Q.Choose the correct alternative:

Showing the 12 most recent of 28 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A solid sphere rolls down without slipping on an inclined plane of angle of inclination $30^\circ$. If the angle of inclination of the plane is increased to $45^\circ$, then the percentage increase in the acceleration of the sphere is nearly (A) 73.2 (B) 41.4 (C) 21.2 (D) 36.6

   ›Reveal solutionSolution

   For a rolling rigid body, acceleration depends on $\sin\theta$ and the moment of inertia factor. The percentage increase in acceleration when $\theta$ goes from $30^\circ$ to $45^\circ$ is about 41.4%, matching option (B).

   The key idea is that for pure rolling down an incline, the acceleration is not simply $g\sin\theta$ (as for sliding without friction) but is reduced by a factor that depends on the body's moment of inertia. For a solid sphere, that factor is $\frac{5}{7}$. So the acceleration is $a = \frac{5}{7} g \sin\theta$. When $\theta$ changes, only $\sin\theta$ changes — the $\frac{5}{7}$ factor stays constant. The percentage increase in acceleration therefore equals the percentage increase in $\sin\theta$ itself.

   Let's work it through.

   1. Acceleration of a rolling sphere For a rigid body of mass $m$, radius $R$, and moment of inertia $I = k m R^2$ (where $k = \frac{2}{5}$ for a solid sphere), the acceleration down an incline of angle $\theta$ under pure rolling is

   $a = \frac{g \sin\theta}{1 + k}$

   For a solid sphere, $k = \frac{2}{5}$, so $1 + k = \frac{7}{5}$, giving

   $a = \frac{g \sin\theta}{7/5} = \frac{5}{7} g \sin\theta$

   2. Acceleration at $\theta = 30^\circ$

   $a_{30} = \frac{5}{7} g \sin 30^\circ = \frac{5}{7} g \cdot \frac{1}{2} = \frac{5g}{14}$

   3. Acceleration at $\theta = 45^\circ$

   $a_{45} = \frac{5}{7} g \sin 45^\circ = \frac{5}{7} g \cdot \frac{1}{\sqrt{2}} = \frac{5g}{7\sqrt{2}}$

   4. Percentage increase The percentage increase from $a_{30}$ to $a_{45}$ is

   $\frac{a_{45} - a_{30}}{a_{30}} \times 100\% = \frac{\frac{5g}{7\sqrt{2}} - \frac{5g}{14}}{\frac{5g}{14}} \times 100\%$

   Cancel $\frac{5g}{7}$ from numerator and denominator (or simply cancel $5g$ everywhere):

   $= \frac{\frac{1}{\sqrt{2}} - \frac{1}{2}}{\frac{1}{2}} \times 100\% = \left( \frac{2}{\sqrt{2}} - 1 \right) \times 100\% = (\sqrt{2} - 1) \times 100\%$

   Since $\sqrt{2} \approx 1.4142$, we get

   $(1.4142 - 1) \times 100\% = 0.4142 \times 100\% = 41.42\%$

   Tip

   Notice that the $\frac{5}{7}$ factor cancels out completely. The percentage increase depends only on $\sin\theta$, not on the shape of the rolling body. So for any object rolling without slipping, the percentage increase in acceleration when $\theta$ goes from $30^\circ$ to $45^\circ$ is $(\sqrt{2} - 1) \times 100\% \approx 41.4\%$.

   Watch out

   A common mistake is to use $a = g\sin\theta$ (the sliding case) and get $(\sin 45^\circ - \sin 30^\circ)/\sin 30^\circ = (\sqrt{2}/2 - 1/2)/(1/2) = \sqrt{2} - 1 \approx 0.414$ — which accidentally gives the same number! But the reasoning is wrong because the actual acceleration is smaller. The percentage increase ratio is the same here only because the inertia factor cancels; don't assume this always works for other angle changes or other physical quantities.

   ✓Final answer

   The percentage increase in acceleration is approximately 41.4%, which corresponds to option (B).

2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Consider the following statements: Statement-I: When two quantities are multiplied, the relative error in the result is the sum of the relative errors in the quantities Statement-II: When two quantities are divided, the relative error in the result is the difference of the relative errors in the quantities The correct answer among the following is (A) Both statements I and II are correct (B) Both statements I and II are not correct (C) Statement I is correct, but statement II is not correct (D) Statement I is not correct, but statement II is correct

   ›Reveal solutionSolution

   The relative error in multiplication adds, but in division it also adds (not subtracts), so Statement I is correct and Statement II is false. The correct option is (C).

   Concept & Intuition

   When we measure quantities, each has an uncertainty (error). The relative error is the ratio of the absolute error to the measured value. For a product or quotient, we want to know how these uncertainties combine. The key rule: relative errors always add when quantities are multiplied or divided. Why? Because division is just multiplication by a reciprocal, and the relative error of a reciprocal is the same as the original relative error (since dividing 1 by a number flips the absolute error proportionally). So subtracting errors would be a mistake — it could even make the total error seem smaller, which is impossible.

   Step-by-step reasoning

   1. Define relative error

      For a quantity $x$ with measured value $\bar{x}$ and absolute error $\Delta x$, the relative error is $\frac{\Delta x}{\bar{x}}$. This is a dimensionless fraction, often expressed as a percentage.

   2. Multiplication case (Statement I)

      Let $z = x \cdot y$. The absolute error in $z$ is approximately

   $\Delta z \approx |y| \Delta x + |x| \Delta y$

   (from calculus: $dz = y\,dx + x\,dy$). Dividing by $z = xy$ gives

   $\frac{\Delta z}{z} \approx \frac{\Delta x}{x} + \frac{\Delta y}{y}$

   So relative errors add. Statement I is correct.

   3. Division case (Statement II) Let $z = \frac{x}{y}$. The absolute error is approximately

   $\Delta z \approx \frac{1}{y} \Delta x + \frac{x}{y^2} \Delta y$

   (from $dz = \frac{dx}{y} - \frac{x}{y^2}dy$; the minus sign is irrelevant for error magnitude because errors add in quadrature or worst-case as absolute values). Dividing by $z = x/y$ yields

   $\frac{\Delta z}{z} \approx \frac{\Delta x}{x} + \frac{\Delta y}{y}$

   The relative errors add, not subtract. Statement II says they subtract — that is false.

   4. Common pitfall

      Watch out

      Students often think the minus sign in the derivative $dz = \frac{dx}{y} - \frac{x}{y^2}dy$ means errors subtract. But errors are magnitudes; we add absolute values (or squares) because uncertainties can’t cancel — they always increase total uncertainty.

   5. Conclusion

      Only Statement I is correct.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The intensive variables among thermodynamic state variables internal energy U, volume V, pressure P, absolute temperature T and total mass M are (A) M, V (B) U, M (C) U, V (D) P, T

   ›Reveal solutionSolution

   Intensive variables are independent of system size; among U, V, P, T, M, only pressure P and temperature T are intensive, so the correct choice is (D).

   Concept & Intuition

   Thermodynamic variables come in two flavors: extensive (scale with the amount of substance) and intensive (independent of system size). If you double the system, extensive quantities double; intensive ones stay the same. Internal energy $U$, volume $V$, and total mass $M$ all clearly double when you take two identical systems and combine them — they are extensive. Pressure $P$ and temperature $T$, however, remain unchanged when you merge two systems at the same $P$ and $T$ — they are intensive. The question asks for the pair that are both intensive.

   Step-by-step reasoning

   1. Identify the nature of each variable

      • Internal energy $U$: If you have two identical blocks of matter, each with energy $U$, the combined system has energy $2U$. So $U$ is extensive.
      • Volume $V$: Similarly, volumes add: two blocks of volume $V$ give total volume $2V$. Extensive.
      • Pressure $P$: If two identical gas samples at pressure $P$ are combined (e.g., by removing a partition), the pressure remains $P$ — it does not double. Intensive.
      • Temperature $T$: Two bodies at the same temperature $T$ in contact stay at $T$; temperature does not add. Intensive.
      • Total mass $M$: Mass is additive: two blocks of mass $M$ give $2M$. Extensive.

   2. Check the options

      • (A) M, V: both extensive → not the answer.
      • (B) U, M: both extensive → not the answer.
      • (C) U, V: both extensive → not the answer.
      • (D) P, T: both intensive → this is the correct pair.

   3. Confirm with a classic test

      Imagine dividing a system into two equal halves. Intensive variables (like $P$ and $T$) have the same value in each half as in the whole. Extensive variables (like $U$, $V$, $M$) are halved. Only option (D) passes this test for both variables.

   Watch out

   A common mistake is to think that “total mass” is intensive because it’s a property of the material. But “total” is the clue: it scales with system size. Specific properties (like density or specific heat) are intensive; total mass is not.

   Tip

   A quick mnemonic: Intensive = “inside” the system doesn’t change with size; extensive = “extends” with size. Pressure and temperature are the two most familiar intensive variables in thermodynamics.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The fundamental force that operates among all the objects in the universe is (A) Electromagnetic force (B) Strong nuclear force (C) Weak nuclear force (D) Gravitational force

   ›Reveal solutionSolution

   The question asks which fundamental force acts among all objects in the universe. The answer is gravitational force, because it is the only one that is universal (affects everything with mass/energy) and has infinite range.

   The key concept here is universality — a force that operates on every object, everywhere. The four fundamental forces (electromagnetic, strong nuclear, weak nuclear, and gravitational) differ in their range and the types of particles they affect. Only one of them is truly universal.

   1. Electromagnetic force acts between charged particles. While it has infinite range, it does not affect neutral objects (like neutrons or neutral atoms). So it is not universal among all objects.
   2. Strong nuclear force holds atomic nuclei together. It only acts over extremely short distances (about the size of a nucleus) and only between quarks (protons and neutrons). It does not affect electrons or distant objects.
   3. Weak nuclear force is responsible for certain types of radioactive decay. It also has a very short range and only affects particles like quarks and leptons in specific interactions — not all objects.
   4. Gravitational force acts between any two objects that have mass or energy. Since all objects in the universe possess mass (or energy, via $E=mc^2$), gravity affects everything. Its range is infinite, and it never switches off. This makes it the only force that operates among all objects.

   Watch out

   A common mistake is to pick electromagnetic force because it seems "everywhere" in daily life. But remember: neutral objects (like a neutron star or a dark matter particle) feel no electromagnetic force, yet they still feel gravity.

   Tip

   Gravity is the weakest force, but it dominates on cosmic scales precisely because it is universal and has no cancellation (positive and negative charges cancel electromagnetism, but mass only adds).

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Two blocks of masses $4.5\, \text{kg}$ and $5.5\, \text{kg}$ are connected to the two ends of a light inextensible string passing over a frictionless pulley fixed to a rigid support. At time $t = 0$, the blocks are released from rest. The distance travelled by the larger block in a time $t = 4\, \text{s}$ is (Acceleration due to gravity $= 10\, \text{ms}^{-2}$) (A) 8 m (B) 4 m (C) 16 m (D) 32 m

   ›Reveal solutionSolution

   The system is an Atwood machine with unequal masses; the net force gives a constant acceleration of $1 \, \text{m/s}^2$, so in $4$ seconds from rest the larger block travels $8 \, \text{m}$.

   The problem is a classic Atwood machine: two masses connected by a light string over a frictionless pulley. The key insight is that the string is inextensible and the pulley is frictionless, so both blocks move with the same magnitude of acceleration — the heavier one goes down, the lighter one goes up. The net force driving the system is the difference in their weights, and the total mass being accelerated is the sum of the two masses.

   Let’s work through it step by step.

   1. Identify the forces and the net driving force The heavier block ($m_2 = 5.5 \, \text{kg}$) wants to fall, the lighter block ($m_1 = 4.5 \, \text{kg}$) wants to rise. The weight of $m_2$ is $m_2 g$ downward, and the weight of $m_1$ is $m_1 g$ downward on its own side. But because they are connected, the net force that actually accelerates the whole system is the difference:

   $F_{\text{net}} = m_2 g - m_1 g = (m_2 - m_1)g$

   2. Apply Newton’s second law to the whole system The total mass being accelerated is $m_1 + m_2$. So:

   $(m_1 + m_2) a = (m_2 - m_1) g$

   Solve for acceleration $a$:

   $a = \frac{(m_2 - m_1) g}{m_1 + m_2}$

   3. Plug in the numbers $m_2 = 5.5 \, \text{kg}$, $m_1 = 4.5 \, \text{kg}$, $g = 10 \, \text{m/s}^2$:

   $a = \frac{(5.5 - 4.5) \times 10}{4.5 + 5.5} = \frac{1.0 \times 10}{10} = 1 \, \text{m/s}^2$

   So both blocks accelerate at $1 \, \text{m/s}^2$ — the heavier one downward, the lighter one upward.

   Watch out

   A common mistake is to think each block has its own independent acceleration. Because the string is inextensible, their accelerations are equal in magnitude. Also, do not forget to add the masses in the denominator — the net force accelerates the whole system, not just one block.

   4. Find the distance travelled by the larger block The larger block starts from rest ($u = 0$) and moves with constant acceleration $a = 1 \, \text{m/s}^2$ for $t = 4 \, \text{s}$. Using the kinematic equation:

   $s = ut + \frac{1}{2} a t^2$

   $s = 0 + \frac{1}{2} \times 1 \times (4)^2 = \frac{1}{2} \times 16 = 8 \, \text{m}$

   Tip

   You could also think of it as: in $4$ seconds at $1 \, \text{m/s}^2$, the average velocity is $2 \, \text{m/s}$ (since $v_{\text{avg}} = \frac{0 + 4}{2}$), giving $s = 2 \times 4 = 8 \, \text{m}$. Same result, less algebra.

   ✓Final answer

   The distance travelled by the larger block is $\boxed{8 \, \text{m}}$, which corresponds to option (A).

6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The ratio of times taken by a freely falling body to travel first 5m, second 5m, third 5m distances is (A) $1 : \sqrt{2} : \sqrt{3}$ (B) $1 : \sqrt{2} - 1 : \sqrt{3} - 2$ (C) $1 : \sqrt{3} : \sqrt{5}$ (D) $1 : \sqrt{2} - 1 : \sqrt{3} - \sqrt{2}$

   ›Reveal solutionSolution

   For a freely falling body starting from rest, the times to cover successive equal distances are in the ratio $1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$. This follows from the relation $s = \frac12 g t^2$ and subtracting cumulative times.

   The key idea is that for uniform acceleration from rest, distance is proportional to the square of time. So the time to fall a total distance $s$ is $t \propto \sqrt{s}$. To get the time for just the second 5 m segment, you subtract the time for the first 5 m from the time for the first 10 m. Similarly for the third segment.

   Let’s work it through.

   1. Set up the relation For a body starting from rest under gravity (acceleration $g$), the distance fallen in time $t$ is

   $s = \frac12 g t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2s}{g}}.$

   So time is proportional to $\sqrt{s}$.

   2. Time for the first 5 m Let $t_1$ be the time to fall the first 5 m.

   $t_1 = \sqrt{\frac{2 \times 5}{g}} = \sqrt{\frac{10}{g}}.$

   3. Time for the first 10 m Let $T_2$ be the time to fall the first 10 m (i.e., first two 5 m segments together).

   $T_2 = \sqrt{\frac{2 \times 10}{g}} = \sqrt{\frac{20}{g}}.$

   Then the time for just the second 5 m segment, call it $t_2$, is

   $t_2 = T_2 - t_1 = \sqrt{\frac{20}{g}} - \sqrt{\frac{10}{g}} = \sqrt{\frac{10}{g}} \left( \sqrt{2} - 1 \right).$

   4. Time for the first 15 m Let $T_3$ be the time to fall the first 15 m.

   $T_3 = \sqrt{\frac{2 \times 15}{g}} = \sqrt{\frac{30}{g}}.$

   Then the time for just the third 5 m segment, call it $t_3$, is

   $t_3 = T_3 - T_2 = \sqrt{\frac{30}{g}} - \sqrt{\frac{20}{g}} = \sqrt{\frac{10}{g}} \left( \sqrt{3} - \sqrt{2} \right).$

   5. Form the ratio The times $t_1 : t_2 : t_3$ are

   $\sqrt{\frac{10}{g}} : \sqrt{\frac{10}{g}}(\sqrt{2} - 1) : \sqrt{\frac{10}{g}}(\sqrt{3} - \sqrt{2}).$

   Cancelling the common factor $\sqrt{10/g}$, we get

   $1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2}).$

   Watch out

   A common mistake is to think the times are in the ratio $1 : \sqrt{2} : \sqrt{3}$ because distances are in the ratio $1 : 2 : 3$. But time is proportional to $\sqrt{\text{distance}}$, not distance itself. Also, the times for successive equal segments are not the same as times for total distances.

   Tip

   This ratio $1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$ is a classic result for any uniformly accelerated motion from rest. It’s worth remembering: the times for equal distances follow the pattern of differences of square roots.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.The fundamental force in nature which operates among heavier elementary particles only is (A) Strong nuclear force (B) Electromagnetic force (C) Gravitational force (D) Weak nuclear force

   ›Reveal solutionSolution

   The strong nuclear force is the only fundamental force that acts exclusively on heavier elementary particles (hadrons like protons and neutrons), binding the atomic nucleus together.

   The question asks which fundamental force operates only among heavier elementary particles. To answer, we need to recall the four fundamental forces of nature and their domains of action.

   1. Gravitational force acts between all objects with mass or energy — it is universal, not restricted to heavier particles. It is the weakest force at the particle scale.

   2. Electromagnetic force acts between all charged particles — electrons, protons, ions, etc. It is also universal for charged matter, not limited to heavier particles.

   3. Weak nuclear force is responsible for certain types of radioactive decay (like beta decay). It acts on both quarks and leptons (including light particles like electrons and neutrinos), so it is not exclusive to heavier particles.

   4. Strong nuclear force is the force that binds quarks together inside protons and neutrons, and holds the nucleus together. It acts only on particles called hadrons — which are the heavier elementary particles made of quarks (protons, neutrons, pions, etc.). It does not act on leptons (like electrons) or photons.

   Watch out

   A common mistake is to think the weak nuclear force also acts only on heavy particles. But the weak force acts on all fermions, including light leptons like electrons and neutrinos — so it is not exclusive to heavier particles.

   Thus, the strong nuclear force is the only fundamental force whose action is confined to heavier elementary particles (hadrons).

   ✓Final answer

   The correct option is (A) Strong nuclear force.

8. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.A body of mass $3\, \text{kg}$ is thrown vertically upward from the ground with a velocity of $10\, \text{ms}^{-1}$. If the maximum height reached by the body is $4.7\, \text{m}$, then the loss of energy due to air resistance is (Acceleration due to gravity $= 10\, \text{ms}^{-2}$) (A) $9\, \text{J}$ (B) $18\, \text{J}$ (C) $30\, \text{J}$ (D) $47\, \text{J}$

   ›Reveal solutionSolution

   The problem asks us to find the energy lost due to air resistance when a body is thrown upwards. We calculate the initial kinetic energy and the potential energy gained at the maximum height. The difference between these two values represents the energy lost to air resistance, which is $\boxed{9\, \text{J}}$.

   When a body is thrown vertically upward, its initial kinetic energy is gradually converted into gravitational potential energy as it rises. In an ideal scenario, where there is no air resistance or any other non-conservative force, the total mechanical energy (kinetic + potential) of the body remains constant. This means that the initial kinetic energy would be entirely converted into potential energy at the maximum height.

   However, in reality, forces like air resistance act against the motion. Air resistance is a non-conservative force, meaning it dissipates mechanical energy, usually converting it into heat and sound. Therefore, when air resistance is present, the final potential energy achieved at the maximum height will be less than the initial kinetic energy. The difference between the initial kinetic energy and the potential energy at the maximum height represents the energy lost due to the work done by air resistance.

   We can express this using the work-energy theorem for non-conservative forces:

   Initial Mechanical Energy + Work done by non-conservative forces = Final Mechanical Energy

   $KE_i + PE_i + W_{nc} = KE_f + PE_f$

   In this problem:

   • The body starts from the ground, so initial potential energy $PE_i = 0$.
   • At the maximum height, the body momentarily stops, so final kinetic energy $KE_f = 0$.
   • The work done by non-conservative forces ($W_{nc}$) is negative, representing the energy lost due to air resistance. Let's call the magnitude of this loss $E_{loss}$. So, $W_{nc} = -E_{loss}$.

   Substituting these into the equation:

   $KE_i + 0 - E_{loss} = 0 + PE_f$

   $KE_i - E_{loss} = PE_f$

   $E_{loss} = KE_i - PE_f$

   This equation tells us that the energy lost is simply the difference between the initial kinetic energy and the potential energy gained at the maximum height.

   1. Calculate the initial kinetic energy of the body.

      The body is thrown upward with an initial velocity. Its kinetic energy at the start is given by the formula $KE = \frac{1}{2}mv^2$.

      $KE = \frac{1}{2}mv^2$

      Given:

      Mass $m = 3\, \text{kg}$

      Initial velocity $u = 10\, \text{ms}^{-1}$

      Initial Kinetic Energy ($KE_i$) $= \frac{1}{2} \times 3\, \text{kg} \times (10\, \text{ms}^{-1})^2$

      $KE_i = \frac{1}{2} \times 3 \times 100\, \text{J}$

      $KE_i = 150\, \text{J}$

   2. Calculate the potential energy of the body at its maximum height.

      As the body rises, its kinetic energy is converted into gravitational potential energy. The potential energy at a certain height is given by $PE = mgh$.

      $PE = mgh$

      Given:

      Mass $m = 3\, \text{kg}$

      Acceleration due to gravity $g = 10\, \text{ms}^{-2}$

      Maximum height reached $h = 4.7\, \text{m}$

      Potential Energy at maximum height ($PE_f$) $= 3\, \text{kg} \times 10\, \text{ms}^{-2} \times 4.7\, \text{m}$

      $PE_f = 30 \times 4.7\, \text{J}$

      $PE_f = 141\, \text{J}$

   3. Determine the loss of energy due to air resistance.

      The initial kinetic energy is the total energy supplied to the body at the start. If there were no losses, this entire energy would be converted into potential energy. However, since the actual potential energy gained is less than the initial kinetic energy, the difference must be the energy lost to air resistance.

      Energy lost due to air resistance ($E_{loss}$) $= KE_i - PE_f$

      $E_{loss} = 150\, \text{J} - 141\, \text{J}$

      $E_{loss} = 9\, \text{J}$

   ✓Final answer

   The loss of energy due to air resistance is $\boxed{9\, \text{J}}$.

9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.In Geiger-Marsden experiment, if the initial speed of $\alpha$ particle is doubled, then the closest distance of approach of the $\alpha$ particle from the gold nucleus (Assume $\alpha$ particle is projected straight towards the gold nucleus) (A) becomes doubled (B) becomes quadrupled (C) becomes half (D) becomes one-fourth

   ›Reveal solutionSolution

   The closest distance of approach is inversely proportional to the kinetic energy of the alpha particle. Doubling the speed quadruples the kinetic energy, so the distance becomes one-fourth. The correct option is (D).

   The Geiger-Marsden experiment (Rutherford's gold foil experiment) is fundamentally about energy conservation. When an alpha particle is fired straight at a gold nucleus, it slows down as it approaches because of the strong electrostatic repulsion between the positively charged alpha particle and the positively charged gold nucleus. At the point of closest approach, the alpha particle momentarily comes to rest — all its initial kinetic energy has been converted into electrostatic potential energy.

   This is a pure energy conversion problem. No forces other than the Coulomb repulsion do work, so mechanical energy is conserved. The key insight: the distance of closest approach $r_0$ is found by equating the initial kinetic energy to the electrostatic potential energy at that distance.

   Let's work through it step by step.

   1. Write the energy conservation equation. At the start, far from the nucleus, the alpha particle has kinetic energy $K = \frac{1}{2} m v^2$ and negligible potential energy. At the closest distance $r_0$, its speed is zero, so all energy is potential:

   $\frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$

   Here $2e$ is the charge of the alpha particle, $Ze$ is the charge of the gold nucleus ($Z=79$), and $m$ is the mass of the alpha particle.

   2. Solve for $r_0$. Rearranging gives:

   $r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{\frac{1}{2} m v^2} = \frac{1}{4\pi\epsilon_0} \frac{4Ze^2}{m v^2}$

   The important thing is the dependence on speed: $r_0 \propto \frac{1}{v^2}$.

   3. Apply the change. If the initial speed is doubled, $v \to 2v$, then:

   $r_0' \propto \frac{1}{(2v)^2} = \frac{1}{4v^2} = \frac{1}{4} r_0$

   The new distance is one-fourth of the original.

   Watch out

   A common mistake is to think $r_0 \propto 1/v$ because kinetic energy is $\frac{1}{2}mv^2$ and you might forget the square. Always write the full energy equation — the $v^2$ in the denominator is unmistakable.

   Tip

   You don't need the actual values of $Z$, $e$, or $\epsilon_0$ for this kind of reasoning. The proportionality $r_0 \propto 1/v^2$ is all you need. This is a hallmark of many "distance of closest approach" problems in electrostatics.

   ✓Final answer

   The closest distance of approach becomes one-fourth, so the correct option is (D).

10. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.A body is falling freely from the top of a tower of height 125 m. The distance covered by the body during the last second of its motion is $x$ % of the height of the tower. Then $x$ is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) 9 (B) 36 (C) 25 (D) 49

    ›Reveal solutionSolution

    The key is to find the total time of fall using $s = \frac12 gt^2$, then compute the distance in the last second as the difference between total distance and distance covered in $(t-1)$ seconds, and finally express that as a percentage of the tower height. The result is $x = 36$, so option (B).

    We start with the concept: for a freely falling body starting from rest, the distance covered in any time interval is given by the equations of motion under constant acceleration. The "last second" distance is tricky because it depends on the total time of fall. Once we know that time, we can subtract the distance covered in all but the last second from the total height.

    ---

    1. Find the total time of fall. The tower height is $h = 125$ m, $g = 10$ m/s², initial velocity $u = 0$. Using $h = \frac12 g t^2$:

    $125 = \frac12 \cdot 10 \cdot t^2 \quad\Rightarrow\quad 125 = 5 t^2 \quad\Rightarrow\quad t^2 = 25 \quad\Rightarrow\quad t = 5 \text{ seconds}.$

    2. Distance covered in the first 4 seconds (i.e., all but the last second).

    $s_{4} = \frac12 g (4)^2 = \frac12 \cdot 10 \cdot 16 = 80 \text{ m}.$

    3. Distance covered in the last second (the 5th second). This is the total height minus the distance covered in the first 4 seconds:

    $\text{last second distance} = 125 - 80 = 45 \text{ m}.$

    4. Express this as a percentage of the tower height.

    $x = \frac{45}{125} \times 100 = \frac{4500}{125} = 36.$

    Tip

    A quick check: the distance covered in the $n$th second of free fall from rest is $g(n - \frac12)$. For $n=5$, that gives $10 \times 4.5 = 45$ m — same result, no subtraction needed.

    Watch out

    A common mistake is to think the last second distance is simply $\frac12 g (1)^2 = 5$ m, but that would only be true if the body fell for exactly one second total. Always find the total time first.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.A body of mass ‘m’ and radius ‘r’ rolling horizontally with a velocity ‘V’, rolls up an inclined plane to a vertical height $\dfrac{V^2}{g}$. The body is (A) a sphere (B) a circular disc (C) a circular ring (D) a solid cylinder

    ›Reveal solutionSolution

    The key idea is that the maximum height reached by a rolling body depends on its moment of inertia via energy conservation. The given height $\frac{V^2}{g}$ corresponds to a body with $k^2/r^2 = 1$, which is a circular ring.

    Concept and Intuition

    When a body rolls without slipping, its total kinetic energy is the sum of translational kinetic energy $\frac12 m V^2$ and rotational kinetic energy $\frac12 I \omega^2$, where $\omega = V/r$ for rolling. As it climbs the incline, this total kinetic energy converts entirely into gravitational potential energy $mgh$. The height $h$ therefore depends on the moment of inertia $I$: a larger $I$ means more rotational energy, so less height for the same initial speed. The problem gives a specific height; we can solve for the required $I$ and match it to the shape.

    ---

    Step-by-step solution

    1. Write the energy conservation equation Initial total kinetic energy = final gravitational potential energy at maximum height $h$:

    $\frac12 m V^2 + \frac12 I \omega^2 = m g h.$

    For rolling without slipping, $\omega = V/r$, so:

    $\frac12 m V^2 + \frac12 I \left(\frac{V}{r}\right)^2 = m g h.$

    2. Factor out $\frac12 m V^2$ Let $I = m k^2$, where $k$ is the radius of gyration. Then:

    $\frac12 m V^2 \left(1 + \frac{k^2}{r^2}\right) = m g h.$

    3. Cancel $m$ and solve for $h$

    $h = \frac{V^2}{2g} \left(1 + \frac{k^2}{r^2}\right).$

    4. Plug in the given height The problem states $h = \dfrac{V^2}{g}$. Substitute:

    $\frac{V^2}{g} = \frac{V^2}{2g} \left(1 + \frac{k^2}{r^2}\right).$

    Cancel $\frac{V^2}{g}$ (nonzero):

    $1 = \frac12 \left(1 + \frac{k^2}{r^2}\right).$

    5. Solve for $\frac{k^2}{r^2}$ Multiply both sides by 2:

    $2 = 1 + \frac{k^2}{r^2} \quad \Rightarrow \quad \frac{k^2}{r^2} = 1.$

    6. Identify the body The condition $k^2/r^2 = 1$ means the moment of inertia is $I = m r^2$. This is exactly the moment of inertia of a circular ring (or thin hoop) about its central axis.
       • Sphere: $k^2/r^2 = 2/5$
       • Circular disc / solid cylinder: $k^2/r^2 = 1/2$
       • Circular ring: $k^2/r^2 = 1$

    Tip

    A quick check: if the body were sliding (no rotation), $h = V^2/(2g)$. The given height is twice that, so the rotational energy equals the translational energy — that happens only when $I = m r^2$, i.e., a ring.

    Watch out

    A common mistake is to forget the rotational kinetic energy and assume $h = V^2/(2g)$, which would incorrectly suggest a sliding object. Always include rotation for a rolling body.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

12. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.The maximum distance between the transmitting and receiving antennas is D. If the heights of both transmitting and receiving antennas are doubled, then the maximum distance between the two antennas is (A) $2D$ (B) $D\sqrt{2}$ (C) $4D$ (D) $\dfrac{D}{\sqrt{2}}$

    ›Reveal solutionSolution

    The maximum line-of-sight distance between two antennas depends on the square root of the antenna heights. Doubling both heights multiplies the distance by $\sqrt{2}$, so the new distance is $D\sqrt{2}$.

    Concept & Intuition

    This problem is about radio horizon — the maximum distance at which two antennas can have a direct line-of-sight, assuming the Earth is a smooth sphere. The key formula comes from geometry: the distance to the horizon from a height $h$ (ignoring atmospheric refraction) is approximately $\sqrt{2Rh}$, where $R$ is the Earth's radius. For two antennas at heights $h_t$ and $h_r$, the total line-of-sight distance is the sum of their individual horizons:

    $D = \sqrt{2R h_t} + \sqrt{2R h_r}$

    If both heights are doubled, each horizon distance increases by a factor of $\sqrt{2}$, so the total also scales by $\sqrt{2}$.

    Step-by-step reasoning

    1. Recall the standard formula for maximum line-of-sight distance between two antennas of heights $h_t$ (transmitter) and $h_r$ (receiver) over a spherical Earth:

    $D = \sqrt{2R h_t} + \sqrt{2R h_r}$

    where $R$ is the Earth's radius. (This assumes standard atmospheric refraction is already accounted for in an effective Earth radius, but the scaling remains the same.)

    2. Apply the given condition: both heights are doubled. So new heights are $h_t' = 2h_t$ and $h_r' = 2h_r$.

    3. Compute the new distance $D'$:

    $D' = \sqrt{2R (2h_t)} + \sqrt{2R (2h_r)} = \sqrt{2} \cdot \sqrt{2R h_t} + \sqrt{2} \cdot \sqrt{2R h_r}$

    Factor out $\sqrt{2}$:

    $D' = \sqrt{2} \left( \sqrt{2R h_t} + \sqrt{2R h_r} \right) = \sqrt{2} \, D$

    4. Interpret the result: The new maximum distance is $D$ multiplied by $\sqrt{2}$, not doubled or quadrupled. This makes physical sense — horizon distance grows slowly with height because of the square-root relationship.

    Watch out

    A common mistake is to think doubling height doubles the distance (option A) or quadruples it (option C). But the square-root dependence means the effect is weaker: only a $\sqrt{2}$ factor.

    Tip

    If both antennas were at the same height $h$, then $D = 2\sqrt{2Rh}$. Doubling $h$ gives $D' = 2\sqrt{2R(2h)} = 2\sqrt{2}\sqrt{2Rh} = \sqrt{2} \cdot (2\sqrt{2Rh}) = \sqrt{2} D$. The same factor appears.

    ✓Final answer

    The correct option is (B).

    ANSWER: B