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 power work and power work, energy and power physics energy and its forms

An engine of $4.9 kW$ power is used to pump water from a well which is $20 m$ deep. What quantity of water in kiloliters can it pump out in $30$ minutes?

  1. $45 kl$
  2. $75 kl$
  3. $25 kl$
  4. $90 kl$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power$=\dfrac{work}{time}=\dfrac{mgh}{t}$

$=\dfrac{4.9\times 10^3\times 30\times 60}{9.8\times 20}=m$
$m=45kl$

Multiple choice power work and power work, energy and power physics energy and its forms

An area of land is an average of $2\ m$ below sea level. To prevent flooding, pumps are used to lift rainwater up to sea level. What is the minimum pump output power required to deal with $1.3 \times 10^9\ Kg$ of rain per day?

  1. $15\ KW$
  2. $30\ KW$
  3. $100\ KW$
  4. $300\ KW$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$m=1.3\times 10^9 kg$
$g=10m/s^2$
$h=2m$
$t=1day$
Power, $P=\dfrac{mgh}{t}$
$P=\dfrac{1.3\times 10^9\times 10\times 2}{1\times 24\times 60\times 60}$
$P=300\times 10^3=300kW$
The correct option is D.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of 20 : 1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 kg ? 

  1. 3.75 kg

  2. 37.5 kg

  3. 7.5 kg

  4. 75 kg
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The answer is A.

Pressure is the amount of force acting per unit area. That is, P=F/A.
where:
p is the pressure,
F is the normal force,
A is the area of the surface on contact. Let us consider A = $\pi { r }^{ 2 }$.
Therefore, $\dfrac { { F } _{ 1 } }{ { \pi { r } _{ 1 } }^{ 2 } } =\dfrac { { F } _{ 2 } }{ { \pi { r } _{ 2 } }^{ 2 } } $.


In this case, $\dfrac { 1500 }{ { 20 }^{ 2 } } =\dfrac { W }{ { 1 }^{ 2 } } ,\quad W\quad =\quad 3.75\quad kg.$
Hence, weight to be placed on the small piston sufficient to lift a car of mass 1500 kg is 3.75 kg.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

Power required to drive a centrifugal pump is directly proportional to __________ of its impeller.

  1. Cube of diameter

  2. Fourth power of diameter

  3. Square of diameter

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

Power required to drive a centrifugal pump is directly proportional to the fourth power of the diameter of its propeller.

$\therefore$  Power     $P \propto  D^4$

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

The area of the piston in hydraulic machine are $10cm^2$ and $225cm^2$. The force required on the smaller piston to support a load of $1000N$ on the larger piston.

  1. 44.44 N

  2. 55.55 N

  3. 33.33 N

  4. 4.44 N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} \dfrac { f }{ { 10 } } =\dfrac { { 1000 } }{ { 225 } }  \ f=\dfrac { { 10000 } }{ { 225 } }  \ f=44.44N \end{array}$

$ \therefore$ Option $A$ is correct.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of $20 : 1$. What weight placed on the small piston will be sufficient to lift a car of mass $1500 kg$ ?

  1. $3.75 kg$
  2. $37.5 kg$
  3. $7.5 kg$
  4. $75 kg$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

Pressure is the amount of force acting per unit area. That is, $P=F/A$.
where:
$p$ is the pressure,
$F$ is the normal force,
$A$ is the area of the surface on contact. Let us consider A = $\pi { r }^{ 2 }$.
Therefore, $\dfrac { { F } _{ 1 } }{ { \pi { r } _{ 1 } }^{ 2 } } =\dfrac { { F } _{ 2 } }{ { \pi { r } _{ 2 } }^{ 2 } } $.
In this case, $\dfrac { 1500 }{ { 20 }^{ 2 } } =\dfrac { W }{ { 1 }^{ 2 } } ,\quad W\quad =\quad 3.75\quad kg.$
Hence, weight to be placed on the small piston sufficient to lift a car of mass 1500 kg is 3.75 kg.

Multiple choice physics machines levers and pulleys pulley pulleys

$1000$ N force is required to lift a hook and $10000$ N force is requires to lift a load slowly. Find power required to lift hook with load with speed $v = 0.5 m/sec$

  1. $5kw$
  2. $5.5kw$
  3. $1.5kw$
  4. $4.5kw$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$P=F.V=(1000+1000)0.5$
$=\dfrac{11000}{2}=5500=5.5kw$

Multiple choice physics machines levers and pulleys pulley pulleys

A pulley system has a velocity ratio $3$ and an efficiency of $80\%$. Calculate the effort required to raise a load of $300N$.

  1. $125 N$
  2. $300 N$
  3. $150 N$
  4. $225 N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$Velocity\ Ratio = 3$

$Efficiency = 80/100$

$Mechanical \ Advantage = efficiency \times Velocity\  ratio$

$\Rightarrow \dfrac{80}{100} \times 3 $

$\Rightarrow 2.4$


$M.A = \dfrac{load}{effort}$

$2.4 = \dfrac{300 N}{Effort}$

$\Rightarrow Effort = \dfrac{300 N}{2.4} = 125N$

Multiple choice physics machines levers and pulleys pulley pulleys

A pulley system has a velocity ratio $3$ and an efficiency of $80\%$. Calculate the mechanical advantage of the system.

  1. $2.0$
  2. $3.0$
  3. $2.4$
  4. $1.4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Velocity\ Ratio = 3$


$Efficiency = 80/100$

$Mechanical \ Advantage = efficiency \times Velocity\  ratio$

$\Rightarrow \dfrac{80}{100} \times 3 $

$\Rightarrow 2.4$

Multiple choice solid waste management environmental management biology

$150$ tonnes of solid waste can generate ________ MW of power.

  1. $12$
  2. $0.12$
  3. $1.2$
  4. $120$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Material recovery is possible from industrial waste. Organic waste can be used for energy production with various processes such as bio-gas production. About $150$ tonnes of solid waste can produce $14000$ cu.cm. Of bio-gas, which generates $1.2$MW of power and $45$ tonnes of manure.

Multiple choice physics energy : forms and sources concept of energy energy for everything forms of energy

What will be the potential energy of a body of mass 5 kg kept at a height of 10 m ?

  1. 50 J

  2. 0.5 J

  3. 500 J

  4. 25 J
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Answer is C.

Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 10 meters per second on earth. The formula for potential energy due to gravity is PE = mgh. As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases. 
In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.
Hence, the potential energy of a body of mass 5 kg kept at a height of 10 m is 500 J.

Multiple choice physics energy : forms and sources concept of energy energy for everything forms of energy

1 kilowatt hour $=$ ___________ .

  1. $10.6 \times 10^6$Joule
  2. $3.6 \times 10^6$Joule
  3. $30.6 \times 10^6$Joule
  4. $3.6 \times 10^5$Joule
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The kilowatt hour is a unit of energy equal to 1,000 watt-hours, or 3.6 megajoule. If the energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time in hours.
1 kilowatt hour $= 1000 \times 3600 joule = 3.6 \times 10^{6}\ J$