A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during th

NCERT Exemplar · Q3

Q. A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is

(a) constant and equal to

mg

in magnitude.

(b) constant and greater than

mg

in magnitude.

mg

(d) at first greater than

mg

, and later becomes equal to

mg

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Concept understanding — Newton Third Law Pairs

Newton's Third Law Pairs

The Intuition: Forces Never Act Alone

Imagine you are standing on a skateboard and you push against a wall. The wall does not move, but you roll backward. You pushed the wall, and the wall pushed back on you — at the very same instant, with the very same strength.

This is the core idea behind Newton's Third Law: whenever one object exerts a force on a second object, the second object exerts an equal and opposite force back on the first. These two forces are called an action-reaction pair (or "Third Law pair").

Note

The two forces in a Third Law pair act at the same instant — there is no delay between "action" and "reaction." They are two sides of a single interaction, not a cause followed by an effect.

The Precise Statement

Newton's Third Law: If object A exerts a force on object B, then object B exerts an equal and opposite force on object A.

\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}

Both forces have the same magnitude and act along the same line, but point in opposite directions — and, crucially, they act on two different objects.

The Key Rule for Spotting a Third Law Pair

A genuine Third Law pair must satisfy two conditions:

The two forces act on different objects (never on the same object).
They arise from the same interaction (the same pair of bodies in contact, or the same gravitational/electric attraction).

If a pair of forces acts on the same object, they are not a Third Law pair — even if they happen to be equal in magnitude and opposite in direction.

Why Equal and Opposite Forces Don't Cancel

This confuses many students: if the two forces are always equal and opposite, why does anything ever move?

Because the two forces act on different objects — they never appear together in the same free-body diagram. When you push against a wall, the wall pushes back on you. That reaction force acts on you, not on the wall, so it can accelerate you backward even though the wall itself does not move (the wall is anchored to the Earth, which is far too massive to notice the force).

A Careful Worked Example: A Book on a Table

A book sits at rest on a table. Two different force pairs are involved here, and it is important not to mix them up:

Pair 1 (gravitational interaction): Earth pulls the book downward with a force equal to the book's weight, $mg$ . By the Third Law, the book pulls the Earth upward with the same magnitude, $mg$ . This pair acts on two different bodies — the book and the Earth.
Pair 2 (contact interaction): The table pushes up on the book with a normal force $N$ . By the Third Law, the book pushes down on the table with a force of the same magnitude, $N$ . This pair also acts on two different bodies — the book and the table.

Watch out

A very common mistake is to say "the normal force equals the weight because of Newton's Third Law." This is wrong. The book's weight and the table's normal force both act on the same object (the book), so they cannot be a Third Law pair — they merely happen to be equal in magnitude here because the book is in equilibrium (Newton's second law with zero acceleration, $N - mg = 0$ ). The true Third Law partner of the book's weight is the pull the book exerts on the Earth; the true Third Law partner of the normal force is the push the book exerts on the table.

More Everyday Examples

Walking: Your foot pushes backward on the ground; the ground pushes forward on your foot. That forward push is what propels you ahead.
Rocket propulsion: The rocket pushes exhaust gases backward; the gases push the rocket forward. This works even in the vacuum of space, since no air is needed — only the exhaust and the rocket exerting forces on each other.
Swimming: A swimmer pushes water backward with their hands and feet; the water pushes the swimmer forward.

How to Use This in Problems

For every force you draw on an object, ask: "What is the Third Law partner of this force, and on what other object does it act?"
Draw a separate free-body diagram for each object. A Third Law pair will show up as one arrow in each of two different diagrams — same length, opposite direction, same type of force (both gravitational, or both contact, or both electric).
Never draw both members of a Third Law pair on the same free-body diagram — they belong to different objects.

The Deeper Reason

Newton's Third Law is closely tied to the conservation of momentum. In any isolated interaction between two objects, whatever momentum one gains, the other loses — because the forces they exert on each other are equal and opposite at every instant. This is why the law holds for every kind of fundamental interaction: gravitational, electromagnetic, and even contact forces, which are ultimately electromagnetic in origin.

Students preparing for boards often pair a search for "Newton Third Law Pairs class 11 physics" with "NCERT Physics syllabus" — Newton Third Law Pairs is a syllabus-aligned topic under Laws of Motion in NCERT Class 11 Physics, making it a natural fit for both board exams and JEE/NEET practice sets. Working through the worked examples above alongside the official NCERT Physics textbook is the most reliable way to turn this understanding into exam-ready recall.

The ground reaction force must first exceed $mg$ to accelerate the man upward from rest, then return to $mg$ once he moves at constant speed (or stops). The correct option is (D).

The key is to think about Newton’s second law — not just the third law. When the man squats and then stands up, his centre of mass does not move at constant velocity. It starts at rest, accelerates upward, then decelerates to rest again at the top. The ground reaction force is the upward normal force $N$ from the floor. The man’s weight $mg$ acts downward. The net force on the man is $N - mg$ , and this equals $ma$ , where $a$ is the acceleration of his centre of mass.

If the man simply stood still, $N = mg$ . But during the act of standing, his centre of mass must gain upward speed, so there must be a period of upward acceleration. That requires $N > mg$ . Later, as he approaches the upright position, he must slow down (decelerate upward), which means $N < mg$ briefly. However, the question’s options only mention “greater than $mg$ ” and “equal to $mg$ ”, so the simplest correct description is that $N$ is first greater than $mg$ , then becomes equal to $mg$ once he is stationary.

Let’s walk through the phases.

Initial state (squatting, at rest)

The man is stationary on the ground. His centre of mass has zero velocity. The net force is zero, so $N = mg$ . But this is only the starting point — the process hasn’t begun yet.
Beginning to stand — upward acceleration

To start moving upward, the man must push harder against the ground. By Newton’s third law, the ground pushes back with an equal and opposite force. So $N$ becomes greater than $mg$ . The net upward force $N - mg$ provides the upward acceleration $a$ .

N - mg = ma \quad \Rightarrow \quad N = m(g + a) > mg

Middle of the motion — possible constant speed

If the man rises at constant speed for a while, then $a = 0$ and $N = mg$ during that interval. But in a natural squat-to-stand movement, the acceleration phase is short and followed by deceleration. The question’s options don’t mention a period where $N < mg$ , so the simplest match is that after the initial acceleration, $N$ returns to $mg$ as the man becomes stationary.
Nearing the top — deceleration

To stop at the upright position, the man must have a downward acceleration (i.e., upward deceleration). That would require $N < mg$ . However, the options given do not include “less than $mg$ ” at any stage. So the intended answer focuses on the fact that $N$ is first greater than $mg$ (to start the motion) and later equal to $mg$ (when at rest or moving at constant speed).

Watch out

A common mistake is to think that because the man is “pushing” on the ground, the reaction is always greater than $mg$ . But once he is moving at constant speed or is stationary, the net force is zero, so $N = mg$ . The extra force is only needed to change his speed.

Tip

Think of standing up as a controlled upward throw of your own body. To throw something upward, you must exert a force greater than its weight initially. Once it’s moving, you can ease off.

Thus, the reaction force is not constant — it varies — and it is greater than $mg$ only during the upward acceleration phase, then equal to $mg$ when the man is at rest (or moving uniformly).

✓Final answer

mg

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 19 on this concept.

TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$\sqrt{\dfrac{T^2}{M^2} - g^2}$
›Reveal solutionSolution
The rope’s tension must provide both the vertical force to balance weight and the horizontal force to accelerate the mass; the maximum acceleration is found from the vector sum of these forces, giving $a_{\text{max}} = \sqrt{(T/M)^2 - g^2}$ .

Concept and intuition

The rope can only pull along its own length. When the aeroplane accelerates horizontally, the hanging mass is no longer vertical — it swings back so that the rope’s tension has both a vertical component (to hold up the weight) and a horizontal component (to provide the acceleration). The rope’s maximum tension $T$ is fixed, so the largest possible horizontal acceleration occurs when the rope is at the critical angle where the tension’s vertical component exactly equals $Mg$ and the horizontal component is as large as possible. This is a classic case of vector resolution under a constraint.

Set up the forces

The mass $M$ experiences two forces: its weight $Mg$ downward, and the rope tension $\vec{T}$ along the rope. The net force causes horizontal acceleration $a$ .

Resolve tension into vertical ( $T \cos\theta$ ) and horizontal ( $T \sin\theta$ ) components, where $\theta$ is the angle the rope makes with the vertical.

Apply Newton’s second law in each direction

Vertical: no vertical acceleration, so $T \cos\theta = Mg$ .

Horizontal: $T \sin\theta = M a$ .

Eliminate the angle

From the vertical equation: $\cos\theta = \frac{Mg}{T}$ .

From the horizontal: $\sin\theta = \frac{Ma}{T}$ .

Use $\sin^2\theta + \cos^2\theta = 1$ :

$\left(\frac{Ma}{T}\right)^2 + \left(\frac{Mg}{T}\right)^2 = 1$
Multiply through by $T^2$ :
$M^2 a^2 + M^2 g^2 = T^2$

Solve for acceleration

$a^2 = \frac{T^2}{M^2} - g^2 \quad\Rightarrow\quad a = \sqrt{\frac{T^2}{M^2} - g^2}$
(Only the positive root is physically meaningful.)

Watch out

A common mistake is to think the tension simply equals $M a$ or $M g$ separately — but tension must simultaneously satisfy both conditions. The vector sum is what matters, not a scalar difference.

Tip

Notice that if the rope were vertical ( $a=0$ ), then $T = Mg$ , which matches the formula: $\sqrt{(Mg/M)^2 - g^2} = 0$ . If $a$ is large, the rope must be nearly horizontal, so $T$ is mostly horizontal — the formula correctly gives $a \approx T/M$ when $g$ is negligible.

✓Final answer
$The correct option is (A) . ANSWER: A$
TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$20~\mathrm{ms}^{-1}$
›Reveal solutionSolution
The key is to find the time when P’s velocity becomes zero, then compute the positions of both balls at that instant. The distance between them is 50 m.

The problem is about two balls thrown vertically upward with different initial speeds. When ball P reaches its highest point, its velocity becomes zero. At that moment, ball Q is still moving upward. The distance between them is simply the difference in their heights at that time.

We use the equations of motion under constant acceleration (gravity). The acceleration is $g = 10 \, \text{m/s}^2$ downward, so we take upward as positive.

Find the time when P’s velocity becomes zero. For ball P: initial velocity $u_P = 20 \, \text{m/s}$ , final velocity $v_P = 0$ , acceleration $a = -g = -10 \, \text{m/s}^2$ . Using $v = u + at$ :

$0 = 20 + (-10)t \quad \Rightarrow \quad t = \frac{20}{10} = 2 \, \text{s}.$

Find the height of ball P at this time. Using $s = ut + \frac{1}{2}at^2$ :

$s_P = 20 \times 2 + \frac{1}{2}(-10)(2)^2 = 40 - 20 = 20 \, \text{m}.$
So P is 20 m above the ground.

Find the height of ball Q at the same time ( $t = 2$ s). For ball Q: $u_Q = 35 \, \text{m/s}$ , $a = -10 \, \text{m/s}^2$ .

$s_Q = 35 \times 2 + \frac{1}{2}(-10)(2)^2 = 70 - 20 = 50 \, \text{m}.$
So Q is 50 m above the ground.

Distance between them is simply the difference in heights:

$s_Q - s_P = 50 - 20 = 30 \, \text{m}.$
Watch out

A common mistake is to think the distance is the difference in their displacements from the thrower, which is correct — but some students mistakenly use the time when Q’s velocity becomes zero instead. Always check which ball’s velocity condition is given.

✓Final answer
$\boxed{30\ \text{m}}$
TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$= 10\,\mathrm{ms}^{-2}$
›Reveal solutionSolution
The relative velocity between two freely falling stones is constant because both accelerate at the same rate $g$ . The first stone has a head start of $1\,\text{s}$ over the third, so their relative velocity is $g \times 1 = 10\,\text{m/s}$ .

The key insight: when two objects are in free fall under the same gravity, their relative acceleration is zero. That means their relative velocity stays constant once both are in motion. So we don’t need to track positions or heights — we just need the velocity difference at the moment the third stone is dropped.

Understand the time gap. The stones are dropped every $0.5\,\text{s}$ . So the first stone is released at $t=0$ , the second at $t=0.5\,\text{s}$ , and the third at $t=1.0\,\text{s}$ . By the time the third stone is just released, the first stone has already been falling for $1$ second.

Velocity of a freely falling body. Starting from rest, after time $t$ the velocity is $v = g t$ , with $g = 10\,\text{m/s}^2$ .

Velocity of the first stone at $t = 1\,\text{s}$ : $v_1 = 10 \times 1 = 10\,\text{m/s}$ downward.

Velocity of the third stone at the instant it is released ( $t=1\,\text{s}$ ): $v_3 = 0$ (it just starts from rest).

Relative velocity at that instant. The relative velocity of the first stone with respect to the third is $v_1 - v_3 = 10 - 0 = 10\,\text{m/s}$ downward.

Why this relative velocity stays constant. After $t=1\,\text{s}$ , both stones are in free fall. Their accelerations are identical ( $g$ downward), so the relative acceleration is $g - g = 0$ . Hence the relative velocity remains $10\,\text{m/s}$ for as long as both are in motion (until the first hits the ground).

Watch out

A common mistake is to compute velocities at different times (e.g., after 2 seconds for the first and 1 second for the third) and subtract — that gives the same answer here, but the reasoning is less clean. The cleanest path: recognise that relative acceleration is zero, so the relative velocity is simply $g$ times the time gap between their releases.

Tip

For any two stones dropped from the same height with a constant time interval $\Delta t$ , the relative velocity is always $g \Delta t$ , independent of how long they’ve been falling. Here $\Delta t = 1\,\text{s}$ (between first and third), so $10 \times 1 = 10\,\text{m/s}$ .

✓Final answer
$10\,\text{m/s}$
TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$= 10\, \mathrm{ms}^{-2}$
›Reveal solutionSolution
The problem splits into two phases: free fall under gravity for 20 m, then uniform motion at the final speed for the remaining 1980 m. Using $g=10\ \mathrm{m/s^2}$ , the free‑fall time is $2\ \mathrm{s}$ and the uniform‑motion time is $99\ \mathrm{s}$ , giving a total of $101\ \mathrm{s}$ . The correct option is (B).

Concept and intuition

The jumper first accelerates from rest under gravity. Once the parachute opens, the problem says he continues “uniformly with the velocity attained” — meaning no further acceleration or deceleration. So the motion is in two clean stages:

Free fall (constant acceleration $g$ ) over a short distance.

Constant‑speed motion over the remaining distance.

The key is to find the speed at the end of free fall, then use it as the constant speed for the second stage. The total time is simply the sum of the times for each stage.

Step‑by‑step solution

Identify the distances

Total height: $2000\ \mathrm{m}$ .

Free‑fall distance: $20\ \mathrm{m}$ .

Remaining distance after parachute opens: $2000 - 20 = 1980\ \mathrm{m}$ .

Find the velocity at the end of free fall

For constant acceleration from rest: $v^2 = u^2 + 2as$ , with $u=0$ , $a=g=10\ \mathrm{m/s^2}$ , $s=20\ \mathrm{m}$ .

$v^2 = 0 + 2 \cdot 10 \cdot 20 = 400 \quad\Rightarrow\quad v = 20\ \mathrm{m/s}.$
So the parachute opens when the jumper is moving at $20\ \mathrm{m/s}$ downward.

Find the time taken during free fall Using $v = u + at$ :

$20 = 0 + 10\,t_1 \quad\Rightarrow\quad t_1 = 2\ \mathrm{s}.$

Find the time taken after the parachute opens Now motion is uniform at $v = 20\ \mathrm{m/s}$ over $1980\ \mathrm{m}$ :

$t_2 = \frac{\text{distance}}{\text{speed}} = \frac{1980}{20} = 99\ \mathrm{s}.$

Total time

$t_{\text{total}} = t_1 + t_2 = 2 + 99 = 101\ \mathrm{s}.$

Watch out

A common mistake is to forget that the free‑fall distance is only 20 m, not the whole 2 km. Another is to use $s = \frac{1}{2}gt^2$ for the whole drop — that would give a much smaller time and a wrong answer.

Tip

Notice that the free‑fall time is very short (2 s) because the distance is small. Almost all the time (99 s) is spent drifting down at constant speed. This makes sense: a parachute drastically limits speed.

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$\sqrt{2}$
›Reveal solutionSolution
The block starts moving when the horizontal component of the applied force just overcomes the maximum static friction. Solving the force balance gives $F = 4~\mathrm{N}$ , so the correct option is (C).

The key idea is that the block is on the verge of slipping, so static friction is at its maximum value $f_{\text{max}} = \mu_s N$ . But the applied force has an upward component, which reduces the normal reaction $N$ — and therefore reduces the friction. You can’t just use $F = \mu_s mg$ ; you must account for the vertical component of $F$ .

Set up the forces

The block has mass $m = \sqrt{2}~\text{kg}$ , so its weight is $mg = \sqrt{2} \times 10 = 10\sqrt{2}~\text{N}$ .

The force $F$ acts at $45^\circ$ above the horizontal. Its components are:

Horizontal: $F \cos 45^\circ = \frac{F}{\sqrt{2}}$

Vertical: $F \sin 45^\circ = \frac{F}{\sqrt{2}}$ (upward)

Normal reaction

The upward pull reduces the normal force from the surface. Vertical equilibrium (no vertical acceleration) gives:

$N + \frac{F}{\sqrt{2}} = mg \quad\Rightarrow\quad N = mg - \frac{F}{\sqrt{2}} = 10\sqrt{2} - \frac{F}{\sqrt{2}}$

Maximum static friction The coefficient of static friction is $\mu_s = 0.25 = \frac14$ . So:

$f_{\text{max}} = \mu_s N = \frac14 \left(10\sqrt{2} - \frac{F}{\sqrt{2}}\right)$

Condition for impending motion The block just starts moving when the horizontal pull equals the maximum friction:

$\frac{F}{\sqrt{2}} = \frac14 \left(10\sqrt{2} - \frac{F}{\sqrt{2}}\right)$

Solve for $F$ Multiply both sides by 4:

$\frac{4F}{\sqrt{2}} = 10\sqrt{2} - \frac{F}{\sqrt{2}}$
Add $\frac{F}{\sqrt{2}}$ to both sides:
$\frac{5F}{\sqrt{2}} = 10\sqrt{2}$
Multiply both sides by $\sqrt{2}$ :
$5F = 10 \times 2 = 20 \quad\Rightarrow\quad F = 4~\text{N}$
Tip

A common mistake is to forget the upward component and write $F = \mu_s mg = 0.25 \times 10\sqrt{2} \approx 3.54~\text{N}$ , which isn’t even one of the options. The upward pull reduces the normal force, so the required $F$ is actually larger than that naive value — here it’s exactly $4~\text{N}$ .

Watch out

Don’t forget that $mg = 10\sqrt{2} \approx 14.14~\text{N}$ , not $10~\text{N}$ ! The mass is $\sqrt{2}~\text{kg}$ , not $1~\text{kg}$ .

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$\sqrt{2}$
›Reveal solutionSolution
The block starts moving when the horizontal component of the applied force just overcomes the maximum static friction. The required force is $F = 4~\mathrm{N}$ , which corresponds to option (C).

The key here is that the applied force is not horizontal — it acts at $45^\circ$ upward. That means part of it reduces the normal reaction, and therefore reduces the friction that opposes motion. Many students forget this and treat friction as $\mu mg$ , which would give a wrong answer. The correct approach is to write the balance of forces in both the vertical and horizontal directions at the instant motion begins.

Draw the free-body diagram. The forces on the block are:

Weight $mg$ downward, where $m = \sqrt{2}~\mathrm{kg}$ and $g = 10~\mathrm{m/s}^2$ , so $mg = 10\sqrt{2}~\mathrm{N}$ .

Normal reaction $N$ upward from the surface.

Applied force $F$ at $45^\circ$ above the horizontal. Its components: horizontal $F\cos45^\circ = F/\sqrt{2}$ , vertical upward $F\sin45^\circ = F/\sqrt{2}$ .

Static friction $f_s$ opposing the impending motion, horizontally. Its maximum value is $f_{s,\max} = \mu_s N$ , with $\mu_s = 0.25$ .

Vertical equilibrium. The block does not lift off, so the net vertical force is zero:

$N + F\sin45^\circ = mg$ $N + \frac{F}{\sqrt{2}} = 10\sqrt{2}$
Hence,
$N = 10\sqrt{2} - \frac{F}{\sqrt{2}}$
Notice: the normal force is less than $mg$ because the vertical component of $F$ helps support the block. This is the crucial point — a smaller $N$ means a smaller maximum friction.

Horizontal condition for impending motion. The block just starts moving when the horizontal pull equals the maximum static friction:

$F\cos45^\circ = \mu_s N$ $\frac{F}{\sqrt{2}} = 0.25 \left(10\sqrt{2} - \frac{F}{\sqrt{2}}\right)$

Solve for $F$ . Multiply both sides by $\sqrt{2}$ to simplify:

$F = 0.25 \left(10\sqrt{2} \cdot \sqrt{2} - F\right)$
Since $\sqrt{2} \cdot \sqrt{2} = 2$ ,
$F = 0.25 \left(20 - F\right)$ $F = 5 - 0.25F$ $F + 0.25F = 5$ $1.25F = 5$ $F = \frac{5}{1.25} = 4~\mathrm{N}$
Watch out

A common mistake is to take $N = mg$ and write $F\cos45^\circ = \mu mg$ , giving $F = \mu mg / \cos45^\circ = 0.25 \times 10\sqrt{2} \times \sqrt{2} = 5~\mathrm{N}$ . That is not among the options — and it is wrong because it ignores the reduction in normal force due to the upward pull.

✓Final answer
$\boxed{4~\mathrm{N}}$
TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$3R_E$
›Reveal solutionSolution
The weight of a body is directly proportional to the acceleration due to gravity, which decreases with the square of the distance from the Earth's center. Using this relationship, the weight of the body at a height of $R_E$ from the surface is $\boxed{360~\mathrm{N}}$ .

The weight of a body is the force exerted on it due to gravity. This force depends on the body's mass and the local acceleration due to gravity ( $g$ ). As we move away from the Earth's surface, the acceleration due to gravity decreases, and consequently, the body's weight also decreases.

The acceleration due to gravity at a height $h$ from the Earth's surface is given by the formula:
$g_h = \frac{GM}{(R_E + h)^2}$
At the Earth's surface (where $h=0$ ), the acceleration due to gravity is $g_0 = \frac{GM}{R_E^2}$ .

Therefore, we can write the acceleration due to gravity at height $h$ in terms of $g_0$ :
$g_h = \frac{GM}{(R_E + h)^2} = \frac{GM}{R_E^2 \left(1 + \frac{h}{R_E}\right)^2} = \frac{g_0}{\left(1 + \frac{h}{R_E}\right)^2}$
And the weight at height $h$ is:
$W_h = m g_h = \frac{m g_0}{\left(1 + \frac{h}{R_E}\right)^2}$
Here, $mg_0$ represents the weight of the body on the Earth's surface.

Let's use this relationship to solve the problem.

Determine the weight on the Earth's surface ( $mg_0$ ). We are given that the weight of the body at a height $h_1 = 3R_E$ from the surface is $W_1 = 90~\mathrm{N}$ . Using the formula for weight at height $h$ :

$W_1 = \frac{mg_0}{\left(1 + \frac{h_1}{R_E}\right)^2}$
Substitute $h_1 = 3R_E$:
$90~\mathrm{N} = \frac{mg_0}{\left(1 + \frac{3R_E}{R_E}\right)^2}$ $90~\mathrm{N} = \frac{mg_0}{(1 + 3)^2}$ $90~\mathrm{N} = \frac{mg_0}{4^2}$ $90~\mathrm{N} = \frac{mg_0}{16}$
Now, we can find the weight of the body on the Earth's surface, $mg_0$:
$mg_0 = 90~\mathrm{N} \times 16 = 1440~\mathrm{N}$
This means the body would weigh $1440~\mathrm{N}$ if it were on the surface of the Earth.

2. Calculate the weight at the new height ( $h_2$ ).

We need to find the weight of the same body at a height $h_2 = R_E$ from the surface of the Earth.

Using the same formula:
$W_2 = \frac{mg_0}{\left(1 + \frac{h_2}{R_E}\right)^2}$
Substitute $h_2 = R_E$:
$W_2 = \frac{mg_0}{\left(1 + \frac{R_E}{R_E}\right)^2}$ $W_2 = \frac{mg_0}{(1 + 1)^2}$ $W_2 = \frac{mg_0}{2^2}$ $W_2 = \frac{mg_0}{4}$

Substitute the value of $mg_0$ to find $W_2$ . From Step 1, we found $mg_0 = 1440~\mathrm{N}$ .

$W_2 = \frac{1440~\mathrm{N}}{4}$ $W_2 = 360~\mathrm{N}$
Watch out

Do not use the approximation $g_h \approx g_0 \left(1 - \frac{2h}{R_E}\right)$ for large heights like $h=R_E$ or $h=3R_E$ . This approximation is only valid when $h \ll R_E$ . For heights comparable to the Earth's radius, the exact inverse square law must be used.

The correct option is (C).

✓Final answer
$R_E$
TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$10~\mathrm{ms}^{-1}$
›Reveal solutionSolution
The net change in momentum equals the sum of the impulses from both forces. Using the impulse-momentum theorem, the mass is found to be 5 kg.

The core idea here is the impulse-momentum theorem: the change in momentum of a body equals the total impulse applied to it. Impulse is force multiplied by time, and since forces act in opposite directions, we must assign signs carefully. The final velocity is given in the direction of the 60 N force, so we take that as positive.

Let’s work through it step by step.

Define the direction.

Let the direction of the 60 N force be positive. Then the 20 N force acts in the negative direction (opposite to the final motion).

Initial velocity $u = 0$ (body at rest).

Final velocity $v = +10~\mathrm{m/s}$ .

Calculate the impulse from each force.

Impulse = force × time.

First force: $F_1 = -20~\mathrm{N}$ , time $t_1 = 2~\mathrm{s}$ Impulse $I_1 = (-20)(2) = -40~\mathrm{N\,s}$

Second force: $F_2 = +60~\mathrm{N}$ , time $t_2 = 1.5~\mathrm{s}$ Impulse $I_2 = (60)(1.5) = +90~\mathrm{N\,s}$

Total impulse = net change in momentum.

Total impulse $I = I_1 + I_2 = -40 + 90 = +50~\mathrm{N\,s}$

By the impulse-momentum theorem:

$I = m v - m u = m(v - u)$
Since $u = 0$ , we have $I = m v$ .

Solve for mass.

$50 = m \times 10$ $m = \frac{50}{10} = 5~\mathrm{kg}$
Watch out

A common mistake is to forget that the 20 N force acts in the opposite direction to the final motion. If you take both forces as positive, you get $m = 12~\mathrm{kg}$ , which is not among the options — a clear sign of the error.

Tip

You can also think of this as: the net impulse is the area under the force-time graph, with sign. Here, the graph has a negative rectangle of area 40 and a positive rectangle of area 90, giving net area 50 N·s.

✓Final answer
$\boxed{5~\mathrm{kg}}$
TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$4\, \text{kg}$
›Reveal solutionSolution
The loss of energy due to air resistance is the difference between the initial and final mechanical energy of the body. The initial mechanical energy is $1200\, \text{J}$ and the final mechanical energy is $600\, \text{J}$ , leading to a loss of $\boxed{600\, \text{J}}$ .

When a body falls under gravity, its mechanical energy (the sum of its kinetic and potential energy) would remain constant if gravity were the only force acting. This is the principle of conservation of mechanical energy. However, in real-world scenarios, forces like air resistance also act on the body. Air resistance is a non-conservative force, meaning it dissipates mechanical energy, usually converting it into heat and sound.

The work done by non-conservative forces, such as air resistance, is equal to the change in the total mechanical energy of the system. Specifically, the loss of mechanical energy is equal to the negative work done by air resistance. In simpler terms, the energy "lost" due to air resistance is the difference between the initial total mechanical energy and the final total mechanical energy.

Here's how we calculate the energy loss:

Identify the initial state and calculate initial mechanical energy:

The body starts falling freely from rest from a height of $h_1 = 30\, \text{m}$ .

Mass of the body, $m = 4\, \text{kg}$ .

Initial height, $h_1 = 30\, \text{m}$ .

Initial velocity, $v_1 = 0\, \text{ms}^{-1}$ (since it falls from rest).

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

The initial potential energy ( $PE_1$ ) is given by $mgh_1$ :

$PE_1 = mgh_1 = (4\, \text{kg})(10\, \text{ms}^{-2})(30\, \text{m}) = 1200\, \text{J}$
The initial kinetic energy ($KE_1$) is given by $\frac{1}{2}mv_1^2$:
$KE_1 = \frac{1}{2}mv_1^2 = \frac{1}{2}(4\, \text{kg})(0\, \text{ms}^{-1})^2 = 0\, \text{J}$
The total initial mechanical energy ($E_1$) is the sum of initial potential and kinetic energy:
$E_1 = PE_1 + KE_1 = 1200\, \text{J} + 0\, \text{J} = 1200\, \text{J}$

Identify the final state and calculate final mechanical energy:

The body is at a height of $h_2 = 10\, \text{m}$ from the ground, and its velocity is $v_2 = 10\, \text{ms}^{-1}$ .

Final height, $h_2 = 10\, \text{m}$ .

Final velocity, $v_2 = 10\, \text{ms}^{-1}$ .

The final potential energy ( $PE_2$ ) is given by $mgh_2$ :

$PE_2 = mgh_2 = (4\, \text{kg})(10\, \text{ms}^{-2})(10\, \text{m}) = 400\, \text{J}$
The final kinetic energy ($KE_2$) is given by $\frac{1}{2}mv_2^2$:
$KE_2 = \frac{1}{2}mv_2^2 = \frac{1}{2}(4\, \text{kg})(10\, \text{ms}^{-1})^2 = 2\, \text{kg} \times 100\, \text{m}^2\text{s}^{-2} = 200\, \text{J}$
The total final mechanical energy ($E_2$) is the sum of final potential and kinetic energy:
$E_2 = PE_2 + KE_2 = 400\, \text{J} + 200\, \text{J} = 600\, \text{J}$

Calculate the loss of energy due to air resistance: The loss of energy due to air resistance is the difference between the initial total mechanical energy and the final total mechanical energy.

$\text{Loss of energy} = E_1 - E_2$ $\text{Loss of energy} = 1200\, \text{J} - 600\, \text{J} = 600\, \text{J}$
Watch out

Do not confuse the work done by gravity with the total mechanical energy. Gravity is a conservative force, and its work changes potential energy. Air resistance is a non-conservative force, and its work changes the total mechanical energy.

✓Final answer
$\boxed{600\, \text{J}}$
TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$35\ \mathrm{ms}^{-1}$
›Reveal solutionSolution
The key is that the body’s speed is the absolute value of its velocity, which changes linearly with time due to gravity. At 3 s the velocity is +5 m/s (upward), at 4 s it is –5 m/s (downward), so both speeds are 5 m/s — ratio 1:1.

Concept & Intuition

When you throw a ball upward, gravity steadily reduces its upward speed by 10 m/s every second. After reaching the top, it falls back down, gaining speed in the downward direction. The speed (magnitude of velocity) at two different times can be the same if one time is on the way up and the other on the way down — symmetric points about the peak. Here, the times 3 s and 4 s are exactly such a pair.

Step-by-step reasoning

Choose a sign convention

Let upward be positive. Initial velocity $u = +35\ \mathrm{m/s}$ . Acceleration due to gravity is downward, so $a = -10\ \mathrm{m/s^2}$ .

Velocity at any time

Using $v = u + at$ :

$v(t) = 35 - 10t \quad (\text{in m/s})$

Velocity at $t = 3\ \mathrm{s}$

$v(3) = 35 - 10(3) = 35 - 30 = +5\ \mathrm{m/s}$
Positive means still moving upward. Speed = $|+5| = 5\ \mathrm{m/s}$ .

Velocity at $t = 4\ \mathrm{s}$

$v(4) = 35 - 10(4) = 35 - 40 = -5\ \mathrm{m/s}$
Negative means moving downward. Speed = $|-5| = 5\ \mathrm{m/s}$ .

Ratio of speeds Speed at 3 s : Speed at 4 s = $5 : 5 = 1 : 1$ .

Tip

The time to reach the highest point is $t = u/g = 35/10 = 3.5\ \mathrm{s}$ . Times 3 s and 4 s are equally spaced 0.5 s before and after the peak — so speeds are equal by symmetry.

Watch out

A common mistake is to use the signed velocities (+5 and –5) directly and get a ratio of 1:–1, forgetting that speed is the absolute value. Always check whether the question asks for velocity or speed.

✓Final answer
$The correct option is (B) . ANSWER: B$
TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$50~\mathrm{ms}^{-1}$
›Reveal solutionSolution
The key idea is Bernoulli’s principle: faster airflow above the wing creates lower pressure, producing a net upward force (lift). Comparing lift to the plane’s weight shows lift > weight, so the plane gains altitude — answer (A).

Concept & Intuition

An aeroplane wing is shaped so that air moves faster over the top surface than below it. Bernoulli’s principle tells us that where speed is higher, pressure is lower. This pressure difference pushes upward on the wing, creating lift. If lift exceeds the plane’s weight, the plane rises; if lift is less, it descends; if equal, it flies level. Here we compute the lift force and compare it to the weight.

Step-by-step reasoning

Apply Bernoulli’s equation For horizontal flow (ignoring small height changes across the wing), Bernoulli’s equation is

$P_{\text{top}} + \frac{1}{2}\rho v_{\text{top}}^2 = P_{\text{bottom}} + \frac{1}{2}\rho v_{\text{bottom}}^2$
where $\rho = 1.3~\text{kg/m}^3$ , $v_{\text{top}} = 50~\text{m/s}$ , $v_{\text{bottom}} = 40~\text{m/s}$ .

Find the pressure difference Rearranging:

$P_{\text{bottom}} - P_{\text{top}} = \frac{1}{2}\rho (v_{\text{top}}^2 - v_{\text{bottom}}^2)$
Substitute values:
$\Delta P = \frac{1}{2} \times 1.3 \times (50^2 - 40^2) = 0.65 \times (2500 - 1600) = 0.65 \times 900 = 585~\text{Pa}$
So the pressure below is 585 Pa higher than above.

Compute the lift force Lift = pressure difference × wing area:

$F_{\text{lift}} = \Delta P \times A = 585 \times 10 = 5850~\text{N}$

Compute the weight of the aeroplane

$W = mg = 500 \times 9.8 = 4900~\text{N}$
(Using $g = 9.8~\text{m/s}^2$ as standard.)

Compare lift and weight

$F_{\text{lift}} = 5850~\text{N} > 4900~\text{N} = W$
Since the upward force exceeds the downward weight, there is a net upward force. Therefore the plane will accelerate upward — it gains altitude.

Watch out

A common mistake is to forget that Bernoulli’s equation applies along a streamline; here we assume the flow is steady and incompressible, which is reasonable for subsonic flight. Also, using $g = 10$ would give weight = 5000 N, still less than 5850 N, so the conclusion remains the same.

Tip

You don’t need to compute lift exactly — just compare $\frac{1}{2}\rho (v_{\text{top}}^2 - v_{\text{bottom}}^2) A$ to $mg$ . The numbers show lift clearly wins.

✓Final answer
$The correct option is (A) . ANSWER: A$
TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$n^{\text{th}}$
›Reveal solutionSolution
The key is to relate the velocity at the end of the nth second to the acceleration, then express the displacement during the last two seconds in terms of that velocity and n. The result is $\frac{2V(n-1)}{n}$ , which corresponds to option (D).

We have a body starting from rest with uniform acceleration. That means its motion is governed by the simple equations of constant acceleration, with initial velocity $u = 0$ . The problem gives us the velocity after the $n^{\text{th}}$ second (the last second of motion) as $V$ , and asks for the displacement in the last two seconds.

Concept & Intuition:

The phrase "after $n^{\text{th}}$ second" means at time $t = n$ seconds (not during the $n$ th second, but at its end). The "last two seconds" are the time interval from $t = n-2$ to $t = n$ . Since acceleration is constant, we can find acceleration from the given velocity, then compute the displacement over any interval using standard kinematic formulas.

Let’s work it through step by step.

Find the acceleration. The body starts from rest ( $u = 0$ ) and has uniform acceleration $a$ . After $n$ seconds, its velocity is $V$ . Using $v = u + at$ :

$V = 0 + a \cdot n \quad \Rightarrow \quad a = \frac{V}{n}.$

Displacement in the last two seconds.

The last two seconds are from $t = n-2$ to $t = n$ . We can find the displacement during this interval by subtracting the displacement up to $t = n-2$ from the displacement up to $t = n$ .

Displacement from rest after time $t$ is $s = \frac{1}{2} a t^2$ .

Displacement up to $t = n$ :

$s_n = \frac{1}{2} a n^2 = \frac{1}{2} \cdot \frac{V}{n} \cdot n^2 = \frac{V n}{2}.$

Displacement up to $t = n-2$ :

$s_{n-2} = \frac{1}{2} a (n-2)^2 = \frac{1}{2} \cdot \frac{V}{n} \cdot (n-2)^2 = \frac{V (n-2)^2}{2n}.$
Therefore, displacement in the last two seconds:
$\Delta s = s_n - s_{n-2} = \frac{V n}{2} - \frac{V (n-2)^2}{2n}.$

Simplify the expression. Factor out $\frac{V}{2n}$ :

$\Delta s = \frac{V}{2n} \left[ n^2 - (n-2)^2 \right].$
Expand $(n-2)^2 = n^2 - 4n + 4$ , so:
$n^2 - (n^2 - 4n + 4) = 4n - 4 = 4(n-1).$
Thus:
$\Delta s = \frac{V}{2n} \cdot 4(n-1) = \frac{2V(n-1)}{n}.$
Tip

A quicker method: The displacement in the last $t$ seconds for uniformly accelerated motion from rest can also be found using the average velocity during that interval. Here, velocity at $t=n-2$ is $a(n-2) = \frac{V}{n}(n-2)$ and at $t=n$ is $V$ . The average velocity over the last 2 seconds is $\frac{V + \frac{V}{n}(n-2)}{2} = \frac{V(2n-2)}{2n} = \frac{V(n-1)}{n}$ , and multiplying by time (2 s) gives the same result.

Watch out

A common mistake is to misinterpret "after $n^{\text{th}}$ second" as "during the $n$ th second" (i.e., the interval from $t=n-1$ to $t=n$ ). That would give a different relation. Always read "after $n$ seconds" as at time $t=n$ .

The expression matches option (D).

✓Final answer
$The correct option is (D). ANSWER: D$

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