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 energy and its forms introduction to work work introduction to work and energy

The correct relation between joule and erg is:

  1. $1\ J = 10^{-5} erg$
  2. $1\ J = 10^{5} erg$
  3. $1\ J = 10^{-7} erg$
  4. $1\ J = 10^{7} erg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.

Thus,

$1 joule = 1 newton \times 1 m\\$

$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$

$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$

$1 J = 10^{7} \ erg$

Thus option D is correct.

Multiple choice physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

Find out the most efficient engine in the following

  1. An engine converts 80 KJ of heat energy into 20 KJ of work

  2. An engine converts 50 KJ of heat energy into 15 KJ of work

  3. An engine converts 30 KJ of heat energy into 6 KJ of work

  4. An engine converts 60 KJ of heat energy into 24 KJ of work
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Efficiency of heat engine= $\frac{work done}{heat input}$ Going by the above options efficiency is maximum in option D, and is equal to 40 percent.

Multiple choice physics heat engine: second law of thermodynamics conversion of heat into work: heat engine and it's efficiency engines and cycles heat engines refrigerators and heat pumps

Which of the following engines is more efficient?

  1. Heat utilised - 80 kilojoules , work done - 32 kilojoules

  2. Heat utilised - 60 kilojoules , work done - 12 kilojoules

  3. Heat utilised - 50 kilojoules , work done - 25 kilojoules

  4. Heat utilised - 90 kilojoules , work done - 27 kilojoules
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Efficiency of engine   $\eta = \dfrac{W}{Q _H}$  

where $W$ is work done and $Q$ is heat taken from source (Heat utilised)
(A) :  $\eta _A = \dfrac{32}{80} = 0.4$
(B) :  $\eta _B = \dfrac{12}{60} = 0.2$
(C) :  $\eta _C = \dfrac{25}{50} = 0.5$
(D) :  $\eta _D = \dfrac{27}{90} = 0.3$
Hence engine C is the most efficient.

Multiple choice perceive colours resolution of optical instruments lenses option c: imaging physics

Using the following data,choose the correct option:
                     C        D        F      
 Crown   1.5145   1.5170  1.5230   
FLINT    1.6444    1.6520  1.6637                  

  1. The dispersive power for crown glass is 0.1644

  2. The dispersive power for flint glass is 0.029601

  3. The dispersive power for crown is 0.01644

  4. The dispersive power for flint glass is 1.29601
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice physics effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

A $1000\Omega$ electric iron is connected to $200v$, $50Hz$ ac source. Calculate average power delivered to iron, peak power and energy spent in one minute?

  1. $400W,\ 800W,\ 12\times 10^{5}\ J$
  2. $400W,\ 900W,\ 1.2\times 10^{5}\ J$
  3. $500W,\ 800W,\ 6\times 10^{5}\ J$
  4. $400W,\ 900W,\ 60\times 10^{5}\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics effects of electric current joule's law of heating thermal effect of electric current heating effect of electric current

An electric iron draws a current of 15 A from a 220 V supply, What is the cost of using iron for 30 min everyday for 15 days if the cost of unit (1 unit =1 kWhr) is 2 rupees ? 

  1. Rs 49.5

  2. Rs 60

  3. Rs 40

  4. Rs 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As we know that power(P)=v$\times$I
so P=15$\times$220=3300 watt
cost per unit=2 rupees 
therefore,$\dfrac {3300\times2\times15\times30}{60\times1000}$=49.5 rupees.

Multiple choice chemistry occurrence of carbon compounds in nature calorific value study of enthalpy chemical thermodynamics

A family consumes 12 kg of LPG in 30 days. Calculate the average energy consumed per day if the calorific value of LPG is 50 kJ/kg.

  1. 10,000 J/day

  2. 15,000 J/day

  3. 20,000 J/day

  4. 25,000 J/day
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy released by 12 kg of LPG =  12 %$\times$ 50 kJ $= $ 600 kJ
$\therefore$ energy consumed per day
$\displaystyle \frac {600\,kJ}{30}\, =\, 20,000$ Joules/day.

Multiple choice physics alternating current power in ac circuits average power in ac circuit and power factor power in ac circuit

If $V=100 \sin 100t$ volt, and $I=100 \sin(100t+\dfrac {\pi}{6})A$. then find the watt less power in watt?

  1. $10^{4}$
  2. $10^{3}$
  3. $10^{2}$
  4. $2.5 \times 10^{3}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P= V _{rms} \times I _{rms} \times \cos \phi$


$\quad= \large\frac{V _0I _0}{\sqrt{2}\times \sqrt{2}}\times \cos \dfrac{\pi}{6}$

$\quad= \large\frac{100 \times 100}{2} \times \frac{\sqrt{3}}{2}=2.5\times 10^3\sqrt{3}W$