Q.A helicopter of mass 2000kg rises with a vertical acceleration of $15$ m s$^{-2}$. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the ($g = 10$ m s$^{-2}$)

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A body of mass M is suspended to the lower end of an aeroplane accelerating horizontally. If the rope used to hang the body can withstand a maximum tension T, then the maximum acceleration of the aeroplane is (g = acceleration due to gravity and neglect the mass of the rope) (A) $\sqrt{\dfrac{T^2}{M^2} - g^2}$ (B) $\dfrac{T}{M} - g$ (C) $\dfrac{T}{M} + g$ (D) $\sqrt{\dfrac{T^2}{M^2} + g^2}$

   ›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.

   ---

   1. 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.

   2. Apply Newton’s second law in each direction

      • Vertical: no vertical acceleration, so $T \cos\theta = Mg$.
      • Horizontal: $T \sin\theta = M a$.

   3. 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 $T^2$:

   $$M^2 a^2 + M^2 g^2 = T^2 …$$

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Two balls P and Q are thrown vertically upwards simultaneously from the ground with velocities $20~\mathrm{ms}^{-1}$ and $35~\mathrm{ms}^{-1}$ respectively. The distance between the two balls when the velocity of the ball P becomes zero is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) 30 m (B) 20 m (C) 50 m (D) 70 m

   ›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.

   ---

   1. 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 $v = u + at$:

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

   2. 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.

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

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Five stones are dropped successively from a height of 50 m from the ground with a time interval of half a second between two successive stones. The relative velocity between the first and third stones when they are in motion is (Acceleration due to gravity $= 10\,\mathrm{ms}^{-2}$) (A) $15\,\mathrm{ms}^{-1}$ (B) $5\,\mathrm{ms}^{-1}$ (C) $10\,\mathrm{ms}^{-1}$ (D) $20\,\mathrm{ms}^{-1}$

   ›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.

   1. 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.

   2. 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).

   3. 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.

   4. 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). …

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4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A person wearing a parachute jumps off a plane from a height of 2 km from the ground and falls freely for 20 m before his parachute opens. After his parachute opens if he continues to move uniformly with the velocity attained due to his freefall, the total time taken by the person to reach the ground is (Acceleration due to gravity $= 10\, \mathrm{ms}^{-2}$) (A) 99 s (B) 101 s (C) 100 s (D) 102 s

   ›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:

   1. Free fall (constant acceleration $g$) over a short distance.
   2. 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

   1. Identify the distances

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

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

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

   2. 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.

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

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

   4. Find the time taken after the parachute opens …

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5. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A block of mass $\sqrt{2}$ kg is placed on a rough horizontal surface. A force ‘F’ acting upwards at an angle of $45^\circ$ with the horizontal causes the block to start motion. If the coefficient of static friction between the surface and the block is $0.25$, the magnitude of the force ‘F’ is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $0.5~\mathrm{N}$ (B) $2~\mathrm{N}$ (C) $4~\mathrm{N}$ (D) $8~\mathrm{N}$

   ›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 $F$.

   ---

   1. 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)

   2. Normal reaction

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

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

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

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

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

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

   5. Solve for $F$ Multiply both sides by 4: $$$$

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6. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A block of mass $\sqrt{2}$ kg is placed on a rough horizontal surface. A force 'F' acting upwards at an angle of $45^\circ$ with the horizontal causes the block to start motion. If the coefficient of static friction between the surface and the block is $0.25$, the magnitude of the force 'F' is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $2~\mathrm{N}$ (B) $0.5~\mathrm{N}$ (C) $4~\mathrm{N}$ (D) $8~\mathrm{N}$

   ›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.

   1. 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 $\mu_s = 0.25$.

   2. 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.

   3. Horizontal condition for impending motion. The block just starts moving when the horizontal pull equals the maximum static friction: $F\cos45^\circ = \mu_s N$ …

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7. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The weight of a body at a height of $3R_E$ from the surface of the earth is $90~\mathrm{N}$, where $R_E$ is radius of the earth. The weight of the same body at a height of $R_E$ from the surface of the earth is (A) $22.5~\mathrm{N}$ (B) $180~\mathrm{N}$ (C) $360~\mathrm{N}$ (D) $90~\mathrm{N}$

   ›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:

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

   where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R_E$ is the radius of the Earth.

   The weight of a body of mass $m$ at this height is $W_h = mg_h$.

   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.

   1. 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 ($h_2$). …

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8. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.A force of 20 N acts on a body at rest for a time of 2 s and then a force of 60 N acts for a time of 1.5 s in the opposite direction. If the final velocity of the body is $10~\mathrm{ms}^{-1}$ in the direction of 60 N force, then the mass of the body is (A) 10 kg (B) 8 kg (C) 5 kg (D) 16 kg

   ›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.

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

   2. 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 $I_2 = (60)(1.5) = +90~\mathrm{N\,s}$

   3. 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 $I = m v$.

   4. Solve for mass. $50 = m \times 10$ …

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9. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.A body of mass $4\, \text{kg}$ is falling freely from rest from a height of $30\, \text{m}$ from the ground. If the velocity of the body when it is at a height of $10\, \text{m}$ from the ground is $10\, \text{ms}^{-1}$, then the loss of energy due to air resistance on the body is (Acceleration due to gravity $= 10\, \text{ms}^{-2}$) (A) $400\, \text{J}$ (B) $600\, \text{J}$ (C) $300\, \text{J}$ (D) $100\, \text{J}$

   ›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:

   1. 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:
   

   $E_1 = PE_1 + KE_1 = 1200\, \text{J} + 0\, \text{J} = 1200\, \text{J}$

   2. 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 $mgh_2$:

      $PE_2 = mgh_2 = (4\, \text{kg})(10\, \text{ms}^{-2})(10\, \text{m}) = 400\, \text{J}$ …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.A body is thrown vertically upwards with a velocity of $35\ \mathrm{ms}^{-1}$ from the ground. The ratio of the speeds of the body at times $3\ \mathrm{s}$ and $4\ \mathrm{s}$ of its motion is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) $3 : 4$ (B) $1 : 1$ (C) $2 : 1$ (D) $3 : 2$

    ›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

    1. Choose a sign convention

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

    2. Velocity at any time

       Using $v = u + at$:

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

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

    $v(3) = 35 - 10(3) = 35 - 30 = +5\ \mathrm{m/s}$

    Positive means still moving upward. Speed = $|+5| = 5\ \mathrm{m/s}$.

    4. Velocity at $t = 4\ \mathrm{s}$ $v(4) = 35 - 10(4) = 35 - 40 = -5\ \mathrm{m/s}$ …

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11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.The velocities of air above and below the surfaces of a flying aeroplane wing are $50~\mathrm{ms}^{-1}$ and $40~\mathrm{ms}^{-1}$ respectively. If the area of the wing is $10~\mathrm{m}^2$ and the mass of the aeroplane is $500~\mathrm{kg}$, then as time passes by (density of air $= 1.3~\mathrm{kg~m}^{-3}$) (A) the aeroplane will gain altitude (B) the aeroplane will experience weightlessness (C) the aeroplane will fly horizontally (D) the aeroplane will loose altitude

    ›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

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

    2. 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.

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

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

    4. 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.) …

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12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.A body starts from rest with uniform acceleration. If its velocity after $n^{\text{th}}$ second (last second) is ‘$V$’ then its displacement in the last two seconds is (A) $\dfrac{2V(n+1)}{n}$ (B) $\dfrac{V(n+1)}{n}$ (C) $\dfrac{V(n-1)}{n}$ (D) $\dfrac{2V(n-1)}{n}$

    ›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.

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

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

    2. 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:

    $\Delta s = s_n - s_{n-2} = \frac{V n}{2} - \frac{V (n-2)^2}{2n}.$

    3. Simplify the expression. Factor out $\frac{V}{2n}$: $\Delta s = \frac{V}{2n} \left[ n^2 - (n-2)^2 \right].$ …

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