell B, $r_B = 4 \text{ cm}$
      • Charge on shell B, $Q_B = 40 \text{ } \mu\text{C}$

   2. Determine the charge on the outer surface of shell A:

      As explained in the concept, the entire charge $Q_A$ given to shell A resides on its outer surface.

      Charge on outer surface of A $= Q_A = 20 \text{ } \mu\text{C}$.

   3. Determine the charge on the outer surface of shell B:

      Due to electrostatic induction from $Q_A$ on shell A, a charge of $-Q_A$ is induced on the inner surface of shell B. Since the total charge given to shell B is $Q_B$, the charge remaining for its outer surface is $Q_B - (-Q_A) = Q_A + Q_B$.

      Charge on outer surface of B $= Q_A + Q_B = 20 \text{ } \mu\text{C} + 40 \text{ } \mu\text{C} = 60 \text{ } \mu\text{C}$.

      Watch out

      A common mistake is to assume the charge on the outer surface of shell B is simply $Q_B$. Remember that the inner shell's charge induces an equal and opposite charge on the inner surface of the outer shell, pushing the net charge to the outer surface of the outer shell.

   4. Recall the formula for surface charge density:

      The surface charge density $\sigma$ on a sphere with charge $Q$ and radius $R$ is given by:

      $\sigma = \frac{Q}{4\pi R^2}$ …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The fundamental force that plays a key role in the large scale phenomena of the universe is (A) electromagnetic force (B) strong nuclear force (C) weak nuclear force (D) gravitational force

   ›Reveal solutionSolution

   The large-scale structure of the universe — galaxies, clusters, and cosmic motion — is governed by gravity, which acts over astronomical distances and dominates the dynamics of massive bodies. The correct answer is (D) gravitational force.

   The question asks which fundamental force is responsible for the large-scale phenomena of the universe. To answer this, you need to think about the range and strength of each fundamental force, and how they manifest on cosmic scales.

   The four fundamental forces are: gravitational, electromagnetic, strong nuclear, and weak nuclear. Each has a very different role.

   1. Strong nuclear force — This is the strongest force, but its range is tiny, about the size of an atomic nucleus ($\approx 10^{-15}$ m). It holds protons and neutrons together inside the nucleus. It simply cannot act over distances larger than a few femtometres, so it plays no role in large-scale phenomena.

   2. Weak nuclear force — Even shorter-ranged than the strong force ($\approx 10^{-18}$ m). It is responsible for certain types of radioactive decay (like beta decay). Again, completely irrelevant on cosmic scales.

   3. Electromagnetic force — This force has infinite range, like gravity, and is much stronger (about $10^{36}$ times stronger than gravity between two protons). However, on large scales, matter is electrically neutral overall — positive and negative charges cancel out. So the net electromagnetic force between astronomical bodies (like stars or galaxies) is essentially zero. Electromagnetism dominates at atomic and molecular scales, but not at the scale of the universe. …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If two bodies A and B of masses 5 kg and 10 kg are thrown vertically upwards from the surface of the earth with velocities $\sqrt{0.5gR}$ and $\sqrt{0.2gR}$ respectively, then the ratio of maximum heights reached by the bodies A and B is (R - Radius of the earth) (A) $3 : 2$ (B) $5 : 2$ (C) $3 : 1$ (D) $2 : 1$

   ›Reveal solutionSolution

   The key idea is that for large velocities (comparable to escape velocity), the maximum height must be found using energy conservation with variable gravity, not constant-g kinematics. The ratio of maximum heights for the two bodies is $5:2$, so option (B) is correct.

   ---

   Concept and Intuition

   When a body is thrown upward with a speed that is a significant fraction of the escape velocity ($\sqrt{2gR}$), the gravitational acceleration is not constant over the ascent. The usual kinematic formula $h = u^2/(2g)$ assumes constant $g$, which fails here because $g$ decreases with height. Instead, we must use conservation of mechanical energy, accounting for the variation of gravitational potential energy with distance from Earth's center. The escape velocity from Earth's surface is $v_e = \sqrt{2gR}$. Both given velocities are comparable to this, so constant-$g$ kinematics would give a wrong answer.

   ---

   Step-by-Step Solution

   1. Write the energy conservation equation At the surface (radius $R$), the body has kinetic energy $\frac{1}{2} m u^2$ and potential energy $-\frac{GMm}{R}$. At the maximum height $h$, its velocity is zero, so only potential energy remains: $-\frac{GMm}{R+h}$. Energy conservation gives:

   $\frac{1}{2} m u^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$

   2. Simplify using $g = GM/R^2$ Substitute $GM = gR^2$:

   $\frac{1}{2} u^2 - \frac{gR^2}{R} = -\frac{gR^2}{R+h}$

   $\frac{1}{2} u^2 - gR = -\frac{gR^2}{R+h}$

   3. Solve for $h$ Rearranging:

   $\frac{gR^2}{R+h} = gR - \frac{1}{2} u^2$

   $R+h = \frac{gR^2}{gR - \frac{1}{2} u^2}$

   $h = \frac{gR^2}{gR - \frac{1}{2} u^2} - R$

   $h = R \left( \frac{gR}{gR - \frac{1}{2} u^2} - 1 \right)$

   $h = R \left( \frac{\frac{1}{2} u^2}{gR - \frac{1}{2} u^2} \right)$

   4. Introduce the ratio $u^2/(gR)$ Let $k = \frac{u^2}{gR}$. Then:

   $h = R \left( \frac{\frac{1}{2} k gR}{gR - \frac{1}{2} k gR} \right) = R \left( \frac{\frac{1}{2} k}{1 - \frac{1}{2} k} \right)$

   So:

   $h = R \cdot \frac{k}{2 - k}$

   5. Apply to each body …

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5. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A cyclotron with dees of radius 50 cm and a magnetic field of 1.5 T is used to accelerate protons and alpha particles separately. The ratio of the maximum kinetic energies acquired by the proton and alpha particle is (A) $1:4$ (B) $1:1$ (C) $1:2$ (D) $1:8$

   ›Reveal solutionSolution

   The maximum kinetic energy a charged particle acquires in a cyclotron depends on its charge-to-mass ratio, the magnetic field, and the cyclotron's radius. For a proton and an alpha particle, the ratio of their maximum kinetic energies in the same cyclotron is $\boxed{1:1}$.

   A cyclotron accelerates charged particles by making them move in a spiral path under a constant magnetic field, while an oscillating electric field provides energy boosts. The magnetic field forces the particles into circular paths, and the electric field between the dees accelerates them each time they cross the gap.

   The particle gains kinetic energy with each acceleration, increasing its speed and thus the radius of its circular path. This process continues until the particle reaches the maximum radius of the dees, at which point it exits the cyclotron. The kinetic energy at this maximum radius is the maximum kinetic energy the particle can acquire.

   To find this maximum kinetic energy, we use the principle that the magnetic force provides the necessary centripetal force for the circular motion.

   1. Derive the formula for maximum kinetic energy: When a charged particle of charge $q$ and mass $m$ moves with velocity $v$ in a magnetic field $B$ perpendicular to its velocity, the magnetic force $F_B = qvB$ acts as the centripetal force $F_c = \frac{mv^2}{r}$. Equating these forces:

   $qvB = \frac{mv^2}{r}$

   From this, we can find the velocity $v$ of the particle at a given radius $r$:
   

   $v = \frac{qBr}{m}$

   The maximum velocity $v_{max}$ is achieved when the particle reaches the maximum radius $R$ of the dees:
   

   $v_{max} = \frac{qBR}{m}$

   The maximum kinetic energy $K_{max}$ is then given by:
   

   $K_{max} = \frac{1}{2}mv_{max}^2$

   Substitute the expression for $v_{max}$:
   

   $K_{max} = \frac{1}{2}m \left(\frac{qBR}{m}\right)^2 = \frac{1}{2}m \frac{q^2B^2R^2}{m^2}$

   > [!FORMULA]
   > The maximum kinetic energy acquired by a particle in a cyclotron is:
   > $$K_{max} = \frac{q^2B^2R^2}{2m}$$
   > Here, $q$ is the charge of the particle, $m$ is its mass, $B$ is the magnetic field strength, and $R$ is the radius of the dees.
   
   Notice that the given values for the radius ($50$ cm) and magnetic field ($1.5$ T) are constant for both particles and will cancel out when we take the ratio.
   

   2. Identify properties of a proton:

   A proton has a charge $q_p = e$ (where $e$ is the elementary charge) and a mass $m_p = m$. …

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6. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The reason for the constancy of binding energy per nucleon of nuclei whose mass number lies between 30 and 170 is (A) Short range of nuclear forces (B) The nuclear force between neutron-neutron is greater than the nuclear force between proton-proton (C) Nuclear forces are weak forces (D) Binding energy per nucleon is lower for these nuclei

   ›Reveal solutionSolution

   The binding energy per nucleon is nearly constant for nuclei with mass number 30–170 because nuclear forces are short-range, so each nucleon only interacts with its immediate neighbours, making the total binding energy roughly proportional to the number of nucleons.

   The key to this question is understanding why the binding energy per nucleon doesn't keep rising as nuclei get bigger. If every nucleon attracted every other nucleon equally, the binding energy would grow roughly like $A^2$ (the number of pairs), and the binding energy per nucleon would keep increasing with $A$. That doesn't happen — it flattens out. The reason is that nuclear forces have a very short range, about the size of a few nucleon diameters.

   1. The saturation of nuclear forces

      A nucleon inside a medium-sized or large nucleus only feels the strong force from its nearest neighbours — nucleons that are, say, 2–3 fm away. Nucleons on the opposite side of the nucleus are too far to contribute any attraction. This is completely different from gravity or electromagnetism, which are long-range and act between every pair.

   2. What this means for binding energy

      Because each nucleon only bonds with a fixed number of nearby nucleons (roughly 12–15), the total binding energy of the nucleus is proportional to the number of nucleons $A$, not to $A^2$. So the binding energy per nucleon becomes roughly constant — about 8 MeV per nucleon — for nuclei where the surface effect is small compared to the volume effect. That happens for $A$ between about 30 and 170.

   3. Why the other options are wrong

      • (B) is false: the nuclear force between a neutron and a proton is actually the strongest of the three pairs (n-p, n-n, p-p), and even if it weren't, that wouldn't explain the constancy of binding energy per nucleon.
      • (C) is false: nuclear forces are the strongest known forces, not weak forces. The weak nuclear force is a different interaction responsible for beta decay. …

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7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Two spherical shells of radii $R$ and $2R$, masses $M$ and $2M$ respectively are arranged concentrically. The net gravitational force acting on a particle of mass '$m$' placed at a distance of $\dfrac{3R}{2}$ from the common centre of the shells is (A) $\dfrac{4GMm}{3R^2}$ (B) $\dfrac{76GMm}{9R^2}$ (C) $\dfrac{4GMm}{9R^2}$ (D) $\dfrac{68GMm}{9R^2}$

   ›Reveal solutionSolution

   The gravitational force on a particle inside a spherical shell is zero, while outside it acts as if all mass is at the center. For the given setup, only the inner shell contributes to the force, resulting in $\boxed{\dfrac{4GMm}{9R^2}}$.

   The problem asks for the net gravitational force on a particle placed at a specific distance from the common center of two concentric spherical shells. To solve this, we need to understand how gravitational force is exerted by a spherical shell on a point mass, which is described by the Shell Theorem.

   Concept and Intuition: The Shell Theorem

   The Shell Theorem is a fundamental result in gravitation that simplifies calculating the gravitational force due to spherical shells. It states two key principles:

   1. For a point outside a spherical shell: The gravitational force exerted by the shell on a particle outside it is the same as if all the mass of the shell were concentrated at its center.
   2. For a point inside a spherical shell: The net gravitational force exerted by the shell on a particle located anywhere inside it is zero. This is because the gravitational pulls from different parts of the shell cancel each other out perfectly.

   When dealing with multiple concentric shells, we apply this theorem to each shell individually and then sum the forces vectorially. Since gravitational force is always attractive and directed towards the center of mass (or the center of the shell in this symmetric case), the forces will be collinear, simplifying the vector sum to an algebraic sum of magnitudes.

   Let's apply this to the given problem.

   Step-by-step Derivation:

   1. Identify the setup:

      • Inner spherical shell: Radius $R$, Mass $M$.
      • Outer spherical shell: Radius $2R$, Mass $2M$.
      • Particle: Mass $m$, placed at a distance $r = \dfrac{3R}{2}$ from the common center.

   2. Analyze the force due to the inner shell:

      The inner shell has radius $R$ and mass $M$. The particle is placed at a distance $r = \dfrac{3R}{2}$ from the center.

      Since $r = \dfrac{3R}{2} = 1.5R$, and $1.5R > R$, the particle is outside the inner spherical shell.

      According to the Shell Theorem, the inner shell behaves as if all its mass $M$ is concentrated at the common center.

      The gravitational force exerted by the inner shell on the particle is given by Newton's Law of Universal Gravitation:

   $F_1 = \frac{GMm}{r^2}$

   Substituting $r = \dfrac{3R}{2}$:
   

   $F_1 = \frac{GMm}{\left(\dfrac{3R}{2}\right)^2} = \frac{GMm}{\dfrac{9R^2}{4}} = \frac{4GMm}{9R^2}$

   This force is directed towards the common center.
   

   3. Analyze the force due to the outer shell: …

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8. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Two charged particles enter a uniform magnetic field normally. If the ratio of the specific charges of the two particles is $2:3$, then the ratio of the times taken by the two particles to complete one revolution is (A) $1:1$ (B) $3:2$ (C) $9:4$ (D) $\sqrt{3}:\sqrt{2}$

   ›Reveal solutionSolution

   The time period for circular motion in a uniform magnetic field depends only on the mass-to-charge ratio, not on speed. Given the specific charge ratio $2:3$, the time period ratio is the inverse, i.e., $3:2$.

   The key concept here is the cyclotron motion of a charged particle in a uniform magnetic field. When a particle of charge $q$ and mass $m$ enters a uniform magnetic field $\mathbf{B}$ perpendicular to its velocity, the magnetic force provides the centripetal force. This force is always perpendicular to velocity, so it changes only the direction, not the speed. The result is uniform circular motion.

   The time taken to complete one full revolution — the time period $T$ — is independent of the particle’s speed. Why? Because a faster particle moves in a larger circle (larger radius) but covers the larger circumference in exactly the same time. Let’s derive that.

   1. Set up the force equation. The magnetic force is $F = qvB$ (since $\mathbf{v} \perp \mathbf{B}$). This equals the centripetal force $mv^2 / r$. So:

   $qvB = \frac{mv^2}{r}$

   Cancel $v$ (non-zero) to get:

   $qB = \frac{mv}{r} \quad \Rightarrow \quad r = \frac{mv}{qB}$

   2. Find the time period. The circumference of the circle is $2\pi r$. The particle’s speed $v$ is constant, so:

   $T = \frac{\text{circumference}}{\text{speed}} = \frac{2\pi r}{v}$

   Substitute $r$ from above:

   $T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$

   $T = \frac{2\pi m}{qB}$

   The time period depends only on mass $m$, charge $q$, and magnetic field $B$ — not on speed or radius.

   3. Interpret the given ratio. The problem gives the ratio of specific charges, which is charge per unit mass: $q/m$. Let’s denote: …

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9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.A body weighs the same on the surfaces of two planets of densities '$\rho_1$' and '$\rho_2$.' The ratio of the radii of the planets is (A) $\dfrac{\rho_2}{\rho_1}$ (B) $\dfrac{\rho_2^2}{\rho_1^2}$ (C) $\dfrac{\rho_2^3}{\rho_1^3}$ (D) $\dfrac{\rho_2^4}{\rho_1^4}$

   ›Reveal solutionSolution

   Weight equality on two planets means their surface gravitational accelerations are equal. Using $g = \frac{4}{3}\pi G \rho R$, the ratio of radii is $\frac{R_1}{R_2} = \frac{\rho_2}{\rho_1}$, so the correct option is (A).

   The key idea is that a body’s weight on a planet’s surface is $mg$, where $g$ is the acceleration due to gravity at that surface. If the weight is the same on both planets, then $g$ must be the same on both — the mass $m$ of the body is unchanged. So the problem reduces to: given two planets of different densities, find the ratio of their radii such that their surface $g$ values are equal.

   For a spherical planet of mass $M$ and radius $R$, the surface gravity is $g = \frac{GM}{R^2}$. But $M$ is not directly given — we know the density $\rho$. Since the planet is a sphere, $M = \rho \cdot \frac{4}{3}\pi R^3$. Substituting this into the expression for $g$ gives a clean relation that eliminates $M$ and ties $g$ directly to $\rho$ and $R$.

   1. Write the surface gravity in terms of density and radius:

   $g = \frac{G}{R^2} \cdot \left( \frac{4}{3}\pi R^3 \rho \right) = \frac{4}{3}\pi G \rho R.$

   Notice that $g$ is proportional to the product $\rho R$ — a larger radius or a larger density both increase surface gravity linearly.

   2. For planet 1 (density $\rho_1$, radius $R_1$) and planet 2 (density $\rho_2$, radius $R_2$), the condition that the body weighs the same on both surfaces is:

   $g_1 = g_2 \quad \Rightarrow \quad \frac{4}{3}\pi G \rho_1 R_1 = \frac{4}{3}\pi G \rho_2 R_2.$

   3. Cancel the common factor $\frac{4}{3}\pi G$ (which is the same for both planets) to get: $\rho_1 R_1 = \rho_2 R_2.$ …

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10. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.Two particles of charges in the ratio 1:2 and masses in the ratio 2:3 moving along a straight line enter a uniform magnetic field at right angles to the direction of the field. If the radii of the circular paths of the particles in the magnetic field are in the ratio 3:4, then the ratio of the initial linear momenta of the two particles is (A) 1:1 (B) 3:4 (C) 3:8 (D) 2:3

    ›Reveal solutionSolution

    The radius of a charged particle’s circular path in a uniform magnetic field is proportional to its linear momentum divided by its charge. Using the given ratios, the momentum ratio is found to be 3:8, which corresponds to option (C).

    Concept & Intuition

    When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circle. The radius $r$ of this circle is given by

    $r = \frac{mv}{qB} = \frac{p}{qB},$

    where $p = mv$ is the linear momentum, $q$ is the charge, and $B$ is the magnetic field strength (same for both particles). So, for a fixed $B$, the radius is directly proportional to momentum and inversely proportional to charge. This relation lets us connect the given ratios.

    Step-by-step solution

    1. Write the radius formula for each particle For particle 1: $r_1 = \frac{p_1}{q_1 B}$ For particle 2: $r_2 = \frac{p_2}{q_2 B}$ Since $B$ is the same, dividing the two equations gives

    $\frac{r_1}{r_2} = \frac{p_1}{p_2} \cdot \frac{q_2}{q_1}.$

    2. Insert the given ratios

       Charges are in the ratio $q_1 : q_2 = 1 : 2$, so $\frac{q_2}{q_1} = \frac{2}{1} = 2$.

       Radii are in the ratio $r_1 : r_2 = 3 : 4$, so $\frac{r_1}{r_2} = \frac{3}{4}$.

    3. Solve for the momentum ratio

       From step 1:

       $\frac{3}{4} = \frac{p_1}{p_2} \times 2.$ …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.Two stars of masses ‘M’ and ‘2M’ that are at a distance ‘d’ apart, are revolving one around another. The angular velocity of the system of two stars is (G-Universal gravitational constant) (A) $\sqrt{\dfrac{4GM}{d^3}}$ (B) $\sqrt{\dfrac{2GM}{d^3}}$ (C) $\sqrt{\dfrac{9GM}{d^3}}$ (D) $\sqrt{\dfrac{3GM}{d^3}}$

    ›Reveal solutionSolution

    The two stars revolve around their common centre of mass, and the gravitational force provides the centripetal force for each star. The angular velocity is found to be $\sqrt{\frac{3GM}{d^3}}$, so the correct option is (D).

    The key concept here is that two bodies orbiting each other do not revolve around the centre of either star. Instead, they both orbit their common centre of mass (the balance point of the system). The gravitational attraction between them supplies the necessary centripetal force for each star’s circular motion around that point. Because the stars have different masses, they orbit at different distances from the centre of mass, but they share the same angular velocity — otherwise they would not stay opposite each other.

    Let’s work through it step by step.

    1. Locate the centre of mass. Place the star of mass $M$ at position $x=0$ and the star of mass $2M$ at $x=d$. The centre of mass (CM) is at

    $x_{\text{CM}} = \frac{M \cdot 0 + 2M \cdot d}{M + 2M} = \frac{2Md}{3M} = \frac{2}{3}d.$

    So the CM is $\frac{2}{3}d$ from the lighter star and $\frac{1}{3}d$ from the heavier star.

    2. Define the orbital radii. Let $r_1$ be the distance from the CM to the star of mass $M$, and $r_2$ the distance to the star of mass $2M$. From step 1:

    $r_1 = \frac{2}{3}d, \quad r_2 = \frac{1}{3}d.$

    Notice $r_1 + r_2 = d$, as expected.

    3. Apply Newton’s law of gravitation and centripetal force for one star. For the star of mass $M$, the gravitational force from the other star is

    $F = \frac{G \cdot M \cdot 2M}{d^2} = \frac{2GM^2}{d^2}.$

    This force provides the centripetal force needed for $M$ to move in a circle of radius $r_1$ with angular velocity $\omega$:

    $M \omega^2 r_1 = \frac{2GM^2}{d^2}.$

    4. Solve for $\omega$. Cancel $M$ from both sides:

    $\omega^2 r_1 = \frac{2GM}{d^2}.$

    Substitute $r_1 = \frac{2}{3}d$:

    $\omega^2 \cdot \frac{2}{3}d = \frac{2GM}{d^2}.$

    Multiply both sides by $\frac{3}{2d}$:

    $$$$

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12. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The ratio of the radii of two planets is ‘r’ and the ratio of accelerations due to gravity on the planets is ‘x’. Then the ratio of the escape velocities from the planets is (A) $xr$ (B) $\sqrt{\dfrac{r}{x}}$ (C) $\sqrt{rx}$ (D) $\sqrt{\dfrac{x}{r}}$

    ›Reveal solutionSolution

    The ratio of escape velocities depends on both the planet’s radius and its surface gravity; combining the formulas gives $\sqrt{rx}$, so the correct option is (C).

    Concept & Intuition

    Escape velocity is the minimum speed needed to break free from a planet’s gravitational pull. It depends on two things: the planet’s mass (which determines how strong the gravity is) and its radius (how far you are from the center when you start). But we aren’t given masses directly — we’re given the ratio of radii ($r$) and the ratio of surface gravities ($x$). Since surface gravity itself depends on mass and radius, we can combine these to find the escape velocity ratio without needing absolute values.

    Step-by-step reasoning

    1. Recall the formula for escape velocity For a planet of mass $M$ and radius $R$, the escape velocity is

    $v_e = \sqrt{\frac{2GM}{R}}$

    where $G$ is the universal gravitational constant.

    2. Express mass in terms of surface gravity The acceleration due to gravity at the surface is

    $g = \frac{GM}{R^2}$

    So $GM = g R^2$.

    3. Substitute into the escape velocity formula

    $v_e = \sqrt{\frac{2(g R^2)}{R}} = \sqrt{2gR}$

    This is a neat form: escape velocity depends only on the planet’s surface gravity and its radius.

    4. Set up the ratios Let planet 1 have radius $R_1$ and gravity $g_1$, planet 2 have $R_2$ and $g_2$. Given: $\frac{R_1}{R_2} = r \quad \text{and} \quad \frac{g_1}{g_2} = x$ …

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