Physics

Power, Energy, and Efficiency

463 Questions

Power, energy, and efficiency are fundamental physics concepts that quantify work and the rate of doing work. Questions cover electrical energy consumption in kilowatt hours, the efficiency of thermodynamic cycles like the Brayton cycle, and mechanical power lifting loads. This topic is essential for general science and engineering exam preparation.

Electrical energy consumptionBrayton cycle efficiencyPower calculation formulasCommercial electricity unitsHydraulic lift mechanics

Power, Energy, and Efficiency Questions

Multiple choice physics heat and energy zeroth law of thermodynamics

If the output energy is$ 1000 J$ and input energy is $200 J $the net output is _____ $J$.

  1. $800$
  2. $1200$
  3. $10$
  4. $500$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The net output $=$ output - input
$= 1000 - 200 = 800 J$

Multiple choice physics forces - vectors and moments the turning of couple couple rotational motion of a rigid body and moment of inertia turning effect of force

An automobile engine develops $100$ $kW$ when rotating at a speed of $1800\ rev/min$. The torque it delivers is

  1. $3.33\ N-m$
  2. $200\ N-m$
  3. $530.5\ N-m$
  4. $2487\ N-m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Power\quad P=100kW\quad =100000W\ w=1800\times \cfrac { 2\pi  }{ 60 } \quad rad/s\ \quad =60\pi \quad rad/s\ P=torque\times w\ torque=530.5\quad Nm$

Multiple choice energy efficiency energy transformations and energy transfers physics

A line having a total resistance of $0.5 \Omega$ delivers $15$kW at $240$ volt to a small factory. The efficiency of transmission will be :

  1. $97\%$
  2. $88.5\%$
  3. $68\%$
  4. $79\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The current delivered is I = P/V = 15000 / 240 = 62.5 A. The power lost in the transmission line is P_loss = I^2 * R = (62.5)^2 * 0.5 = 1953.125 W. The input power is P_in = P_out + P_loss = 15000 + 1953.125 = 16953.125 W, yielding an efficiency of (15000 / 16953.125) * 100 = 88.5 percent.

Multiple choice energy efficiency energy transformations and energy transfers physics

A crane can lift up $10,000$ kg of coal in $1$ hour from a mine of $180$m depth. If the efficiency of the crane is $60\%$, its input power must be? $(g=10ms^{-2})$

  1. $5$ kW
  2. $8.3$ kW
  3. $50$ kW
  4. $62.5$ kW
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Useful work done per second or output power is P_out = mgh / t = (10000 * 10 * 180) / 3600 = 5000 W = 5 kW. Given that the efficiency is 60 percent, input power is P_in = P_out / efficiency = 5 kW / 0.6 = 8.33 kW.

Multiple choice energy efficiency energy transformations and energy transfers physics

An engine pump $400$kg of water through height of $10m$ in $40s$. Find the power of the engine if its efficiency is $80\%$ (Taken $g=10ms^{-2}$).

  1. 800W

  2. 900W

  3. 600W

  4. 500W
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m=400kg$
$H=10m$
$t=40 \sec$
Input energy $=400\times g \times H=mgn$
$E _i=400 \times 10\times 10=40000 J$
Energy per $\sec , E _1=\dfrac{40000}{40}=1000 J/s$
Power input $=1000 J/s$
output =$1000\times 80\%$
$800J/s$=$800W$

Multiple choice energy efficiency energy transformations and energy transfers physics

Among the following the correct expression of efficiency is 

  1. $ \dfrac {Output \ energy}{Input\ energy} $
  2. $ \dfrac {Work\ done\ by \ the\ machine}{Work\ done\ on\ the\ machine} $
  3. $ \dfrac {Load}{Effort} $

  4. All the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$effieciency = \dfrac{{energy\,\,output}}{{energy\,\,input}}$

Hence,
option $(A)$ is correct answer.

Multiple choice energy efficiency energy transformations and energy transfers physics

The power of a heart which pumps $5\times{10}^{3}cc$ of blood per minute at a pressure of $120mm$ of mercury ($g=10m{s}^{-2}$ and density of $Hg=13.6\times{10}^{3}kg/m$) is

  1. $1.36W$
  2. $13.6W$
  3. $0.136W$
  4. $136W$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Pressure P = h * rho * g = 0.12m * 13600 * 10 = 16320 Pa. Volume V = 5 * 10^3 cc = 5 * 10^-3 m^3. Work = P * V = 16320 * 5 * 10^-3 = 81.6 J. Power = Work / time = 81.6 J / 60s = 1.36 W.

Multiple choice energy efficiency energy transformations and energy transfers physics

A motor has an electrical input of $30 kJ$ and is used to raise $100 kg$ load to a height of $25 m$ when fired to a crane winch. What is the efficient of winch ? ($g = 10 \ ms^{-1}$)

  1. $0.75 %$
  2. $83.3 %$
  3. $75 %$
  4. $17.5 %$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Efficiency is calculated as (useful work output / total energy input) * 100. The useful work is potential energy mgh = 100 * 10 * 25 = 25,000 J or 25 kJ. Efficiency = (25 kJ / 30 kJ) * 100 = 83.33%.

Multiple choice energy efficiency energy transformations and energy transfers physics

A machine which is 75% efficient, uses 12 J of energy in lifting 1 kg mass through a certain distance. The mass is then allowed to fall through the same distance. The velocity at the end of its fall is:

  1. $\sqrt{12} $ m/s
  2. $\sqrt{18} $ m/s
  3. $\sqrt{24} $ m/s
  4. $\sqrt{32} $ m/s
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Efficiency = 75%, Input energy = 12 J
$\therefore \displaystyle \frac{75}{100} = \frac{\text{Output energy}}{\text{Input energy}}$
$\Rightarrow $ Output energy $= \displaystyle \frac{75}{100} \times 12 = 9 J$
$\therefore$ P.E. of the mass = 9 J
At the end of the fall it will be converted to K.E.
$\therefore \displaystyle \frac{1}{2} mv^2 = 9$
$\Rightarrow \displaystyle \frac{1}{2} \times 1 \times v^2 = 9$
$\Rightarrow v^2 = 18$
$\Rightarrow v = \sqrt{18} m/s$

Multiple choice energy efficiency energy transformations and energy transfers physics

An installation consisting of an electric motor driving a water pump left $75 L$ of water per second to a height of $4.7 m$. If the motor consumes a power of $5 kW$, then the efficiency of the installation is

  1. $39$%
  2. $69$%
  3. $93$%
  4. $96$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power consumed by motor$5kW= 5 \times 10^3 W= 5000W$

power used i lifting water = $\dfrac{mgh}{t}=7.5 \times 9.8 \times 4.7= 3454.5 W$
Efficiency = $\dfrac{\text{Power used}}{\text{Power consumed}} \times 100$% = $\dfrac{3454.5}{5000} \times 100$% = 69 %

Multiple choice power transmission household electricity household circuits electricity and magnetism physics

An electric fan marked 60 watt consumes 3 units for the duration:

  1. 50 hours

  2. 150 hours

  3. 10 hours

  4. 15 hours
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The total energy consumed is 3 units = 3 kWh = 3000 Wh.
The total enrgy dissipated is given as Power * Time.
Therefore, time = Energy dissipated / Power.
Time taken = 3000 Wh / 60 W = 50 hours.
Hence, an electric fan marked 60 watt consumes 3 units for the duration of 50 hours.

Multiple choice power transmission household electricity household circuits electricity and magnetism physics

The heart does $1.5\ J$ of work in each heart beat. How many times per minute does it beat if its power is $2W$?

  1. $20$
  2. $40$
  3. $60$
  4. $80$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Power is defined as work done per unit time, or P = W_total / t. Total work per minute is number of beats (n) multiplied by work per beat (1.5 J). Thus, 2 W = (n * 1.5 J) / 60 s, which yields n = (2 * 60) / 1.5 = 80 beats per minute.