Q.The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.

Showing the 12 most recent of 43 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The power P (in watt) acting on a body of mass 2 kg is given by $4.5P = 8t^2 + 14t + 9$, where ‘t’ is time in second. If the body starts from rest at $t = 0$, then the velocity of the body at time $t = 3$ s is (A) $15\ \mathrm{ms}^{-1}$ (B) $12\ \mathrm{ms}^{-1}$ (C) $6\ \mathrm{ms}^{-1}$ (D) $9\ \mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Power is the rate of change of kinetic energy. Integrate power with respect to time to find the change in kinetic energy, then solve for velocity. The velocity at $t = 3$ s is $9\ \mathrm{ms}^{-1}$.

   The key idea here is that power is not force times velocity in a simple way when the force is not constant — but power is always the rate at which work is done, which equals the rate of change of kinetic energy. That’s the clean path: $P = \frac{dK}{dt}$, where $K = \frac12 m v^2$.

   Since the body starts from rest, its initial kinetic energy is zero. So the kinetic energy at any time $t$ is just the integral of power from $0$ to $t$. Once we have $K(t)$, we get $v(t)$.

   1. Write the given power equation clearly. We have $4.5P = 8t^2 + 14t + 9$. So

   $P = \frac{8t^2 + 14t + 9}{4.5}.$

   It’s often easier to work in fractions: $4.5 = \frac{9}{2}$, so

   $P = \frac{8t^2 + 14t + 9}{9/2} = \frac{2}{9}(8t^2 + 14t + 9).$

   2. Relate power to kinetic energy. For a body of constant mass,

   $P = \frac{dK}{dt}, \quad \text{where } K = \frac12 m v^2.$

   Here $m = 2$ kg, so $K = \frac12 \cdot 2 \cdot v^2 = v^2$.

   That’s a nice simplification: the kinetic energy in joules is numerically equal to $v^2$ when $v$ is in m/s.

   3. Integrate to find $K(t)$. Since the body starts from rest at $t=0$, $K(0) = 0$.

   $K(t) = \int_0^t P \, dt = \int_0^t \frac{2}{9}(8t^2 + 14t + 9) \, dt.$

   Factor out the constant:

   $K(t) = \frac{2}{9} \int_0^t (8t^2 + 14t + 9) \, dt.$

   Integrate term by term:

   $\int (8t^2 + 14t + 9) \, dt = \frac{8t^3}{3} + \frac{14t^2}{2} + 9t = \frac{8t^3}{3} + 7t^2 + 9t.$

   So

   $K(t) = \frac{2}{9} \left( \frac{8t^3}{3} + 7t^2 + 9t \right).$

   4. Evaluate at $t = 3$ s. First compute the bracket at $t=3$: …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A body is thrown vertically upwards from the surface of the earth with a velocity K times the orbital velocity of a satellite near the surface of the earth. If the maximum height reached by the body is 200% more than the radius of the earth, then the value of $K^2$ is (A) 4.5 (B) 2.5 (C) 1.5 (D) 3.5

   ›Reveal solutionSolution

   With $h=3R$ and $v_o^2=gR$, energy conservation gives $K^2=\dfrac{2h}{R+h}=1.5$.

   Orbital velocity near the surface: $v_o=\sqrt{gR}$, so $v_o^2=gR$. Launch speed $v=Kv_o$, hence $v^2=K^2gR$.

   "Maximum height $200\%$ more than $R$" means $h=R+2R=3R$.

   Since $h$ is comparable to $R$, use energy conservation (not $v^2=2gh$):

   $\frac12 mv^2-\frac{GMm}{R}=-\frac{GMm}{R+h}\;\Rightarrow\;\frac12v^2=GM\left(\frac1R-\frac1{R+h}\right)=\frac{GM\,h}{R(R+h)}.$ …

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3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.If the displacement of a body moving with constant acceleration in a straight line path in the first four seconds of time is 56 m and its displacement in the fourth second of the motion is 17 m, then the average velocity of the body during sixth second of its motion is (A) $18\ \mathrm{ms}^{-1}$ (B) $21\ \mathrm{ms}^{-1}$ (C) $11\ \mathrm{ms}^{-1}$ (D) $26\ \mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Using equations of motion for constant acceleration, we find the initial velocity and acceleration from the given data, then compute the average velocity during the sixth second as the velocity at 5.5 s, which is $21\ \mathrm{ms}^{-1}$.

   The problem gives two pieces of information about a body moving with constant acceleration: the total displacement in the first 4 seconds (56 m) and the displacement during the 4th second alone (17 m). From these, we can determine the initial velocity $u$ and acceleration $a$. Once we have those, the average velocity during any time interval under constant acceleration is simply the instantaneous velocity at the midpoint of that interval. For the sixth second (from $t = 5$ s to $t = 6$ s), the midpoint is $t = 5.5$ s, so we need $v$ at that instant.

   Let’s work through it step by step.

   1. Displacement in the first 4 seconds For constant acceleration, displacement in time $t$ is $s = ut + \frac{1}{2}at^2$. With $t = 4$ s and $s = 56$ m:

   $56 = u(4) + \frac{1}{2}a(4)^2 = 4u + 8a$

   Divide through by 2:

   $28 = 2u + 4a \quad \text{(Equation 1)}$

   2. Displacement in the 4th second The displacement during the $n$th second is $s_n = u + \frac{a}{2}(2n - 1)$. For $n = 4$, $s_4 = 17$ m:

   $17 = u + \frac{a}{2}(2 \times 4 - 1) = u + \frac{a}{2}(7) = u + 3.5a$

   Multiply by 2 to avoid decimals:

   $34 = 2u + 7a \quad \text{(Equation 2)}$

   Tip

   The formula $s_n = u + \frac{a}{2}(2n-1)$ is derived from the difference in displacements up to $n$ seconds and $n-1$ seconds. It gives the distance covered in the $n$th second directly — a real time-saver in constant-acceleration problems.

   3. Solve for $u$ and $a$ Subtract Equation 1 from Equation 2:

   $(2u + 7a) - (2u + 4a) = 34 - 28$

   $3a = 6 \quad \Rightarrow \quad a = 2\ \mathrm{m/s^2}$

   Substitute $a = 2$ into Equation 1:

   $28 = 2u + 4(2) = 2u + 8$

   $2u = 20 \quad \Rightarrow \quad u = 10\ \mathrm{m/s}$

   4. Average velocity during the sixth second …

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.A body initially at rest at a certain height from the ground is falling freely under gravity. At a time of one second, the body is at a height of 120 m from the ground. At a time $t = 4.5\ \mathrm{s}$, if the potential energy of the body is $K$ times its total energy, then the value of $K$ is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) 0.19 (B) 0.81 (C) 0.37 (D) 0.63

   ›Reveal solutionSolution

   The problem uses the relation between potential energy and total energy during free fall. At $t = 4.5\ \mathrm{s}$, the fraction $K = \frac{\text{PE}}{\text{Total energy}} = 0.19$, so option (A) is correct.

   The key idea is that for a body falling freely from rest, total mechanical energy (sum of kinetic and potential) remains constant. At any instant, the potential energy is $mgh$, where $h$ is the height above ground, and total energy equals $mgH$, where $H$ is the initial height. So $K = \frac{h}{H}$ — the ratio of current height to initial height. Find $H$ from the given data, then find $h$ at $t = 4.5\ \mathrm{s}$, and take the ratio.

   1. Find the initial height $H$. The body starts from rest at height $H$. After $1\ \mathrm{s}$ of free fall, distance fallen is

   $s = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 1^2 = 5\ \mathrm{m}.$

   So its height above ground at $t = 1\ \mathrm{s}$ is $H - 5$. This is given as $120\ \mathrm{m}$:

   $H - 5 = 120 \quad\Rightarrow\quad H = 125\ \mathrm{m}.$

   2. Find the height at $t = 4.5\ \mathrm{s}$. Distance fallen in $4.5\ \mathrm{s}$:

   $s = \frac{1}{2} \times 10 \times (4.5)^2 = 5 \times 20.25 = 101.25\ \mathrm{m}.$

   Height above ground:

   $h = H - s = 125 - 101.25 = 23.75\ \mathrm{m}.$ …

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5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A body is at a point P, at some height above the surface of a planet of mass M and radius R. If the potential energy of the body at point P is half of its potential energy on the surface of the planet, then the difference between the escape velocity of the body from point P and its escape velocity from the surface of the planet is (G - Universal gravitational constant) (A) $\sqrt{\dfrac{GM}{R}(\sqrt{3}-\sqrt{2})}$ (B) $\sqrt{\dfrac{GM}{R}(\sqrt{3}-1)}$ (C) $\sqrt{\dfrac{GM}{R}(\sqrt{2}-1)}$ (D) $\sqrt{\dfrac{GM}{R}(2-\sqrt{2})}$

   ›Reveal solutionSolution

   The problem asks for the difference in escape velocities from a point P and the planet's surface, given a condition on potential energy. By using the potential energy condition, we find that point P is at a height $R$ above the surface. Then, calculating the escape velocities at the surface and at point P, the difference is found to be $\boxed{\sqrt{\frac{GM}{R}(\sqrt{2}-1)}}$.

   The core concepts here are gravitational potential energy and escape velocity. Both are derived from the universal law of gravitation and represent different aspects of a body's interaction with a gravitational field. Understanding their definitions and how they depend on distance is key to solving this problem.

   Gravitational potential energy is the energy a body possesses due to its position in a gravitational field. It's defined as the work done by an external agent to bring the body from infinity (where potential energy is conventionally zero) to that position without acceleration. Because gravity is an attractive force, work is done by the field as the body approaches, so potential energy is negative and becomes more negative closer to the planet.

   Escape velocity is the minimum speed an object needs to completely escape the gravitational pull of a massive body, meaning it will reach an infinite distance with zero kinetic energy remaining. This is a direct application of the principle of conservation of mechanical energy.

   Let's break down the problem:

   1. Define Gravitational Potential Energy:

      The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of a planet of mass $M$ is given by:

      $U = -\frac{GMm}{r}$

      Here, $G$ is the universal gravitational constant. The negative sign indicates that the force is attractive and that potential energy is zero at infinite separation.

      • Potential energy on the surface ($U_S$): On the surface of the planet, the distance from the center is $R$.

   $U_S = -\frac{GMm}{R}$

   *   **Potential energy at point P ($U_P$):**
       Let the height of point P above the surface be $h$. The distance of point P from the center of the planet will be $r_P = R+h$.
   

   $U_P = -\frac{GMm}{R+h}$

   2. Use the given condition to find the height of P: The problem states that the potential energy of the body at point P is half of its potential energy on the surface of the planet.

   $U_P = \frac{1}{2} U_S$

   Substitute the expressions for $U_P$ and $U_S$:
   

   $-\frac{GMm}{R+h} = \frac{1}{2} \left(-\frac{GMm}{R}\right)$

   We can cancel out the common terms $-GMm$ from both sides:
   

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

   This implies:
   

   $R+h = 2R$

   Solving for $h$:
   

   $h = 2R - R = R$

   So, point P is at a height $R$ above the surface of the planet. This means its distance from the center of the planet is $r_P = R+h = R+R = 2R$.
   

   3. Define Escape Velocity: …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the moment of inertia of a uniform solid cylinder about the axis of the cylinder is $\dfrac{1}{n}$ times its moment of inertia about an axis passing through its midpoint and perpendicular to its length, then the ratio of the length and radius of the cylinder is (A) $\sqrt{2(3n+1)}$ (B) $\sqrt{2(3n-1)}$ (C) $\sqrt{3(2n+1)}$ (D) $\sqrt{3(2n-1)}$

   ›Reveal solutionSolution

   The problem equates two moments of inertia of a uniform solid cylinder — about its symmetry axis and about a perpendicular axis through its centre — and solves for the ratio of length to radius. The answer is $\sqrt{3(2n-1)}$, which corresponds to option (D).

   We start by recalling the relevant formulas for the moment of inertia of a uniform solid cylinder of mass $M$, radius $R$, and length $L$.

   Concept and intuition:

   The moment of inertia about the cylinder’s own axis (the symmetry axis) is easy — it’s just like a solid disk, $I_{\text{axis}} = \frac{1}{2} M R^2$.

   The moment of inertia about an axis through the midpoint and perpendicular to the length is trickier: it’s like a rod rotating about its centre, but the mass is spread out in a cylinder, not a thin rod. We use the perpendicular axis theorem and the parallel axis theorem, or directly integrate, to get $I_{\perp} = \frac{1}{4} M R^2 + \frac{1}{12} M L^2$.

   The problem says $I_{\text{axis}} = \frac{1}{n} I_{\perp}$, so we set up an equation and solve for $L/R$.

   ---

   Step-by-step solution:

   1. Write the moment of inertia about the cylinder’s symmetry axis. For a solid cylinder of mass $M$ and radius $R$, rotating about its central axis:

   $I_{\text{axis}} = \frac{1}{2} M R^2.$

   2. Write the moment of inertia about a perpendicular axis through the centre. For a cylinder of length $L$ and radius $R$, the moment about an axis through its midpoint, perpendicular to its length, is the sum of two contributions:
      • The moment of a thin rod of length $L$ and mass $M$ about its centre: $\frac{1}{12} M L^2$.
      • The moment due to the “disk-like” cross-section: each thin slice perpendicular to the length is a disk of radius $R$, and for rotation about a diameter of that disk, the moment is $\frac{1}{4} M_{\text{slice}} R^2$. Summing over all slices gives $\frac{1}{4} M R^2$. Hence:

   $I_{\perp} = \frac{1}{4} M R^2 + \frac{1}{12} M L^2.$

   3. Apply the given condition. The problem states:

   $I_{\text{axis}} = \frac{1}{n} I_{\perp}.$

   Substitute the expressions: …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A body of mass 0.5 kg is supplied with a power 'P' (in watt) which varies with time 't' (in second) as $P = 3t^2 + 3$. If the velocity of the body at time $t = 0$ is zero, then the velocity of the body at time $t = 3\,\mathrm{s}$ is (A) $12\,\mathrm{ms}^{-1}$ (B) $24\,\mathrm{ms}^{-1}$ (C) $18\,\mathrm{ms}^{-1}$ (D) $36\,\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The key idea is that power equals force times velocity, and also equals the rate of change of kinetic energy. By integrating power over time, we find the change in kinetic energy, and from that the velocity. The velocity at $t=3\,\mathrm{s}$ is $12\,\mathrm{m/s}$, so the correct option is (A).

   We start with the concept: Power is the rate at which work is done, and for a body, that work changes its kinetic energy. Since the body starts from rest, the total work done by the power up to time $t$ equals the kinetic energy at that time. Mathematically, $P = \frac{dK}{dt}$, where $K = \frac{1}{2}mv^2$. This avoids needing to find force or acceleration directly — we go straight from power to energy.

   1. Relate power to kinetic energy. Power is defined as $P = \frac{dW}{dt}$, and by the work-energy theorem, $dW = dK$. So

   $P = \frac{dK}{dt}.$

   Given $P = 3t^2 + 3$ and mass $m = 0.5\,\mathrm{kg}$, we have

   $\frac{d}{dt}\left(\frac{1}{2} m v^2\right) = 3t^2 + 3.$

   2. Integrate to find the change in kinetic energy. Since $v=0$ at $t=0$, the kinetic energy at $t=0$ is zero. Integrating from $t=0$ to $t=3$:

   $\int_0^3 \frac{dK}{dt}\,dt = \int_0^3 (3t^2 + 3)\,dt.$

   The left side is simply $K(3) - K(0) = K(3)$. So

   $K(3) = \left[ t^3 + 3t \right]_0^3 = (27 + 9) - 0 = 36\,\mathrm{J}.$

   3. Convert kinetic energy to velocity. Kinetic energy is $K = \frac{1}{2} m v^2$. With $m = 0.5\,\mathrm{kg}$: $$$$

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A body of mass 0.5 kg is supplied with a power 'P' (in watt) which varies with time 't' (in second) as $P = 3t^2 + 3$. If the velocity of the body at time $t = 0$ is zero, then the velocity of the body at time $t = 3$ s is (A) $36\ \mathrm{ms}^{-1}$ (B) $12\ \mathrm{ms}^{-1}$ (C) $18\ \mathrm{ms}^{-1}$ (D) $24\ \mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The key idea is that power is the rate of change of kinetic energy. Integrating $P = dK/dt$ gives the change in kinetic energy, from which velocity follows. The velocity at $t=3$ s is $12\ \mathrm{m/s}$, so the correct option is (B).

   We start with the concept: Power is defined as the rate at which work is done, or equivalently, the rate of change of kinetic energy for a body with no other forms of energy storage. For a particle of mass $m$, kinetic energy $K = \frac12 m v^2$, so

   $P = \frac{dK}{dt}.$

   Since we are given $P(t)$ and the initial condition $v(0)=0$, we can integrate to find $K(t)$, then solve for $v(t)$. This avoids dealing with forces or acceleration directly — a neat shortcut.

   1. Write the relation between power and kinetic energy. For a body of constant mass,

   $P = \frac{dK}{dt}.$

   Here $P = 3t^2 + 3$ (in watts) and $m = 0.5$ kg.

   2. Integrate to find the change in kinetic energy. Since $K(0)=0$ (because $v(0)=0$),

   $K(t) = \int_0^t P \, dt = \int_0^t (3t^2 + 3)\, dt.$

   Compute:

   $K(t) = \left[ t^3 + 3t \right]_0^t = t^3 + 3t.$

   At $t=3$ s,

   $K(3) = 3^3 + 3\cdot 3 = 27 + 9 = 36\ \text{J}.$

   3. Relate kinetic energy to velocity. Kinetic energy is $K = \frac12 m v^2$, so $v = \sqrt{\frac{2K}{m}}.$ …

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9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The vertical displacement ($y$ in metre) of a projectile in terms of its horizontal displacement ($x$ in metre) is given by $y = (\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $\sqrt{3} s$ (B) $0.2 s$ (C) $0.5\sqrt{3} s$ (D) $0.2\sqrt{3} s$

   ›Reveal solutionSolution

   Comparing $y = \sqrt{3}\,x - 0.2x^2$ with the projectile path gives $\tan\theta=\sqrt3$ ($\theta=60^\circ$) and $u=10\ \text{m s}^{-1}$, so the time of flight is $T=\dfrac{2u\sin\theta}{g}=\sqrt3\ \text{s}$.

   Match the trajectory

   A projectile launched from the origin follows

   $y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}.$

   Comparing coefficients with $y = \sqrt3\,x - 0.2x^2$:

   • Linear term: $\tan\theta = \sqrt3 \implies \theta = 60^\circ$.
   • Quadratic term: $\dfrac{g}{2u^2\cos^2\theta} = 0.2$.

   Find the launch speed

   With $g=10$ and $\cos60^\circ = \tfrac12$ (so $\cos^2\theta=\tfrac14$):

   $\frac{10}{2u^2\cdot\frac14} = \frac{20}{u^2} = 0.2 \implies u^2 = 100 \implies u = 10\ \text{m s}^{-1}.$

   Time of flight …

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10. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.A proton and an alpha particle enter a uniform electric field perpendicular to the direction of the field. The ratio of distances travelled by the proton and the alpha particle in the direction of the field after a time of 't' seconds is (A) $1:2$ (B) $1:4$ (C) $2:1$ (D) $4:1$

    ›Reveal solutionSolution

    Both particles experience the same electric force per unit charge, but the acceleration depends on mass. The distance travelled in the field direction is proportional to $q/m$, giving a ratio of $2:1$ for proton vs alpha particle.

    The key here is to separate the motion into two independent directions. The particles enter perpendicular to the field, so their initial velocity is entirely along the perpendicular direction. The electric field acts only along the other axis, so the motion along the field is purely accelerated from rest.

    Concept: In a uniform electric field $E$, the force on a charge $q$ is $F = qE$. By Newton’s second law, acceleration $a = F/m = qE/m$. Both particles start with zero initial velocity in the field direction, so the distance travelled along the field in time $t$ is $s = \frac{1}{2} a t^2 = \frac{1}{2} \frac{qE}{m} t^2$. The ratio of distances depends only on $q/m$ because $E$ and $t$ are the same for both.

    1. Identify the charges and masses.

       • Proton: charge $+e$, mass $m_p$.
       • Alpha particle (helium nucleus): charge $+2e$, mass $4m_p$.

    2. Write the distance formula for each.

       For the proton:

    $s_p = \frac{1}{2} \frac{eE}{m_p} t^2$

    For the alpha particle:

    $s_\alpha = \frac{1}{2} \frac{(2e)E}{4m_p} t^2 = \frac{1}{2} \frac{eE}{2m_p} t^2$

    3. Take the ratio. …

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11. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.A body is projected from the surface of the earth with a velocity of $10\sqrt{3}$ ms$^{-1}$ such that its range is maximum. The velocity of the body at half of the maximum height is (Acceleration due to gravity = $10$ ms$^{-2}$) (A) $10\sqrt{3}$ ms$^{-1}$ (B) $15$ ms$^{-1}$ (C) $15\sqrt{2}$ ms$^{-1}$ (D) $30$ ms$^{-1}$

    ›Reveal solutionSolution

    For maximum range, the launch angle is $45^\circ$. We find the initial horizontal and vertical velocity components, calculate the maximum height, then determine the vertical velocity at half that height. Combining this with the constant horizontal velocity gives the final speed. The velocity of the body at half of the maximum height is $\boxed{15 \text{ ms}^{-1}}$.

    When a body is projected, its motion can be analyzed by separating it into independent horizontal and vertical components.

    • The horizontal velocity remains constant throughout the flight (assuming no air resistance) because there is no horizontal force acting on the body.
    • The vertical velocity changes due to the constant downward acceleration of gravity ($g$). It decreases as the body rises, becomes zero at the maximum height, and then increases in the downward direction as the body falls.

    The range of a projectile is the horizontal distance it covers. For a given initial speed, the maximum range is achieved when the launch angle is $45^\circ$. This is a fundamental result in projectile motion.

    To find the velocity of the body at any point, we need to determine both its horizontal and vertical velocity components at that specific point. The magnitude of the velocity (speed) is then the vector sum of these components: $v = \sqrt{v_x^2 + v_y^2}$.

    Here's how we solve the problem step-by-step:

    1. Determine the launch angle and initial velocity components.

       The problem states that the range is maximum. For a projectile launched from the surface of the earth, the range is maximum when the angle of projection, $\theta$, is $45^\circ$.

       The initial velocity of projection is given as $u = 10\sqrt{3}$ ms$^{-1}$.

       We resolve this initial velocity into its horizontal and vertical components:

       • Initial horizontal velocity: $u_x = u \cos \theta = (10\sqrt{3}) \cos 45^\circ$ $u_x = (10\sqrt{3}) \left(\frac{1}{\sqrt{2}}\right) = \frac{10\sqrt{3}}{\sqrt{2}} = \frac{10\sqrt{6}}{2} = 5\sqrt{6}$ ms$^{-1}$.
       • Initial vertical velocity: $u_y = u \sin \theta = (10\sqrt{3}) \sin 45^\circ$ $u_y = (10\sqrt{3}) \left(\frac{1}{\sqrt{2}}\right) = \frac{10\sqrt{3}}{\sqrt{2}} = \frac{10\sqrt{6}}{2} = 5\sqrt{6}$ ms$^{-1}$.
       Important

       For maximum range, the launch angle is $45^\circ$. At this angle, the initial horizontal and vertical velocity components are equal: $u_x = u_y = u/\sqrt{2}$.

    2. Calculate the maximum height ($H_{max}$).

       The maximum height reached by a projectile is given by the formula:

       $H_{max} = \frac{u^2 \sin^2 \theta}{2g}$

       Substituting the values $u = 10\sqrt{3}$ ms$^{-1}$, $\theta = 45^\circ$, and $g = 10$ ms$^{-2}$:

       $H_{max} = \frac{(10\sqrt{3})^2 (\sin 45^\circ)^2}{2 \times 10}$

       $H_{max} = \frac{(100 \times 3) \left(\frac{1}{\sqrt{2}}\right)^2}{20}$

       $H_{max} = \frac{300 \times \frac{1}{2}}{20}$

       $H_{max} = \frac{150}{20} = 7.5$ m.

    3. Determine half of the maximum height ($h$). …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.A plane electromagnetic wave of electric and magnetic fields $E_0$ and $B_0$ respectively incidents on a surface. If the total energy transferred to the surface in a time of 't' is 'U', then the magnitude of the total momentum delivered to the surface for complete absorption is (A) $\dfrac{UE_0}{B_0}$ (B) $\dfrac{UB_0}{E_0}$ (C) $\dfrac{U}{E_0 B_0}$ (D) $\dfrac{UB_0}{E_0^2}$

    ›Reveal solutionSolution

    For complete absorption, the momentum delivered equals the energy absorbed divided by the speed of light. Using the relation $c = E_0/B_0$ for an electromagnetic wave, the momentum is $U B_0 / E_0$, so the correct option is (B).

    The key idea is that electromagnetic waves carry both energy and momentum, and for complete absorption, the momentum transferred is simply the energy divided by the speed of light. But here we are given the fields $E_0$ and $B_0$, not $c$ directly — so we must use the fundamental relation between them in a plane wave: $E_0 = c B_0$. This lets us express $c$ in terms of the given quantities.

    1. Recall the momentum-energy relation for light For any electromagnetic wave, the momentum $p$ delivered to a perfectly absorbing surface is related to the absorbed energy $U$ by

    $p = \frac{U}{c},$

    because each photon of energy $E$ carries momentum $E/c$, and absorption transfers all of it.

    2. Relate the speed of light to the given fields In a plane electromagnetic wave in vacuum, the magnitudes of the electric and magnetic fields satisfy

    $E_0 = c B_0.$

    This is a direct consequence of Maxwell’s equations — the ratio of the field strengths is always the speed of light.

    3. Solve for $c$ and substitute From $E_0 = c B_0$, we get

    $c = \frac{E_0}{B_0}.$

    Plugging this into the momentum expression:

    $$$$

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