Physics

Power, Energy, and Efficiency

443 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

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

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 physics pressure in fluids and atmospheric pressure examples of hydraulic press applications of pascal's law pascal's law and its applications

The radii of the press plunger and the pump plunger are in the ratio 30 : 4. If an effort of 32 kgf acts on the pump plunger. Find the maximum effort the press plunger can overcome.

  1. 1600 kgf

  2. 1700 kgf

  3. 1800 kgf

  4. 1900 kgf
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac {\text {Radius of the press plunger}}{\text {Radius of pump plunger}}=\dfrac {R}{r}$


$=\dfrac {30}{4}$

According to the principal of hydraulic machine,

$\dfrac {L}{E}=\dfrac {\pi R^2}{\pi r^2}$

$\Rightarrow \dfrac {L}{32 kgf}=(\dfrac {30}{4})^2$


$\Rightarrow \dfrac {L}{32}=\dfrac {900}{16}$

$\Rightarrow L=\dfrac {900\times 32}{16}=1800 kgf$