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

A force applied by the engine of a train of mass 2.05 x $10^{6} kg$ changes its velocity from 5 $ms^{-1}$ to 25  $ms^{-1}$ in 5 minutes. The power of the engine is then

  1. 1.025 MW

  2. 2.05 MW

  3. 5 MW

  4. 6 MW
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that,

Mass of train, $m=2.05\times {{10}^{6}}\,Kg$

Time, $t=5\,\min utes=300\,s$

$ v=25\,m/s $

$ u=5\,m/s $

Acceleration,

$ a=\dfrac{v-u}{t} $

$ a=\dfrac{25-5}{300} $

$ a=\dfrac{2}{30}=\dfrac{1}{15}\,m/{{s}^{2}} $

Using equation of motion,

$ {{v}^{2}}-{{u}^{2}}=2as $

$ {{(25)}^{2}}-{{(5)}^{2}}=2\times \dfrac{1}{15}\times s $

$ s=4500\,m $

Power,

$ P=\dfrac{work\,\,done}{time} $

$ P=\dfrac{F\times d}{t} $

$ P=\dfrac{m\times a\times s}{t} $

$ P=\dfrac{2.05\times {{10}^{6}}\times 1\times 4500}{15\times 300} $

$ P=2.05\times {{10}^{6}}\,W $

$ P=2.05\,MW $

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

A motor lifts $100 kg$ of water in $2 min$ from a well of $60m$ depth then the electric power of the motor is$(Taken g=10 m/s^2)$

  1. $1000 W$
  2. $750 W$
  3. $1200 W$
  4. $500W$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A motor lifts $=100kg$ of water

Time $=2min=2\times60=120s$
Depth$=60m$ depth then,
electric power of the motor$=?$
Taking $=10m/s^2$
$P=Power=\cfrac{mgh}{t}\ \quad=\cfrac{100\times10\times60}{120}\ \quad=500W$

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

Your uncle pushes a $60.0 kg$ crate along a floor with average speed $v=0.65 {m}/{s}$ for $5.0$ seconds as he moves furniture to clean up the garage.
If the coefficient of friction between the floor and the crate is $\mu=0.340$, what is the average power output of your uncle during this time?

  1. $26.0 W$
  2. $130 W$
  3. $383 W$
  4. $650 W$

  5. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given :  $\mu = 0.340$             $v = 0.65$  m/s                $m =60.0$  kg

As the crate moves with constant speed, thus the force applied by uncles must be equal to the frictional force.
$\therefore$   $F  = \mu mg = 0.340 \times 60.0 \times 9.8  = 199.92$  N
Average power output        $P = F v = 199.92 \times 0.65  \approx 130$  W

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

Find the power of a pump which takes $10 s$ to draw $100 kg$ of water from a tank situated at a height of $20 m$.

  1. $2\times { 10 }^{ 4 }W$
  2. $2\times { 10 }^{ 3 }W$
  3. $200 W$
  4. $1 kW$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power = $\dfrac{work done}{time}$


             =  $\dfrac{mgh}{time}$

            =  $\dfrac{100 \ast 10 \ast 20}{10}$

             =  $\dfrac{20000}{10}$

               = 2000W  i.e  2 $\times$ 10$^{3}$W

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

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