Power MCQ Quiz - Objective Question with Answer for Power - Download Free PDF

Calculation:

Power Consumed in heater is, P = V²/R 

 P_A/ P_B =  R_B/ R_A   

 For Series Combination  P_S = V ²/3 R _B 

 For Parallel Combination   P_P = 3 V²/ 2 R_B 

  PS/  PP =2/9

∴ The correct option is 2)

The correct answer is 746 Watt.

Key Points

• One Horsepower (HP) is defined as 746 watts (W).
• The term "horsepower" was coined by James Watt, the inventor of the steam engine, to compare the output of steam engines with the power of draft horses.
• Horsepower is a unit of measurement of power, or the rate at which work is done, commonly used in reference to the output of engines and motors.
• 1 HP is equivalent to lifting 550 pounds (lb) one foot (ft) in one second.

Additional Information

• Watts:
  • A watt is a unit of power in the International System of Units (SI) equivalent to one joule per second.
  • It is used to quantify the rate of energy transfer.
• James Watt:
  • James Watt was a Scottish inventor and mechanical engineer whose improvements to the steam engine were fundamental to the changes brought by the Industrial Revolution.
  • The unit of power "watt" was named in his honor.
• Power Measurement:
  • Power is the rate at which work is done or energy is transferred over time.
  • Other common units of power include kilowatts (kW), megawatts (MW), and British thermal units per hour (BTU/h).
• Engine and Motor Ratings:
  • Horsepower is commonly used to rate the power output of engines, especially in automotive and industrial contexts.
  • Electrical motors are often rated in both horsepower and watts or kilowatts.

Calculation:

\(\text { Power }=\vec{\mathrm{F}} \cdot \vec{\mathrm{V}} \)

\(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}[\mathrm{p}=\mathrm{mv}] \)

\(\mathrm{F}=\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right) \mathrm{v}=\mathrm{C}(\sqrt{\mathrm{v}}) \mathrm{v} \)

\(\mathrm{F}=\mathrm{Cv}^{\frac{3}{2}} \)

Power = C(v^3/2)v = Cv^5/2

P2 ∝ v5

Hence, the correct answer is option (4).

Instantaneous power: p(t) = V(t) × I(t) = V₀ sin(ωt) × I₀ sin(ωt − π/2)

Using sin(ωt − π/2) = −cos(ωt):

p(t) = −V₀ I₀ sin(ωt) cos(ωt) = −(V₀ I₀ / 2) × sin(2ωt)

Average power: P_avg = (1/T) ∫ p(t) dt = 0

∴ The power consumed per cycle is zero. Option 1 is correct.

Concept:

Gravitational potential energy (GPE) can be converted into electrical energy. The formula for GPE is:

GPE = mgh

The electrical power generated can be found by converting the GPE to electrical energy and then dividing by time. Since only 50% of the GPE is converted, the formula becomes:

Electrical Power = 0.5 × (mgh / t)

Calculation:

We have,

⇒ m = 9 × 10⁴ kg per hour

⇒ h = 40 m

⇒ g = 10 m/s²

⇒ Time (t) = 3600 s (since 1 hour = 3600 seconds)

Calculate the gravitational potential energy:

⇒ GPE = mgh

⇒ GPE = 9 × 10⁴ kg × 10 m/s² × 40 m

⇒ GPE = 36 × 10⁶ J

Since only 50% of GPE is converted to electrical energy:

⇒ Electrical Energy = 0.5 × 36 × 10⁶ J

⇒ Electrical Energy = 18 × 10⁶ J

Now, calculate the electrical power generated:

⇒ Electrical Power = Electrical Energy / t

⇒ Electrical Power = 18 × 10⁶ J / 3600 s

⇒ Electrical Power = 5 × 10³ W

Each lamp is 100 W, so the number of 100 W lamps that can be lit:

⇒ Number of lamps = Total Power / Power per lamp

⇒ Number of lamps = 5 × 10³ W / 100 W

⇒ Number of lamps = 50

∴ The correct answer is option 2 (50).

CONCEPT:

Power:

• The rate of doing work is called power.
• SI unit of power is the watt.

\(⇒ P=\frac{W}{t}\)

.Where P = power, W = work done and t = time

CALCULATION:

Given m = 20 kg, h = 5 m, t = 20 sec and g = 10 m/sec²

• We know that the work done in lifting a body of mass m to height h is given as,

⇒  W = mgh    -----(1)

By equation 1,

⇒  W = 20 × 10 × 5 = 1000 J 

• So power required,

\(⇒ P=\frac{W}{t}\)

\(⇒ P=\frac{1000}{20}\)

\(⇒ P=50\,watt\)

• Hence, option 1 is correct.

CONCEPT:

• Power is defined as the rate of doing work.
• It is a scalar quantity that has only magnitude but no direction.
• The SI unit of power is Watt.
• There is another unit of power i.e. horsepower.
• 1 h.p. = 746 Watts or 0.75 kW.
• When work is done, an equal amount of energy is consumed.
• 1 Watt is the power of an appliance that consumes energy at the rate of 1 Joule per second.
  • 1 Watt = 1 Joule per second

EXPLANATION:

The power of an electrical appliance tells us the rate at which electrical energy is consumed or doing work by it. Its unit is Watt. It is a scalar quantity.

• There is another unit of power i.e. horsepower.
• 1 h.p. = 746 Watts or 0.75 kW.
•  1 horsepower is not equal to 746 kW. So statement 3 is incorrect.

So option 3 is correct.

• 'Watt' is named after the Scottish inventor, engineer, and designer James Watt who became famous for improving the design of the steam engine.

Concept

Work done = mgh

where, m = mass of body; g = acceleration due to gravity; h = height of fall

Power = \(\frac{W}{t}\)

Calculation

Given 

mass (m) = 80 kg

g = 10 m/sec2

h = 4 metre

t = 10 seconds

Work done \(=mgh=80\times10\times4=3200J\)

Power used, \(P= \frac {W}{t}=\frac {3200}{10}=320 W\)

The power spent by the man is 320 watts.

CONCEPT:

• Work done (W): When a force is applied on a body and there is a displacement of the body in the direction of force then the work is said to be done by the force. The SI unit of work is Joule.

Work (W) = Force (F) × Displacement (S)

• Potential energy (PE): The energy of a body due to its position is called potential energy.

PE = m g h

Where m is mass, g is the acceleration due to gravity and h is height.

• Power (P): The rate of work done is called power.
  • The SI unit of power is the watt (W).

Power (P) = work done (W) / time (t)   (P = W / t = J / s)

CALCULATION:

Given that:

Mass (m) = 40 kg

Number of stairs (N) = 50

Height of each stairs = 20 cm

Total height (h) = Number of stairs × Height of each stairs = 50 × 20 = 1000 cm = 10 m

Potential energy gained by the girl = PE = m g h = 40 × 10 × 10 = 4000 J

Work done = PE = 4000 J

Power (P) = Work done/Time taken = 4000/50 = 80 Watt

Hence option 3 is correct.

CONCEPT:

Work done (W): When a force is applied on a body and there is a displacement of the body in the direction of force then the work is said to be done by the force.

• The SI unit of work is Joule.
• 1 Joule (J) = 1 Nm

Work (W) = Force (F) × Displacement (S)

Power (P): The rate of work done is called power.

• The SI unit of power is the watt (W).

Power (P) = work done (W) / time (t)   

P = W / t = J / s

CALCULATION:

Given that:

Work (W) = 120 Joules

Time (t)  = 3 seconds

The formula of Power,

P = W/t = 120/3 = 40 Watts

Power of Engine = 40 Watts

Option 2 is correct.

CONCEPT:

• Power (P): The rate of work done is called power.
  • The SI unit of power is the watt (W).

Power (P) = work done (W) / time taken (t)   (P = W / t = J / s)

• Potential energy (PE): The energy possessed by a body by virtue of its position or configuration is called potential energy.

\(PE = mgh\)

Where, m = mass of the body, g = acceleration due to gravity and h = height of the body

CALCULATION:

Given that:

Mass of the object (m) = 5 kg 

Height (h) = 4 m

Time taken (t) = 2 sec

Potential energy (PE) = m g h = 5 × 10 × 4 = 200 J

• Since work done by the gravity will be equal to potential energy acquired by the object.

Work done (W) = PE = 200 J

Then power is given by:

Power (P) = work done (W)/time taken (t) = 200/2 = 100 W.

So option 2 is correct.

CONCEPT:

• Work: When a force is applied on a body and there is a displacement of the body in the direction of force then the work is said to be done by the force.
  • The SI unit of work is Joule. 1 Joule (J) = 1 Nm

Work (W) = Force (F) × Displacement (S)

• Power: The rate of work done is power.

Power = work / time   (P = W / t = J / s)

CALCULATION:

Given that

The power output of an engine = 40 kW

 Time taken = 20 seconds 

Power = Work done/ Time taken

Work done (W) = P × t = 40 × 20

  ⇒  W = 800 kJ (since power is in kilowatts).

the correct answer is Power

Key Points 

• Power - Power is defined as the rate at which work is done or energy is transferred over time. It is measured in Watts (W) in the International System of Units (SI), where one Watt equals one Joule per second.

Additional Information 

• Power quantifies the speed at which work is completed or energy is converted.
• Joule: This is a unit of work or energy in the SI system, not a rate. One Joule is the energy transferred to an object when a force of one newton acts on it over a distance of one meter. -
• Force: Force is a physical cause that changes or may tend to change the state of rest or motion of an object. It is measured in Newtons (N) in the SI system. Force itself is not a rate but a vector quantity causing acceleration.
• Speed: Speed is a scalar quantity that refers to how fast an object is moving. It is the rate of change of distance with respect to time but does not directly relate to the concepts of work or energy transfer.

Option 3 is correct.

CONCEPT:

• Power (P): The rate of work done is called power.
  • The SI unit of power is the watt (W).
• Force: It is the push or pulls off an object. Push and pull come from the objects interacting with one another.
• Energy: An object having the capability to do work is said to possess energy.
  • Energy has the same unit as that of work.
  • Energy exists in nature in several forms such as kinetic energy, potential energy, heat energy, chemical energy, etc. The sum of the kinetic and potential energies of an object is called mechanical energy.
  • Energy is the capacity to do work or to produce heat.

EXPLANATION:

• The rate of doing work is called power. So option 3 is correct.

CONCEPT:

• Power: The rate of work done is called power.
  • It is denoted by P.
  • Mathematically it is written as

\(Power\;\left( P \right) = \frac{{\;W}}{{t}}\)

CALCULATION:

Given - Power (P) = 2 HP = 1492 J/sec and time (t) = 10 minutes = 600 sec

• The electrical energy consumed by it is 

⇒ W = Pt = 1492 × 600 = 895200 J = 8.952 × 105 J