Q.A family uses $8\ \text{kW}$ of power.

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The blades of a wind mill sweep out a circle of area $A = 2\ m^2$. The wind is flowing at velocity $V = 6\ ms^{-1}$ perpendiculars to circle and the density of air is $1.2\ kg.m^{-3}$. Then the power of the mill is (A) 160.8 W (B) 259.2 W (C) 302.5 W (D) 239.2 W

   ›Reveal solutionSolution

   The power available in wind flowing through area $A$ at speed $V$ is $\tfrac12\rho A V^3$, which evaluates to 259.2 W here.

   Concept and Intuition

   The mass of air passing through the swept area per second is $\dot m=\rho A V$ (a cylinder of air of cross-section $A$ and length $V$ passes through each second). The kinetic energy carried by this mass per second — the power available to the windmill — is $P=\tfrac12\dot m V^2=\tfrac12\rho A V^3$.

   Step-by-Step Solution

   1. Mass flow rate of air: $\dot m=\rho A V$.
   2. Power (kinetic energy delivered per second) $P=\tfrac12\dot m V^2=\tfrac12\rho A V^3$.
   3. Substitute $\rho=1.2\ kg/m^3$, $A=2\ m^2$, $V=6\ ms^{-1}$: $V^3=216$. …

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2. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.A machine gun with power $15\ \mathrm{kW}$ can fire $360$ bullets per minute. If the mass of each bullet is $5\ \mathrm{g}$, then the velocity of the bullets is (A) $1500\ \mathrm{m\,s^{-1}}$ (B) $1000\ \mathrm{m\,s^{-1}}$ (C) $3600\ \mathrm{m\,s^{-1}}$ (D) $500\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   Power equals the kinetic energy given to the bullets per second; solving for $v$ from the given firing rate and bullet mass gives 1000 m/s.

   Concept and Intuition

   The machine gun's power output is the rate at which it delivers kinetic energy to bullets. If $n$ bullets, each of mass $m$ and speed $v$, are fired per second, then

   $P = n\times\left(\frac12 m v^2\right).$

   This lets us solve for the muzzle velocity $v$ once the firing rate, bullet mass, and power are known.

   Step-by-Step Solution

   1. Firing rate: $360$ bullets per minute $= \dfrac{360}{60} = 6$ bullets per second.
   2. Bullet mass: $5\ \text{g} = 0.005\ \text{kg}$.
   3. Power equation: $P = n\cdot\frac12 m v^2 \Rightarrow 15000 = 6\times\frac12\times0.005\times v^2$. …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.A wind mill converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. The electrical power output P is related to the velocity of wind v as (A) $v^3$ (B) $v^2$ (C) $\mathrm{v}^{3 / 2}$ (D) $\mathrm{V}^{5 / 2}$

   ›Reveal solutionSolution

   The power available from wind is proportional to the cube of the wind speed, because kinetic energy flux depends on $v^3$; the correct option is (A).

   Concept & Intuition

   Wind power comes from the kinetic energy of moving air. When wind passes through the area swept by the turbine blades, the mass of air hitting the blades per second depends on the wind speed, and the kinetic energy per unit mass also depends on the wind speed. Multiplying these gives a cubic relationship. The problem says the turbine converts a fixed fraction of the intercepted wind energy — so the output power is simply proportional to the total power in the wind.

   ---

   1. Kinetic energy of a moving mass A parcel of air of mass $m$ moving at speed $v$ has kinetic energy

   $KE = \frac{1}{2} m v^2.$

   2. Mass flow rate through the turbine In time $\Delta t$, the air that passes through the swept area $A$ of the blades is a cylinder of length $v \Delta t$ and cross-section $A$. Its volume is $A v \Delta t$, so the mass is

   $m = \rho \cdot (A v \Delta t),$

   where $\rho$ is the air density.

   The mass flow rate (mass per second) is therefore

   $\frac{m}{\Delta t} = \rho A v.$

   3. Power in the wind Power is energy per unit time. The kinetic energy arriving per second is

   $P_{\text{wind}} = \frac{1}{2} \left( \frac{m}{\Delta t} \right) v^2 = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} \rho A v^3.$

   4. Turbine output …

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4. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The kinetic energy of a body of mass $4\,\text{kg}$ moving with a velocity of $(2\hat{i} - 4\hat{j} - \hat{k})\,\text{ms}^{-1}$ is (A) $84\,\text{J}$ (B) $63\,\text{J}$ (C) $42\,\text{J}$ (D) $21\,\text{J}$

   ›Reveal solutionSolution

   Kinetic energy is a scalar given by $\frac12 m |\vec{v}|^2$; here $m=4\,\text{kg}$ and $|\vec{v}| = \sqrt{2^2+(-4)^2+(-1)^2} = \sqrt{21}$, so $KE = \frac12 \cdot 4 \cdot 21 = 42\,\text{J}$.

   The key idea is that kinetic energy depends only on the magnitude of velocity, not its direction. The velocity vector $(2\hat{i} - 4\hat{j} - \hat{k})$ tells us the speed through its components. We square each component, sum them, and take the square root to get speed — but since kinetic energy uses speed squared, we can skip the square root entirely.

   1. Recall the formula for kinetic energy For a body of mass $m$ moving with speed $v$,

   $KE = \frac12 m v^2.$

   Here $v$ is the magnitude of the velocity vector $\vec{v}$.

   2. Find the magnitude squared of the velocity Given $\vec{v} = 2\hat{i} - 4\hat{j} - \hat{k}$,

   $|\vec{v}|^2 = (2)^2 + (-4)^2 + (-1)^2 = 4 + 16 + 1 = 21.$

   So $v^2 = 21$ (units: $\text{m}^2/\text{s}^2$).

   3. Plug into the kinetic energy formula Mass $m = 4\,\text{kg}$, so $KE = \frac12 \cdot 4 \cdot 21 = 2 \cdot 21 = 42\,\text{J}.$ …

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5. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.In a hydroelectric power station, the water is flowing at $2\ \mathrm{ms}^{-1}$ in the river which is 100 m wide and 5 m depth. The maximum power output from the river is (A) 1.5 MW (B) 2 MW (C) 2.5 MW (D) 3 MW

   ›Reveal solutionSolution

   This tests computing the maximum power extractable from flowing water as the rate of kinetic energy delivery, $P=\tfrac12\dot m v^2=\tfrac12\rho A v^3$.

   Concept and Intuition

   The "maximum power" a hydroelectric station could ever extract from moving water equals the rate at which kinetic energy is carried past a cross-section by the flow — i.e. the KE of the water passing per second, assuming (ideally) all of it could be converted. Mass flow rate is $\dot m=\rho A v$ (density × cross-sectional area × speed), and the power carried is $P=\tfrac12\dot m v^2=\tfrac12\rho A v^3$.

   Step-by-Step Solution

   1. Cross-sectional area of the river $A = \text{width}\times\text{depth} = 100\times5=500\ \text{m}^2$.
   2. Mass flow rate $\dot m = \rho A v = 1000\times500\times2=1{,}000{,}000\ \text{kg/s}$. …

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6. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If the vertical component of earth’s magnetic field is $0.5 \times 10^{-4}$ T at a point. When an aeroplane of wing span 4 m is moving horizontally at this place at 360 kmh$^{-1}$, then the motional emf formed across the ends of the wings is (A) $20 \times 10^{-4}$ V (B) $20 \times 10^{-2}$ V (C) $20 \times 10^{-3}$ V (D) $2 \times 10^{-4}$ V

   ›Reveal solutionSolution

   The motional emf across the wings is given by $\mathcal{E} = B v L$, where $B$ is the vertical component of Earth’s magnetic field, $v$ is the plane’s speed, and $L$ is the wingspan. Substituting values gives $\mathcal{E} = 20 \times 10^{-3}$ V, so the correct option is (C).

   Concept & Intuition

   When a conductor moves through a magnetic field, the free charges inside experience a magnetic force, causing them to separate until an electric field balances it. This separation creates a potential difference — the motional emf. For a straight conductor of length $L$ moving perpendicular to a uniform magnetic field $B$ with speed $v$, the emf is $\mathcal{E} = B v L$. Here, the wings act as the conductor, and only the vertical component of Earth’s field matters because the plane moves horizontally — the horizontal component is parallel to the motion and produces no emf.

   Step-by-step solution

   1. Identify the relevant quantities

      • Vertical component of Earth’s magnetic field: $B = 0.5 \times 10^{-4} \, \text{T}$
      • Wingspan (length of conductor): $L = 4 \, \text{m}$
      • Speed of the aeroplane: $v = 360 \, \text{km/h}$

   2. Convert speed to SI units

      Since $1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}$,

   $v = 360 \times \frac{5}{18} = 100 \, \text{m/s}$

   3. Apply the motional emf formula The emf induced across the wingtips is

   $\mathcal{E} = B v L$

   Substitute the values:

   $\mathcal{E} = (0.5 \times 10^{-4}) \times 100 \times 4$

   4. Calculate step by step $0.5 \times 10^{-4} \times 100 = 0.5 \times 10^{-2} = 5 \times 10^{-3}$ …

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7. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.A bomb of mass 16 kg explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 m s$^{-1}$. The kinetic energy of the second piece is (A) 144 J (B) 192 J (C) 96 J (D) 288 J

   ›Reveal solutionSolution

   Using conservation of momentum, the velocity of the 4 kg piece is found to be 12 m/s opposite to the 12 kg piece, giving it a kinetic energy of 288 J. The correct option is (D).

   The key idea here is that in an explosion, the total momentum before and after is the same — and since the bomb starts at rest, the total momentum after the explosion must be zero. That means the two fragments fly off in opposite directions with momenta that cancel. Once we know the velocity of one piece, we can find the other’s velocity, and then its kinetic energy.

   1. Set up conservation of momentum. The bomb is initially at rest, so initial momentum = 0. After explosion, let the 4 kg piece have velocity $v$ (we’ll determine direction later). The 12 kg piece has velocity $4 \, \text{m/s}$ in some direction. Momentum conservation:

   $m_1 v_1 + m_2 v_2 = 0$

   $4 \cdot v + 12 \cdot 4 = 0$

   (We take the 12 kg’s velocity as positive; the sign of $v$ will come out negative, meaning opposite direction.)

   2. Solve for the unknown velocity.

   $4v + 48 = 0 \quad \Rightarrow \quad 4v = -48 \quad \Rightarrow \quad v = -12 \, \text{m/s}$

   The negative sign tells us the 4 kg piece moves in the opposite direction to the 12 kg piece, at 12 m/s.

   3. Compute the kinetic energy of the 4 kg piece. Kinetic energy formula: $KE = \frac{1}{2} m v^2$ …

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8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.A machine gun fires 300 bullets per minute each with a velocity of $500\ \mathrm{ms}^{-1}$. If the mass of each bullet is 4 g, the power of the machine gun is (A) 3.6 kW (B) 3 kW (C) 5.4 kW (D) 2.5 kW

   ›Reveal solutionSolution

   Power delivered by the gun is the kinetic energy imparted per bullet times the firing rate; this comes out to $2.5$ kW.

   Concept and Intuition

   Power is energy delivered per unit time. Each bullet leaves with a certain kinetic energy; the gun delivers this energy at the rate bullets are fired (bullets per second), so multiplying KE-per-bullet by the firing rate (in bullets/second) gives the average power output.

   Step-by-Step Solution

   1. Firing rate: $300$ bullets/min $= \dfrac{300}{60}=5$ bullets/s.
   2. Mass per bullet: $m = 4\text{ g} = 0.004\text{ kg}$; speed $v=500\text{ m/s}$.
   3. KE per bullet: $\dfrac12 mv^2 = \dfrac12(0.004)(500)^2 = \dfrac12(0.004)(250000) = 500\text{ J}$. …

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9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.A boat of mass $1000\, \mathrm{kg}$ goes from rest to speed $20.0\, \mathrm{m/s}$ in $5.0\, \mathrm{s}$. The water exerts a constant drag force and the acceleration of the boat is constant. If the average power required by the boat is $45000\, \mathrm{W}$, then the magnitude of the drag force is (A) $500\, \mathrm{N}$ (B) $750\, \mathrm{N}$ (C) $250\, \mathrm{N}$ (D) $1000\, \mathrm{N}$

   ›Reveal solutionSolution

   The extra power the engine supplies over and above the rate of gaining kinetic energy is spent against drag. This gives a drag force of $500\,\mathrm{N}$, option (A).

   1. Acceleration and net force.

   $a = \frac{\Delta v}{\Delta t} = \frac{20.0}{5.0} = 4.0\ \mathrm{m/s^2}, \qquad F_{\text{net}} = ma = 1000 \times 4.0 = 4000\ \mathrm{N}.$

   2. Distance travelled (from rest, constant acceleration):

   $d = \tfrac{1}{2}a t^2 = \tfrac{1}{2}(4.0)(5.0)^2 = 50\ \mathrm{m}.$

   3. Total work done by the engine (average power $\times$ time):

   $W_{\text{engine}} = P_{\text{avg}}\,t = 45000 \times 5.0 = 225000\ \mathrm{J}.$

   4. Work that becomes kinetic energy (work–energy theorem):

   $\Delta K = \tfrac{1}{2}m v^2 = \tfrac{1}{2}(1000)(20.0)^2 = 200000\ \mathrm{J}.$

   5. Work done against drag is the difference: $W_{\text{drag}} = W_{\text{engine}} - \Delta K = 225000 - 200000 = 25000\ \mathrm{J}.$ …

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10. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.The Bernoulli equation can be presented as $P + \frac{1}{2} \rho V^2 + \rho gz = C$ where $P$ is pressure, $\rho$ is fluid density, $V$ is fluid velocity, $g$ is gravitational acceleration, $z$ is distance and $C$ is the constant. The dimension of the constant $C$ is (A) $\left\{ \frac{m}{TL} \right\}$ (B) $\left\{ \frac{m}{TL^2} \right\}$ (C) $\left\{ \frac{m}{T^2 L^2} \right\}$ (D) $\left\{ \frac{m}{T^2 L} \right\}$

    ›Reveal solutionSolution

    The principle of dimensional homogeneity states that all terms in a valid physical equation must have the same dimensions. By finding the dimensions of any term in the Bernoulli equation, we find the dimension of the constant $C$ to be $ML^{-1}T^{-2}$, which corresponds to option (D).

    Concept and Intuition

    In physics, every physical quantity has a dimension, which tells us what fundamental quantities (like mass, length, and time) it is composed of. The principle of dimensional homogeneity is a fundamental rule: for any physically meaningful equation, the dimensions of all terms on both sides of the equation must be identical. This means you cannot add or subtract quantities that have different dimensions (e.g., you can't add a length to a time).

    The Bernoulli equation is given as $P + \frac{1}{2} \rho V^2 + \rho gz = C$. Since $C$ is a constant that is equal to the sum of the terms on the left side, its dimension must be the same as the dimension of each individual term on the left side. If they weren't, the equation would be dimensionally inconsistent and thus physically incorrect. Therefore, to find the dimension of $C$, we only need to find the dimension of any one of the terms on the left side.

    1. Identify the terms and their components:

       The Bernoulli equation is $P + \frac{1}{2} \rho V^2 + \rho gz = C$.

       We need to find the dimension of $C$. According to the principle of dimensional homogeneity, the dimension of $C$ must be equal to the dimension of $P$, or $\frac{1}{2} \rho V^2$, or $\rho gz$. Let's choose the pressure term $P$ as it is often the most straightforward.

    2. Determine the dimensions of pressure ($P$):

       Pressure is defined as force per unit area.

       • Force ($F$): From Newton's second law, $F = ma$ (mass $\times$ acceleration).
         • Dimension of mass ($m$) = $[M]$
         • Dimension of acceleration ($a$) = $[LT^{-2}]$ (length per time squared)
         • So, dimension of force ($F$) = $[M][LT^{-2}] = [MLT^{-2}]$
       • Area ($A$): Area is length squared.
         • Dimension of area ($A$) = $[L^2]$
       • Pressure ($P$):
         • Dimension of $P = \frac{\text{Dimension of Force}}{\text{Dimension of Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$

    3. Verify with other terms (optional but good practice):

       Let's quickly check the dimensions of the other terms to confirm our understanding.

       • Term 2: $\frac{1}{2} \rho V^2$
         • The constant $\frac{1}{2}$ is dimensionless.
         • Dimension of density ($\rho$) = $\frac{\text{Dimension of Mass}}{\text{Dimension of Volume}} = \frac{[M]}{[L^3]} = [ML^{-3}]$ …

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11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.A body of mass 0.6 kg is moving along a circular path of radius 1 m. If the body moves with $\dfrac{900}{\pi}$ revolutions per minute, its kinetic energy is (A) 120 J (B) 270 J (C) 360 J (D) 240 J

    ›Reveal solutionSolution

    Convert the given rpm to angular speed, get the linear speed from

    $v=\omega r$, then apply the usual translational kinetic-energy formula —

    the answer is 270 J.

    Concept and Intuition

    A particle moving on a circular path still has ordinary translational kinetic

    energy $\tfrac12mv^2$, where $v$ is its instantaneous linear (tangential)

    speed. The awkward-looking $900/\pi$ rpm is a deliberate hint: converting rpm

    to rad/s always brings in a factor of $2\pi/60$, and the given number is

    engineered to cancel that $\pi$ exactly, leaving a clean $\omega$.

    Step-by-Step Solution

    1. Angular speed: $\omega = \dfrac{2\pi N}{60}$ with $N = \dfrac{900}{\pi}$ rpm.
    2. $\omega = \dfrac{2\pi}{60}\times\dfrac{900}{\pi} = \dfrac{2\times900}{60} = 30\ \mathrm{rad/s}$ (the $\pi$'s cancel).
    3. Linear (tangential) speed: $v = \omega r = 30\times1 = 30\ \mathrm{ms^{-1}}$. …

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12. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.Water is falling on the blades of a turbine from a height of 25 m. $3 \times 10^3$ kg of water pours on the blade per minute. If the whole of energy is transferred to the turbine, power delivered is: (A) 12250 W (B) 16250 W (C) 8250 W (D) 20250 W

    ›Reveal solutionSolution

    The turbine converts the gravitational PE lost by falling water each second into mechanical power; $P=mgh/t = 12250$ W.

    Concept and Intuition

    Water falling through a height $h$ loses gravitational potential energy $mgh$. If ALL of this energy is transferred to the turbine blades (ideal, no losses), the power delivered equals the rate at which this potential energy is lost, i.e. $P = \dfrac{mgh}{t}$, where $m$ is the mass falling in time $t$.

    Step-by-Step Solution

    1. Mass flow rate: $m = 3\times 10^3$ kg every $t = 1$ min $= 60$ s.
    2. Height of fall: $h = 25$ m.
    3. Energy delivered per minute: $E = mgh = 3\times10^3 \times 9.8 \times 25 = 7.35\times10^5\ \text{J}$. …

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