Which of the following options are correct? (Note: more than one of the given options may be correct.) (a) Acc

NCERT Exemplar · Q9

Q. Which of the following options are correct? (Note: more than one of the given options may be correct.)

(a) Acceleration due to gravity decreases with increasing altitude.

(b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

(d) Acceleration due to gravity is independent of the mass of the earth.

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Concept understanding — Variation Of Gravity

Variation of Gravity: Why Your Weight Changes Even When You Don't

Imagine you step on a weighing scale at sea level in Mumbai, then carry that same scale to the top of Mount Everest. The scale would show a smaller number — you'd weigh less. But you haven't lost any mass. What changed?

The force pulling you down — gravity — is not constant everywhere on Earth. It varies. That's what we mean by variation of gravity.

The Core Idea

Gravity is the force with which the Earth pulls objects toward its centre. The strength of this pull depends on two things: the mass of the Earth and your distance from its centre. Since the Earth is not a perfect sphere and it spins, that distance and the effective pull change from place to place.

The acceleration due to gravity, denoted by $g$ , is approximately $9.8 \, \text{m/s}^2$ at sea level. But that's an average. The actual value can be slightly higher or lower depending on where you are.

Why Does Gravity Vary? Three Main Reasons

1. Altitude (Height Above Sea Level)

This is the most intuitive one. As you go higher, you move farther from the Earth's centre. Gravity follows an inverse-square law: double the distance, and the force becomes one-fourth.

The formula for $g$ at a height $h$ above the Earth's surface (where $R$ is Earth's radius, about 6400 km) is:

g_h = \frac{GM}{(R+h)^2}

For small heights compared to $R$ , we can approximate:

g_h \approx g \left(1 - \frac{2h}{R}\right)

Note

This means for every kilometre you go up, $g$ decreases by roughly $0.003 \, \text{m/s}^2$ . That's why at the top of a tall mountain, you weigh about 0.5% less than at sea level.

2. Depth (Going Underground)

What happens if you go down a mine or into the Earth's crust? Intuition might say gravity increases because you're closer to the centre. But the opposite happens.

Inside the Earth, the mass above you pulls upward, partially cancelling the pull from below. For a uniform Earth, only the mass inside the sphere of radius $r$ (your distance from the centre) contributes to gravity at that point.

g_d = \frac{GM'}{r^2}

Where $M'$ is the mass of the sphere of radius $r$ . If Earth had uniform density $\rho$ , then $M' = \frac{4}{3}\pi r^3 \rho$ , giving:

g_d = \frac{4}{3}\pi G \rho r

This means gravity decreases linearly as you go deeper. At the centre of the Earth, $g = 0$ — you'd be weightless, pulled equally in all directions.

Watch out

This linear decrease assumes uniform density. The real Earth has a dense iron core, so the actual variation is more complicated — gravity actually increases slightly as you go down through the crust before eventually decreasing.

3. Rotation of the Earth (Latitude Effect)

The Earth spins once every 24 hours. This rotation creates a centrifugal force that acts outward, away from the axis of rotation. This force effectively reduces the weight you feel.

The effect is strongest at the equator (where the rotational speed is highest, about 1670 km/h) and zero at the poles (where you're on the axis of rotation).

The effective $g$ at latitude $\phi$ is:

g_{\text{eff}} = g - \omega^2 R \cos^2 \phi

Where $\omega$ is Earth's angular speed ( $7.3 \times 10^{-5} \, \text{rad/s}$ ) and $R$ is Earth's radius.

At the equator ( $\phi = 0^\circ$ ), the reduction is about $0.034 \, \text{m/s}^2$ — roughly 0.35% of $g$ .

Tip

| Location | Approximate $g$ (m/s²) | Why? |

|----------|------------------------|------|

| Equator (sea level) | 9.78 | Fastest rotation + bulging equator |

| 45° latitude | 9.81 | Intermediate |

| North Pole | 9.83 | No rotation effect + closer to centre |

4. Shape of the Earth (Oblateness)

The Earth is not a perfect sphere. Because of its rotation, it bulges at the equator and flattens at the poles. The equatorial radius is about 21 km larger than the polar radius.

This means:

At the poles, you're closer to the Earth's centre → stronger gravity
At the equator, you're farther from the centre → weaker gravity …

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Acceleration due to gravity changes with altitude, depth, and latitude. It decreases with increasing altitude and increases with increasing latitude.

The correct options are (A) and (C).

The acceleration due to gravity, denoted by $g$ , is not constant across the Earth's surface or within its interior. It varies depending on several factors, including altitude, depth, the Earth's rotation (which depends on latitude), and the distribution of mass within the Earth. Understanding these variations is crucial for many applications in physics and geophysics.

Let's analyze each option based on the fundamental principles governing gravitational acceleration.

1. Analyzing Option (A): Acceleration due to gravity decreases with increasing altitude.

The acceleration due to gravity at the Earth's surface is given by $g = \frac{GM}{R^2}$ , where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.

When an object is at an altitude $h$ above the Earth's surface, its distance from the center of the Earth becomes $r = R+h$ . Assuming the Earth is a perfect sphere and its mass is concentrated at its center, the acceleration due to gravity at altitude $h$ , denoted as $g_h$ , is:

g_h = \frac{GM}{(R+h)^2}

We can rewrite this in terms of $g$ :

g_h = \frac{GM}{R^2 \left(1 + \frac{h}{R}\right)^2} = g \left(1 + \frac{h}{R}\right)^{-2}

From this formula, it is clear that as $h$ increases, the denominator $(R+h)^2$ increases, and thus $g_h$ decreases.

Tip

For altitudes much smaller than the Earth's radius ( $h \ll R$ ), we can use the binomial approximation $(1+x)^n \approx 1+nx$ .

$g_h \approx g \left(1 - \frac{2h}{R}\right)$

This approximation clearly shows the decrease in $g$ with increasing altitude.

Therefore, option (A) is correct.

2. Analyzing Option (B): Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

When we consider a point at a depth $d$ below the Earth's surface, its distance from the center of the Earth is $r = R-d$ . To calculate the acceleration due to gravity at this depth, we only consider the mass of the Earth contained within a sphere of radius $r$ . This is because the gravitational forces due to the spherical shell of matter outside this radius cancel out.

Let $\rho$ be the uniform density of the Earth.

The total mass of the Earth is $M = \frac{4}{3}\pi R^3 \rho$ .

The mass of the inner sphere of radius $r = R-d$ is $M' = \frac{4}{3}\pi r^3 \rho$ .

The acceleration due to gravity at depth $d$ , denoted as $g_d$ , is:

g_d = \frac{GM'}{r^2} = \frac{G \left(\frac{4}{3}\pi r^3 \rho\right)}{r^2} = G \frac{4}{3}\pi \rho r

Substitute $r = R-d$ :

g_d = G \frac{4}{3}\pi \rho (R-d)

We know that $g = \frac{GM}{R^2} = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = G \frac{4}{3}\pi R \rho$ .

So, we can write $G \frac{4}{3}\pi \rho = \frac{g}{R}$ .

Substituting this into the expression for $g_d$ :

g_d = \frac{g}{R} (R-d) = g \left(1 - \frac{d}{R}\right)

This formula shows that $g_d$ decreases linearly with increasing depth $d$ .

At the surface ( $d=0$ ), $g_d = g$ .
At the center of the Earth ( $d=R$ ), $g_d = 0$ .

Watch out

A common misconception is that gravity increases as one goes deeper into the Earth because one is getting closer to the center. However, the mass contributing to the gravitational force also decreases, and this effect dominates, leading to a decrease in $g$ .

Therefore, option (B) is incorrect.

3. Analyzing Option (C): Acceleration due to gravity increases with increasing latitude.

The Earth rotates about its axis. Due to this rotation, objects on the surface experience a centrifugal force directed outwards from the axis of rotation. This centrifugal force reduces the effective acceleration due to gravity. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 28 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$30^\circ$
›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.

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

Acceleration at $\theta = 30^\circ$

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

Acceleration at $\theta = 45^\circ$

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

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

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

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.

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.

Division case (Statement II) Let $z = \frac{x}{y}$ . The absolute error is approximately $\Delta z \approx \frac{1}{y} \Delta x + \frac{x}{y^2} \Delta y$ …
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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

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.

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.

Confirm with a classic test …
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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.

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.

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.

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. …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$4.5\, \text{kg}$
›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.

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:

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

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

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. …
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TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$1 : \sqrt{2} : \sqrt{3}$
›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.

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 $\sqrt{s}$ .

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

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

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

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

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

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.

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

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. …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$3\, \text{kg}$
›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.

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

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

$KE_i = 150\, \text{J}$ …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$\alpha$
›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.

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.

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: $r_0 \propto \frac{1}{v^2}$ . …
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TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$x$
›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.

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

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

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

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

Distance covered in the last second (the 5th second). This is the total height minus the distance covered in the first 4 seconds: $\text{last second distance} = 125 - 80 = 45 \text{ m}.$ …
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$\dfrac{V^2}{g}$
›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

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

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

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

Cancel $m$ and solve for $h$

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

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

Solve for $\frac{k^2}{r^2}$ Multiply both sides by 2: $2 = 1 + \frac{k^2}{r^2} \quad \Rightarrow \quad \frac{k^2}{r^2} = 1. …$
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$2D$
›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

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.)

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

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 $\sqrt{2}$ :
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Q.Which of the following options are correct? (Note: more than one of the given options may be correct.)

1. Analyzing Option (A): Acceleration due to gravity decreases with increasing altitude.

2. Analyzing Option (B): Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).

3. Analyzing Option (C): Acceleration due to gravity increases with increasing latitude.

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