Q.An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively, in a uniform magnetic field $\vec{B} = B_0\hat{i}$, each with an equal momentum of magnitude $p = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

Showing the 12 most recent of 71 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A thin conducting wire of length $L$ carrying a current of $2\,\mathrm{A}$ is bent into a square loop of $2$ turns and another thin conducting wire of length $2L$ carrying a current of $3\,\mathrm{A}$ is bent into a circular loop of $3$ turns. If the magnetic moment of the circular loop is $\frac{4}{\pi}\,\mathrm{Am}^2$, then the magnetic moment of the square loop is (A) $0.5\pi\,\mathrm{Am}^2$ (B) $0.5\,\mathrm{Am}^2$ (C) $0.25\pi\,\mathrm{Am}^2$ (D) $0.25\,\mathrm{Am}^2$

   ›Reveal solutionSolution

   The magnetic moment of a planar loop is $M = N I A$, where $N$ is the number of turns, $I$ the current, and $A$ the area. Using the given data for the circular loop to find $L$, then computing the square loop’s moment gives $0.25\,\mathrm{Am}^2$, which is option (D).

   The magnetic moment of any current-carrying coil is the product of the number of turns, the current, and the area enclosed per turn. The key here is that both loops are made from wires of fixed total length, so the side length of the square and the radius of the circle are determined by $L$ and the number of turns. Once we find $L$ from the circular loop’s given moment, the square loop’s moment follows directly.

   1. Circular loop: find $L$ from its magnetic moment

      The wire length is $2L$ and it is bent into 3 turns. Each turn is a circle, so the circumference of one turn is $\frac{2L}{3}$.

      Radius of one turn: $2\pi r = \frac{2L}{3} \;\Rightarrow\; r = \frac{L}{3\pi}$.

      Area of one turn: $A_c = \pi r^2 = \pi \left(\frac{L}{3\pi}\right)^2 = \frac{L^2}{9\pi}$.

      Magnetic moment of the circular loop: $M_c = N_c I_c A_c = 3 \times 3 \times \frac{L^2}{9\pi} = \frac{L^2}{\pi}$.

      Given $M_c = \frac{4}{\pi}\,\mathrm{Am}^2$, we have $\frac{L^2}{\pi} = \frac{4}{\pi} \;\Rightarrow\; L^2 = 4 \;\Rightarrow\; L = 2\,\mathrm{m}$ (length in metres).

   2. Square loop: compute its magnetic moment …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the radius of $_{13}^{27}\mathrm{Al}$ nucleus is $3.6$ fm, then the number of neutrons in a nucleus of atomic number $29$ and radius $4.8$ fm is (A) $64$ (B) $35$ (C) $42$ (D) $49$

   ›Reveal solutionSolution

   Nuclear radius follows $R = R_0 A^{1/3}$, so the ratio of radii gives the ratio of mass numbers. From the given data, the unknown nucleus has $A = 64$; with atomic number $29$, neutrons $= 64 - 29 = 35$.

   The key idea is that nuclear radius depends only on the total number of nucleons (mass number $A$), not on the proton count separately. The formula $R = R_0 A^{1/3}$ is an empirical result — it comes from the fact that nucleons are packed at nearly constant density, so volume is proportional to $A$, and radius goes as $A^{1/3}$.

   We are given one known nucleus: $_{13}^{27}\mathrm{Al}$ has $A = 27$ and $R = 3.6$ fm. Another nucleus has atomic number $Z = 29$ and $R = 4.8$ fm. We need its neutron number $N = A - Z$.

   1. Write the radius relation for both nuclei.

      For aluminium: $R_1 = R_0 A_1^{1/3}$ with $A_1 = 27$, $R_1 = 3.6$ fm.

      For the unknown: $R_2 = R_0 A_2^{1/3}$ with $R_2 = 4.8$ fm.

   2. Take the ratio to eliminate $R_0$:

   $\frac{R_2}{R_1} = \frac{A_2^{1/3}}{A_1^{1/3}} = \left( \frac{A_2}{A_1} \right)^{1/3}.$

   3. Substitute the numbers:

   $\frac{4.8}{3.6} = \frac{4}{3} = \left( \frac{A_2}{27} \right)^{1/3}.$

   4. Cube both sides:

   $\left( \frac{4}{3} \right)^3 = \frac{A_2}{27} \quad\Rightarrow\quad \frac{64}{27} = \frac{A_2}{27}.$

   So $A_2 = 64$. …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If the excess pressure inside a spherical mercury drop is $1240\, \mathrm{Nm}^{-2}$, then the radius of the mercury drop is (Surface tension of mercury $= 0.465\, \mathrm{Nm}^{-1}$) (A) 0.75 mm (B) 1.5 mm (C) 0.375 mm (D) 2.25 mm

   ›Reveal solutionSolution

   The excess pressure inside a spherical drop is given by $\Delta P = \frac{2T}{r}$. Solving for $r$ with $\Delta P = 1240\, \mathrm{N/m^2}$ and $T = 0.465\, \mathrm{N/m}$ gives $r = 0.75\, \mathrm{mm}$, which corresponds to option (A).

   The key concept here is the Laplace pressure for a spherical liquid drop. When a liquid forms a small sphere, surface tension pulls inward, creating a higher pressure inside than outside. For a spherical drop, the excess pressure $\Delta P$ is inversely proportional to the radius: the smaller the drop, the larger the pressure difference. This is why tiny droplets are harder to deform and why blowing a soap bubble requires more effort at first.

   The formula is:

   $\Delta P = \frac{2T}{r}$

   where $T$ is the surface tension and $r$ is the radius. Notice that for a soap bubble (which has two surfaces), the factor is $4T/r$, but for a simple liquid drop, it’s $2T/r$.

   Now, let’s solve step by step.

   1. Write down the given values

      Excess pressure: $\Delta P = 1240\, \mathrm{N/m^2}$

      Surface tension: $T = 0.465\, \mathrm{N/m}$

      We need the radius $r$.

   2. Use the Laplace pressure formula for a spherical drop

   $\Delta P = \frac{2T}{r}$

   3. Rearrange to solve for $r$ Multiply both sides by $r$:

   $r \cdot \Delta P = 2T$

   Then divide by $\Delta P$:

   $r = \frac{2T}{\Delta P}$

   4. Substitute the numbers $r = \frac{2 \times 0.465}{1240}$ …

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4. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The emissivities of the surfaces of two spheres P and Q of radii $2R$ and $3R$ are $0.35$ and $0.7$ respectively. The ratio of the powers radiated by the spheres P and Q is $9:8$. If the wavelength at which sphere P emits radiations of maximum intensity is $4000$ Å, then the wavelength at which sphere Q emits radiations of maximum intensity is (A) $5000$ Å (B) $3000$ Å (C) $4500$ Å (D) $6000$ Å

   ›Reveal solutionSolution

   Using the Stefan–Boltzmann law for radiated power and Wien’s displacement law for peak wavelength, the ratio of powers and emissivities gives the temperature ratio, from which the peak wavelength of sphere Q is found to be 4500 Å.

   Concept & Intuition

   Two key physical laws govern this problem:

   • Stefan–Boltzmann law: The total power radiated by a blackbody is $P = \varepsilon \sigma A T^4$, where $\varepsilon$ is emissivity, $A$ is surface area, and $T$ is absolute temperature.
   • Wien’s displacement law: The wavelength $\lambda_{\text{max}}$ at which the intensity is maximum is inversely proportional to temperature: $\lambda_{\text{max}} T = \text{constant}$.

   We are given the ratio of powers, the radii, and the emissivities. From these we can find the ratio of temperatures. Then, knowing $\lambda_{\text{max}}$ for sphere P, we can compute $\lambda_{\text{max}}$ for sphere Q.

   Step-by-step solution

   1. Write the power radiated by each sphere For sphere P: radius $r_P = 2R$, emissivity $\varepsilon_P = 0.35$, temperature $T_P$. Surface area $A_P = 4\pi (2R)^2 = 16\pi R^2$. Power:

   $P_P = \varepsilon_P \sigma A_P T_P^4 = 0.35 \cdot \sigma \cdot 16\pi R^2 \cdot T_P^4.$

   For sphere Q: radius $r_Q = 3R$, emissivity $\varepsilon_Q = 0.7$, temperature $T_Q$.

   Surface area $A_Q = 4\pi (3R)^2 = 36\pi R^2$.

   Power:

   $P_Q = \varepsilon_Q \sigma A_Q T_Q^4 = 0.7 \cdot \sigma \cdot 36\pi R^2 \cdot T_Q^4.$

   2. Use the given power ratio

   $\frac{P_P}{P_Q} = \frac{9}{8}.$

   Substitute the expressions:

   $\frac{0.35 \cdot 16\pi R^2 \cdot T_P^4}{0.7 \cdot 36\pi R^2 \cdot T_Q^4} = \frac{9}{8}.$

   Cancel $\pi R^2$ and simplify the constants:

   $\frac{0.35}{0.7} = 0.5, \quad \frac{16}{36} = \frac{4}{9}.$

   So:

   $0.5 \cdot \frac{4}{9} \cdot \frac{T_P^4}{T_Q^4} = \frac{9}{8}.$

   $\frac{2}{9} \cdot \frac{T_P^4}{T_Q^4} = \frac{9}{8}.$

   Multiply both sides by $9/2$:

   $\frac{T_P^4}{T_Q^4} = \frac{9}{8} \cdot \frac{9}{2} = \frac{81}{16}.$

   Take the fourth root:

   $\frac{T_P}{T_Q} = \left(\frac{81}{16}\right)^{1/4} = \frac{81^{1/4}}{16^{1/4}} = \frac{3}{2}.$ …

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The lengths of two copper wires A and B are 180 cm and 270 cm respectively. If the mass of wire A is twice the mass of wire B and the electrical resistance of wire A is $200~\Omega$, then the electrical resistance of wire B is (A) $900~\Omega$ (B) $400~\Omega$ (C) $600~\Omega$ (D) $300~\Omega$

   ›Reveal solutionSolution

   The resistance of a wire depends on its length, cross‑sectional area, and material. Using the given mass relation to link the areas, we find that wire B has resistance $600~\Omega$.

   We are given two copper wires (same material, so same resistivity $\rho$ and density $d$).

   Wire A: length $L_A = 180$ cm, mass $m_A$, resistance $R_A = 200~\Omega$.

   Wire B: length $L_B = 270$ cm, mass $m_B$, and $m_A = 2\,m_B$.

   We need $R_B$.

   Concept and intuition:

   Resistance of a wire is $R = \rho \frac{L}{A}$, where $A$ is cross‑sectional area. Mass is $m = d \cdot (\text{volume}) = d \cdot A L$. Since both wires are copper, $\rho$ and $d$ are the same. The mass relation gives a relation between $A_A$ and $A_B$, which then lets us compare resistances.

   ---

   1. Write expressions for mass and resistance. For wire A:

   $m_A = d\, A_A L_A, \qquad R_A = \rho \frac{L_A}{A_A}.$

   For wire B:

   $m_B = d\, A_B L_B, \qquad R_B = \rho \frac{L_B}{A_B}.$

   2. Use the mass condition to relate areas. Given $m_A = 2 m_B$, substitute:

   $d\, A_A L_A = 2\, (d\, A_B L_B).$

   Cancel $d$:

   $A_A L_A = 2 A_B L_B.$

   So

   $\frac{A_A}{A_B} = 2 \frac{L_B}{L_A}.$

   3. Find the ratio of resistances.

   $\frac{R_B}{R_A} = \frac{\rho L_B / A_B}{\rho L_A / A_A} = \frac{L_B}{L_A} \cdot \frac{A_A}{A_B}.$

   Substitute the area ratio from step 2:

   $\frac{R_B}{R_A} = \frac{L_B}{L_A} \cdot \left( 2 \frac{L_B}{L_A} \right) = 2 \left( \frac{L_B}{L_A} \right)^2.$

   4. Plug in the lengths. $L_A = 180$ cm, $L_B = 270$ cm, so $$$$

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6. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A current of $\dfrac{1}{3\pi}$ A passes through an ideal toroid of 900 turns per metre. If the relative permeability of the material of the core of the toroid is 400, then the magnetic field inside the core of the toroid is (A) 48 mT (B) 24 mT (C) 72 mT (D) 96 mT

   ›Reveal solutionSolution

   The magnetic field inside a toroid is given by $B = \mu_0 \mu_r n I$. Substituting the given values yields $B = 48 \,\text{mT}$, so the correct option is (A).

   Concept & Intuition

   A toroid is essentially a solenoid bent into a doughnut shape. For an ideal toroid, the magnetic field is confined entirely to the core and is uniform along the circular path inside it. The field depends on three things: the number of turns per unit length ($n$), the current ($I$), and the magnetic properties of the core material (permeability $\mu = \mu_0 \mu_r$). The formula $B = \mu n I$ comes directly from Ampère’s law, where the line integral of $\mathbf{B}$ around a closed loop inside the toroid equals $\mu$ times the total current enclosed. Since the core has a relative permeability $\mu_r = 400$, the effective permeability is $400$ times that of vacuum.

   Step-by-step solution

   1. Identify the given quantities

      • Current: $I = \dfrac{1}{3\pi} \,\text{A}$
      • Turns per metre: $n = 900 \,\text{turns/m}$
      • Relative permeability: $\mu_r = 400$
      • Permeability of free space: $\mu_0 = 4\pi \times 10^{-7} \,\text{H/m}$

   2. Write the formula for the magnetic field inside an ideal toroid

      For a toroid (or a long solenoid) with a core of permeability $\mu = \mu_0 \mu_r$, the magnetic field is

   $B = \mu n I = \mu_0 \mu_r n I$

   This holds because the field lines are circular and the line integral $\oint \mathbf{B} \cdot d\mathbf{l} = B \cdot (2\pi r)$ equals $\mu$ times the total current $n(2\pi r)I$, giving $B = \mu n I$.

   3. Substitute the numbers

   $B = (4\pi \times 10^{-7}) \times 400 \times 900 \times \frac{1}{3\pi}$

   4. Simplify step by step
      • Cancel $\pi$: $4\pi \times \frac{1}{\pi} = 4$
      • So $B = 4 \times 10^{-7} \times 400 \times 900 \times \frac{1}{3}$
      • Multiply $400 \times 900 = 360\,000$ …

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7. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The range of the weak nuclear force is of the order of (A) $10^{26}$ \AA (B) $10^{-6}$ \AA (C) $10^{18}$ \AA (D) $10^{-2}$ \AA

   ›Reveal solutionSolution

   The weak nuclear force is a short-range force mediated by massive bosons; its range is roughly $10^{-18}$ m, which converts to $10^{-8}$ Å, so the closest given option is $10^{-6}$ Å.

   The weak nuclear force is one of the four fundamental forces, but unlike electromagnetism or gravity, it has a very short range. Why? Because its force carriers—the W and Z bosons—are massive. In quantum field theory, the range of a force is inversely proportional to the mass of the exchange particle: roughly $R \approx \hbar c / m c^2$. For the W boson, $m c^2 \approx 80$ GeV, giving a range around $10^{-18}$ meters. That’s about $10^{-8}$ angstroms (since 1 Å = $10^{-10}$ m). Among the options, $10^{-6}$ Å is the closest order of magnitude.

   Let’s walk through the reasoning step by step.

   1. Recall the relation between mass and range In quantum mechanics, a force mediated by a particle of mass $m$ has a range given by the Compton wavelength of that particle:

   $R \approx \frac{\hbar}{m c}$

   For the weak force, the mediators are the W and Z bosons, with masses around $80$–$90$ GeV/$c^2$.

   2. Plug in numbers Use $\hbar c \approx 197$ MeV·fm (where 1 fm = $10^{-15}$ m). For $m c^2 \approx 80$ GeV = $80\,000$ MeV:

   $R \approx \frac{197 \text{ MeV·fm}}{80\,000 \text{ MeV}} \approx 0.0025 \text{ fm} = 2.5 \times 10^{-18} \text{ m}$

   3. Convert to angstroms 1 Å = $10^{-10}$ m, so:

   $R \approx 2.5 \times 10^{-18} \text{ m} = 2.5 \times 10^{-8} \text{ Å}$

   4. Compare with the options
      • (A) $10^{26}$ Å — that’s larger than the observable universe. …

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8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.A long charged cylinder of linear charge density $\lambda$ is surrounded by a hollow coaxial conducting cylinder. The magnitude of electric field in the space between the two cylinders at a distance $R$ from the common axis of the cylinders is (A) $\dfrac{\lambda}{4\pi\epsilon_0 R}$ (B) $0$ (C) $\dfrac{\lambda}{2\pi\epsilon_0 R}$ (D) $\dfrac{\lambda}{\pi\epsilon_0 R}$

   ›Reveal solutionSolution

   The electric field between the cylinders is due only to the inner charged cylinder, because the outer conducting cylinder’s induced charge cancels its own field inside its hollow region. Using Gauss’s law, the field magnitude is $\frac{\lambda}{2\pi\epsilon_0 R}$, so the correct option is (C).

   Concept and intuition

   The problem involves a long charged cylinder (linear charge density $\lambda$) and a coaxial conducting shell. The key idea: inside a conductor in electrostatic equilibrium, the electric field is zero. The outer cylinder is a conductor, so the field inside its material must be zero. But we are asked for the field between the two cylinders — that is, in the hollow region inside the conducting shell but outside the inner cylinder.

   In that region, the only charge that contributes to the field is the inner cylinder’s charge. Why? Because the conducting shell will have induced charges on its inner surface, but those induced charges produce a field that cancels the inner cylinder’s field inside the conductor itself — not in the hollow region. In fact, for a cylindrical Gaussian surface that lies entirely in the gap, the net enclosed charge is just $\lambda$ times the length, because the induced charge on the inner surface of the shell is outside that Gaussian surface. So the field is exactly the same as if the outer cylinder weren’t there.

   Step-by-step reasoning

   1. Choose a Gaussian surface

      Because the system has cylindrical symmetry, we use a cylindrical Gaussian surface of radius $R$ (where $a < R < b$, with $a$ the inner cylinder’s radius and $b$ the inner radius of the conducting shell) and length $L$, coaxial with the cylinders.

   2. Apply Gauss’s law

      Gauss’s law states:

   $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$

   By symmetry, $\vec{E}$ is radial and constant in magnitude over the curved surface of the Gaussian cylinder. The flux through the ends is zero (field is parallel to the ends). So the flux is $E \cdot (2\pi R L)$.

   3. Find the enclosed charge The only charge inside the Gaussian surface is the charge on the inner cylinder: $Q_{\text{enc}} = \lambda L$ …

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9. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The maximum kinetic energy of a proton ejected from a cyclotron with dees of radius 70 cm and oscillating frequency 5 MHz (in MeV) is (Mass of proton = $1.67 \times 10^{-27}$ kg) (A) 2.53 (B) 3.87 (C) 1.46 (D) 4.84

   ›Reveal solutionSolution

   The maximum kinetic energy of a proton in a cyclotron is given by $K_{\text{max}} = \frac{q^2 B^2 R^2}{2m}$, where the magnetic field $B$ is found from the cyclotron frequency $f = \frac{qB}{2\pi m}$. Using $R = 0.70\ \text{m}$, $f = 5 \times 10^6\ \text{Hz}$, and proton mass $m = 1.67 \times 10^{-27}\ \text{kg}$, we get $K_{\text{max}} \approx 2.53\ \text{MeV}$, so the correct option is (A).

   ---

   Concept and Intuition

   A cyclotron accelerates charged particles by applying an alternating electric field across two D-shaped electrodes (dees), while a uniform magnetic field bends them into circular paths. The key insight: the cyclotron frequency $f$ (the rate at which the electric field reverses) must match the particle’s orbital frequency so that it gets a “kick” each time it crosses the gap. This frequency depends only on the charge-to-mass ratio and the magnetic field, not on speed — that’s why the cyclotron works for non-relativistic particles.

   The maximum kinetic energy is reached when the particle’s circular path has the largest possible radius — the radius of the dees themselves. At that point, the magnetic force provides the centripetal force:

   $q v B = \frac{m v^2}{R}$

   so the speed is $v = \frac{q B R}{m}$, and kinetic energy is $K = \frac{1}{2} m v^2 = \frac{q^2 B^2 R^2}{2m}$.

   We are given the frequency, not the magnetic field directly. But the cyclotron frequency formula gives us $B$ from $f$.

   ---

   Step-by-Step Solution

   1. Relate the cyclotron frequency to the magnetic field. The cyclotron (orbital) frequency for a non-relativistic particle is

   $f = \frac{q B}{2 \pi m}$

   where $q$ is the charge, $B$ the magnetic field, and $m$ the mass. For a proton, $q = 1.6 \times 10^{-19}\ \text{C}$.

   Rearranging:

   $B = \frac{2 \pi m f}{q}$

   2. Plug in the numbers to find $B$. Given $m = 1.67 \times 10^{-27}\ \text{kg}$, $f = 5 \times 10^6\ \text{Hz}$:

   $B = \frac{2 \pi (1.67 \times 10^{-27}) (5 \times 10^6)}{1.6 \times 10^{-19}}$

   Compute stepwise:

   • $2 \pi \approx 6.2832$
   • $6.2832 \times 1.67 \times 10^{-27} \approx 1.049 \times 10^{-26}$
   • Multiply by $5 \times 10^6$: $1.049 \times 10^{-26} \times 5 \times 10^6 = 5.245 \times 10^{-20}$
   • Divide by $1.6 \times 10^{-19}$: $B \approx 0.3278\ \text{T}$

   3. Now compute the maximum kinetic energy. The maximum radius is $R = 70\ \text{cm} = 0.70\ \text{m}$.

   $K_{\text{max}} = \frac{q^2 B^2 R^2}{2m}$

   Substitute values:

   $K_{\text{max}} = \frac{(1.6 \times 10^{-19})^2 \times (0.3278)^2 \times (0.70)^2}{2 \times 1.67 \times 10^{-27}}$

   Compute numerator stepwise:

   • $q^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$
   • $B^2 = (0.3278)^2 \approx 0.1075$
   • $R^2 = (0.70)^2 = 0.49$
   • Product: $2.56 \times 10^{-38} \times 0.1075 \approx 2.752 \times 10^{-39}$ …

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10. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.The number of gas molecules per unit volume in a vessel is $2 \times 10^{25}~\mathrm{m}^{-3}$ and the surface area of each gas molecule is $12.5\sqrt{2} \times 10^{-20}~\mathrm{m}^2$. If all the gas molecules are in motion, then the mean free path of the gas molecules (in $\mathrm{\AA}$) is (A) 500 (B) 1000 (C) 2000 (D) 4000

    ›Reveal solutionSolution

    The mean free path is found from $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$, using the given surface area to get the molecular diameter. The result is $\mathbf{2000 \, \text{Å}}$, option (C).

    The mean free path is the average distance a molecule travels between collisions. For a gas of identical hard-sphere molecules, the formula is

    $\lambda = \frac{1}{\sqrt{2} \, \pi d^2 \, n}$

    where $d$ is the molecular diameter and $n$ is the number density (molecules per unit volume). The $\sqrt{2}$ factor accounts for the relative motion of all molecules — if only one molecule moved and the rest were stationary, the factor would be $1$, but since all are in motion, the collision rate is higher, shortening the mean free path.

    Here we are told the surface area of each molecule. For a sphere, the surface area is $4\pi r^2 = \pi d^2$. So the given area directly gives $\pi d^2$, which is exactly what appears in the denominator of the mean free path formula. That is the key insight — we don’t need to find $d$ separately.

    1. Extract the given data

       Number density: $n = 2 \times 10^{25}~\mathrm{m}^{-3}$

       Surface area of one molecule: $A = 12.5\sqrt{2} \times 10^{-20}~\mathrm{m}^2$

       For a sphere, $A = \pi d^2$.

    2. Write the mean free path formula

    $\lambda = \frac{1}{\sqrt{2} \, (\pi d^2) \, n}$

    Notice $\pi d^2$ is exactly the given surface area $A$.

    3. Substitute directly

    $\lambda = \frac{1}{\sqrt{2} \times (12.5\sqrt{2} \times 10^{-20}) \times (2 \times 10^{25})}$

    4. Simplify the numbers The $\sqrt{2}$ in the numerator and the $\sqrt{2}$ in the denominator cancel:

    $\sqrt{2} \times \sqrt{2} = 2$

    So the denominator becomes:

    $12.5 \times 2 \times 10^{-20} \times 2 \times 10^{25} = 12.5 \times 4 \times 10^{5} = 50 \times 10^{5}$

    Thus …

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11. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.A magnet suspended in the horizontal plane makes 24 oscillations per minute at a place A where the dip angle is $30^\circ$ and ‘$n$’ oscillations per minute at another place B where the dip angle is $60^\circ$. If the ratio of the horizontal components of the earth’s magnetic field at places A and B is $16:9$, then the value of ‘$n$’ is (A) 18 (B) 12 (C) 24 (D) 36

    ›Reveal solutionSolution

    The oscillation frequency of a suspended magnet depends on the horizontal component of Earth’s field. Using the time-period formula and given ratios, we find $n = 18$ oscillations per minute.

    The key idea is that a magnet suspended in the horizontal plane oscillates under the restoring torque due only to the horizontal component $B_H$ of Earth’s magnetic field. The vertical component does not affect the oscillation because the magnet is free to rotate horizontally. The time period of small oscillations is $T = 2\pi \sqrt{\frac{I}{m B_H}}$, where $I$ is the moment of inertia and $m$ the magnetic moment of the magnet. Since the same magnet is used at both places, $I$ and $m$ are constant. So the oscillation frequency (oscillations per unit time) is proportional to $\sqrt{B_H}$.

    Let’s work through the steps.

    1. Relate frequency to $B_H$ At place A, frequency $f_A = 24$ oscillations per minute. At place B, frequency $f_B = n$ oscillations per minute. The time period $T \propto 1/\sqrt{B_H}$, so frequency $f \propto \sqrt{B_H}$. Hence,

    $\frac{f_A}{f_B} = \sqrt{\frac{B_{H_A}}{B_{H_B}}}.$

    2. Use the given ratio of horizontal components The problem states $\frac{B_{H_A}}{B_{H_B}} = \frac{16}{9}$. Therefore,

    $\frac{24}{n} = \sqrt{\frac{16}{9}} = \frac{4}{3}.$

    3. Solve for $n$ Cross-multiplying: $24 \times 3 = 4n \quad \Rightarrow \quad 72 = 4n \quad \Rightarrow \quad n = 18.$ …

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12. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
    Q.The acceleration due to gravity at a point A at certain height from surface of the earth is $\dfrac{g}{4}$, where ‘g’ is acceleration due to gravity on the surface of the earth. At a point B which lies vertically at certain height above point A, the acceleration due to gravity is $\dfrac{g}{9}$. The distance between points A and B is (R - Radius of the earth) (A) $5R$ (B) $3R$ (C) $R$ (D) $2R$

    ›Reveal solutionSolution

    The key idea is that gravity varies with distance from Earth’s centre as $g \propto 1/r^2$. Using the given values, we find the distances of A and B from the centre, then subtract to get the height difference. The distance between A and B is $R$.

    The problem is about how gravitational acceleration changes with altitude. On the Earth’s surface, gravity is $g$. At a height $h$ above the surface, the distance from the Earth’s centre becomes $R + h$, and the acceleration is $g' = \frac{GM}{(R+h)^2}$. Since $g = \frac{GM}{R^2}$, we have the simple inverse-square relation:

    $g' = g \cdot \frac{R^2}{(R+h)^2}$

    So if you know $g'$ as a fraction of $g$, you can directly solve for $R+h$, and then for the height $h$. The distance between two points at different heights is just the difference in their heights.

    Let’s work it out.

    1. Find the distance of point A from Earth’s centre. At A, $g_A = \frac{g}{4}$. Using the formula:

    $\frac{g}{4} = g \cdot \frac{R^2}{(R + h_A)^2}$

    Cancel $g$ (non-zero) and rearrange:

    $\frac{1}{4} = \frac{R^2}{(R + h_A)^2}$

    Take square roots (positive distances):

    $\frac{1}{2} = \frac{R}{R + h_A}$

    So:

    $R + h_A = 2R \quad \Rightarrow \quad h_A = R$

    2. Find the distance of point B from Earth’s centre. At B, $g_B = \frac{g}{9}$. Similarly:

    $\frac{g}{9} = g \cdot \frac{R^2}{(R + h_B)^2}$

    $\frac{1}{9} = \frac{R^2}{(R + h_B)^2}$

    $\frac{1}{3} = \frac{R}{R + h_B}$

    Hence:

    $R + h_B = 3R \quad \Rightarrow \quad h_B = 2R$

    3. Distance between A and B. …

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