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 biology modern agriculture modern equipments in agriculture soil quality methods of crop production

Agricultural tractors generally have a horsepower (HP) of 

  1. 20-50

  2. 50-70

  3. 10-15

  4. 20-25
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Tractors are generally categorised as based on the size, weight and horsepower. It provides high torque and slow speed and thus is very useful in agricultural fields. In India, four wheel tractors for agricultural operations are fitted with 25-80 hp. Walking type tractors are fitted with 8-12 hp engines
Clutch. Thus , the correct answer is option A.
Multiple choice biology modern agriculture modern equipments in agriculture soil quality methods of crop production

The working efficiency per day of deshi plough is

  1. 0.3 ha

  2. 0.4 ha

  3. 0.6 ha

  4. 0.8 ha
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Plough is a tool which is used by the farmers for cultivation of the soil, planting for the sowing of seeds and turning over the soil and cut furrows. The tool is pulled by oxen or tractors through the ground in order to prepare furrows on the land. 

So the correct option is '0.4 ha'.

Multiple choice luminous intensity measurements physics

The luminous efficiency of a lamp is $8.8$ lumen/watt and its luminous intensity is $700\ Cd$. The power of the lamp will be 

  1. $10^{1}$ $W$
  2. $10^{2}$ $W$
  3. $10^{3}$ $W$
  4. $10^{4}$ $W$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$Luminous \ flux = Luminous \ intensity * solid \ angle $

Total solid angle = 4 $\pi$

So, Luminous flux = 700 x 4 $\pi$

Luminous flux = Luminous efficiency x power
=> 700 x 4 $\pi$ = 8.8 x power
=> Power of lamp = 1000 W

Answer. C) 10$^3$ W
Multiple choice luminous intensity measurements physics

The luminous efficiency of a lamp is $5$ lm $W^{-1}$ and its luminous intensity is $30$ candela. The power of the lamp will be 

  1. $6\pi$ $W$
  2. $12\pi$ $W$
  3. $24\pi$ $W$
  4. $48\pi$ $W$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Luminous \ flux = luminous \ intensity * 4 \pi$

$\longrightarrow$ F = $30 * 4 \pi   lm$

Luminous efficiency = $ 5  lm / W$

Power = $ F / Power$
            = $ 30 * 4 \pi / 5 = 24 \pi$

Answer. C) $24 \pi  W$

Multiple choice luminous intensity measurements physics

A lamp of $250$ candle power is hanging at a distance of $6$m from a wall. The illuminace at a point on the wall at a minimum distance from the lamp will be 

  1. $9.64$ lux
  2. $4.69$ lux
  3. $6.94$ lux

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

1 Candle power = 0.981 Candela

Luminous intensity = 250*0.981 Candela

Minimum distance of lamp from wall = 6 m

Illuminance = $L/d^{2}$
= $250*0.981/6^{2}$
= 6.8125 lux

Answer. D) none of these

Multiple choice luminous intensity measurements physics

The lumen efficiency, if an electric bulb emit $68.5\dfrac{lumen}{watt}$ is:

  1. $2.5\%$
  2. $5\%$
  3. $10\%$
  4. $20\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Luminous efficiency is the ratio of luminous flux to power. A standard 100% efficient source (monochromatic green light at 555 nm) produces 685 lumens/watt. Efficiency = (68.5 / 685) * 100% = 10%.

Multiple choice luminous intensity measurements physics

The luminous efficiency of the bulb in lumen/watt, if luminous intensity of a $100$ watt unidirectional bulb is $100$ candela, is

  1. $12$
  2. $12.56$
  3. $13$
  4. $15$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$F=4\pi l$
$F=(4\times 3.14)\times 100=1256$ lumen
$\therefore$ Luminous efficiency $=\dfrac{luminous\ flux}{electric\ power}=\dfrac{1256}{100}=12.56$

Multiple choice physics solar equipment production of electricity from solar energy solar power alternate sources of energy solar power plant solar energy and its applications renewable and non-renewable resources renewable and non-renewable sources of energy generation of electricity

The efficiency of a solar cell is 20% and its surface area is $4  {cm}^{2}$. Find the electrical energy generated in one second, by 3000 cells, connected in a solar panel, if $640  J$ of solar energy is incident on one meter square area, in one second.

  1. 183.6 J

  2. 153.6 J

  3. 193.6 J

  4. 123.6 J
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total surface area of solar cells on which solar energy is falling = area of each cell $\times$ number of cells $= 4 \times 3000 {cm}^{2} = 1.2  {m}^{2}$
The solar energy falling on 3000 cells 
${Q} _{1} = 1.2 \times 640  J = 768  J$
The energy converted into electric energy $= \displaystyle\frac{20}{100} \times 768 = 153.6  J$
The electric energy generated in one second is $153.6  J$

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

Power of a water pump is 2 kW. If $g=m/{ sec }^{ 2 }$, The amount of water it can raise in one minute to a height of 10 m/s 

  1. 100 Litre

  2. 1200 Litre

  3. 1000 Litre

  4. 2000 Litre
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = 2 kW = 2000 W. Work needed to lift mass m to height h in time t: W = mgh. P = mgh/t, so m = Pt/(gh) = (2000×60)/(10×10) = 120000/100 = 1200 kg. This equals 1200 liters of water.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A pump motor is used to deliver water at a certain rate from the given pipe. To obtain 'n' times water from the same pipe in the same time, the amount of power of the motor should be increased to:

  1. np
  2. ${n^3}$
  3. ${n^2}$

  4. 2np
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power P = (1/2)mv^2/t = (1/2)(rho*A*v*t)v^2/t = (1/2)rho*A*v^3. If flow rate Q = Av is increased by n, then v must increase by n (assuming area constant). Thus P is proportional to v^3, so power increases by n^3.

Multiple choice commercial applications management of natural resources unequal distribution of resources protecting our environment rules are for everyone

A biogas plant works to its maximum capacity when

  1. Conditions are aerobic and sewage is supplied
  2. Conditions are anaerobic and temperature $40^oC$
  3. Conditions are aerobic and temperature $40^oC$

  4. Conditions are anaerobic and sewage is supplied
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Biogas refers to a mixture of different gases produced by the breakdown of organic matter in the absence of oxygen. Biogas is produced by anaerobic digestion with methanogen or anaerobic organisms, which digest material inside a closed system or fermentation of biodegradable materials. There are two key processes: mesophilic and thermophilic digestion which is dependent on temperature. Research in biogas production indicates that the optimal temperature for mesophilic bacteria is around 37°C and around 55°C for thermophilic bacteria.


So, the correct answer is 'Conditions are anaerobic and temperature 40C'.

Multiple choice chemistry rocks and minerals coal and its products study of coal coal

How much coal is needed to supply enough electricity to light ten 100-watt bulbs for about an hour?

  1. 1 pound

  2. 10 pounds

  3. 50 pounds

  4. 20 pounds
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

100 watts = 0.1 KW


For 1 bulb 0.1 KW of power is required, for 10 bulbs that turn out to be 1KW.
Now 10 bulbs have to light for 1 hour that will be 1 KW multiplied by 1 hour = 1 KW

The thermal energy of coal is 6,150 KW/ton, only 40% of the thermal energy of the coal is converted to electricity. so the electricity converted per ton of coal is 0.4 * 6,150 = 2,460 KW/ton.

Number of tons of coal burned for 10 light bulbs for one hour = 1 KW / 2,460KW/ton = 0.0004065 tons 

Multiplying by 2000 pound/ton we get 0.81300 $\approx$ 1 pounds.