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 lever common machines terms related to machines introduction to simple machines

A load will be lifted by an effort of 12 N, the velocity ratio is 18 and the efficiency of the machine at this load is 60%, if the machine has a constant frictional resistance, determine the law of machine.

  1. P = W/18 + 4.8

  2. P = W/8 + 6.4

  3. P = W/14 + 8.2

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Efficiency = (Mechanical Advantage) / (Velocity Ratio). Given efficiency = 0.60 and Velocity Ratio = 18, the Mechanical Advantage (MA) = 0.60 * 18 = 10.8. Since MA = W/P, then P = W/10.8. The law of machine is P = mW + c. Using the provided option A, P = W/18 + 4.8, this does not match the calculated slope. However, assuming the question implies a standard linear form, the provided answer is likely based on specific machine parameters not fully defined.

Multiple choice physics lever common machines terms related to machines introduction to simple machines

In a simple machine, whose velocity ratio is $30$, a load of $2400$ N is lifted by an effort of $150$ N and a load of $3000$ N is lifted by an effort of $180$ N. Find the law of machine -

  1. $P = 0.50W + 144$
  2. $P = 0.60W + 25$
  3. $P = 0.05 W + 144$
  4. $P = 0.05 W + 30$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The law of machine is P = mW + c. We have two points: (2400, 150) and (3000, 180). Slope m = (180 - 150) / (3000 - 2400) = 30 / 600 = 0.05. Substituting into P = 0.05W + c: 150 = 0.05(2400) + c => 150 = 120 + c => c = 30. Thus, P = 0.05W + 30.

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

The height of water in a dam reduces by $20  m$, which is used to generate electricity. The water further fell by $10  m$ into the tunnel to strike the turbine plates. If the volume of water is ${10}^{4}  {m}^{3}$, find the hydel energy generated. Assume all the potential energy of the water being converted into electricity. Take $g = 10  m{s}^{-2}$.

  1. 2000 MJ

  2. 200 MJ

  3. 1000 MJ

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

The potential energy of the stored water is converted into kinetic energy which is then converted into electrical energy.
The potential energy of water $= mgh$        ...... (1)
The volume of water $= 10000  {m}^{3}$.
The mass of water $=$ volume $\times$ density $= 10000 \times 1000 = {10}^{7}  kg$.
Also, the depth of water $= 20  m$.
$\therefore$ The height of CG of stored water $= 10  m$.
The total height through which water falls $= 20  m$.
Substituting it in (1), we get potential energy of water $= {10}^{7} \times 20 \times 10 = 2000  MJ$.

Multiple choice physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

A house is fitted with 10 tubes of 40 W. If all tubes are lighted for 10 hours and if the cost of one unit of electricity energy is Rs. 2.50 the total cost of electricity consumption is ...

  1. Rs. $100$
  2. Rs. $20$
  3. Rs. $25$
  4. Rs. $10$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total power = 10 tubes * 40 W = 400 W = 0.4 kW. Energy = 0.4 kW * 10 hours = 4 kWh. Cost = 4 kWh * 2.50 Rs/kWh = 10 Rs.

Multiple choice physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

An electric lamp is marked $60\ W,\ 220\ V$. The cost of kilo watt hour of electricity is $Rs.\ 1.25$. The cost of using thing lamp on $220\ V$ for $8\ hrs$ is:

  1. $Rs.\ 0.25$
  2. $Rs.\ 0.60$
  3. $Rs.\ 1.2$
  4. $Rs.\ 4.00$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:

Power, $P=60\ W$
Time, $t=8\ hrs$

Energy consumed per day, $E=Pt=60\times 8=480\ Whr=\dfrac{480}{1000}=0.48\ kW hr$
Hence total cost, $C=0.48\times 1.25=Rs. 0.60 $

Multiple choice physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

An electric bulb of $60\ W$ is used for $6\ hrs/day$. Calculate the units of energy consumed in one day by the bulb.

  1. $0.18\ \text{units}$
  2. $0.36\ \text{units}$
  3. $0.54\ \text{units}$
  4. $0.72\ \text{units}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given :    $t = 6\ hrs$             

                $P = 0.06\ kW$
Thus energy consumed per day is:

$E=  Pt = 0.06\times 6  = 0.36\ kWh$$ = 0.36\ \text{units}$

Multiple choice physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

An electric bulb of $30\ W$ consumes $0.72$ units of energy in a day. Find the number of hours it is working in a day?

  1. $6\ hrs$
  2. $12\ hrs$
  3. $18\ hrs$
  4. $24\ hrs$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given :   $P =0.030$ kW
Energy consumed per day       $E= 0.72\ \text{units}$$ = 0.72\ kWh$
Using   $E = P\times t$
$\therefore$   $0.72 = 0.03\times t$               
$\implies t = 24\ hrs$
Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

How much units of energy is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

  1. 100 kwh

  2. 150 kwh

  3. 160 kwh

  4. 500 kwh
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given :    $P = 0.05$ kW               $t = 10$ hours per day 
Number of bulbs     $n = 10$
Energy consumed per day       $E= nPt = 10 \times 0.05\times 10 = 5$ kwh  per day
Total energy consumed in 30 days    $E _t = 5\times 30 = 150$ kWh
Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

  1. 1500

  2. 5000

  3. 15

  4. 150
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy consumed $=10\times 50\times 10\times 30\times 3600 J$
$[1 Kwh=3600\times 1000J]$
$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$


Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A geyser of $2.5 kw$ is used for $8$ hours daily. Calculate the monthly consumption ( 30 days) of electrical energy units. Also calculate the cost of electricity units consumed in a month if rate per unit is $3.50$

  1. $Rs. 2100.00$
  2. $Rs. 155.00$
  3. $Rs. 150.00$
  4. $Rs. 30.00$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Daily usage of electricity   $E _1 = 2.5 \ kw\times 8 = 20 \ kwh = 20 \ units$

Energy used in one month   $E = 30E _1 = 30\times 20 = 600\ units$
Cost    $ = Rs. 600\times 3.5 = Rs 2100.00$