Power and Commercial Unit of Energy Questions in English

(B) Power is defined as the rate of doing work or the rate of transfer of energy.
If $W$ is the amount of work done in time $t$,then power $P$ is given by the expression:
$P = \frac{W}{t} = \frac{\text{Energy}}{\text{Time}}$
The $SI$ unit of power is the watt $(W)$,which is defined as $1 \text{ joule per second } (J/s)$.

(N/A) body is said to have a power of $1$ watt if it does work at the rate of $1$ joule per second.
Mathematically,it is expressed as:
$1\,W = 1\,J/1\,s$
This means that $1$ watt is the power consumed when $1$ joule of work is done in $1$ second.

(C) Power is defined as the rate of doing work or the rate of energy consumption.
The formula for power $(P)$ is given by:
$P = \frac{W}{t}$
Where:
$W$ (Work done or Energy consumed) $= 1000\, J$
$t$ (Time taken) $= 10\, s$
Substituting the values into the formula:
$P = \frac{1000\, J}{10\, s} = 100\, J/s$
Since $1\, J/s = 1\, W$ (Watt):
$P = 100\, W$
Therefore,the power of the lamp is $100\, W$.

(N/A) Average power is defined as the total work done by an object divided by the total time taken to perform that work.
Since a body may perform different amounts of work in different time intervals,calculating the average power provides a better representation of the rate of work done over a specific duration.
Mathematically,it is expressed as:
$\text{Average Power} = \frac{\text{Total work done}}{\text{Total time taken}}$

(A) We know that $1$ unit of electrical energy is equal to $1 \text{ kilowatt-hour (kWh)}$.
$1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ h} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$.
Given that the household consumed $250$ units.
Therefore,total energy in joules $= 250 \times 3.6 \times 10^6 \text{ J}$.
$= 900 \times 10^6 \text{ J}$.
$= 9 \times 10^8 \text{ J}$.
Hence,the energy consumed is $9 \times 10^8 \text{ J}$.

(B) We know that,$\text{Energy} = \text{Power} \times \text{time}$.
Given:
$\text{Power} = 1500 \, W$
$\text{Time} = 10 \, \text{hours} = 10 \times 60 \times 60 \, \text{seconds} = 36000 \, \text{seconds}$.
Therefore,
$\text{Energy} = 1500 \, W \times 36000 \, \text{s} = 54,000,000 \, J$.
In scientific notation,this is $5.4 \times 10^7 \, J$.

(A) The power of four devices $= 4 \times 500\, W = 2000\, W$.
Time $= 10$ hours.
Energy consumed $= \text{Power} \times \text{Time}$.
Energy $= 2000\, W \times 10\, h = 20000\, Wh$.
Since $1\, kWh = 1000\, Wh$, we divide by $1000$.
Energy $= 20000 / 1000 = 20\, kWh$.

(B) Energy is defined as the capacity to do work. The $SI$ unit of energy is the $Joule$ $(J)$.
$1$ $Joule$ is equal to $1$ $Newton$ $metre$ $(N \cdot m)$.
$Kilowatt$ $hour$ $(kWh)$ is a commercial unit of electrical energy,where $1$ $kWh = 3.6 \times 10^6$ $J$.
$Kilowatt$ $(kW)$ is a unit of power,not energy. Power is defined as the rate of doing work $(Power = Work / Time)$.
Therefore,$kilowatt$ is not a unit of energy.

(A) Power is defined as the rate of doing work,calculated as $P = F \times v$,where $F$ is the force and $v$ is the velocity.
For Avinash:
$P_{A} = F_{A} \times v_{A} = 10 \, N \times 8 \, m \, s^{-1} = 80 \, W$.
For Kapil:
$P_{K} = F_{K} \times v_{K} = 25 \, N \times 3 \, m \, s^{-1} = 75 \, W$.
Comparing the two,$80 \, W > 75 \, W$. Therefore,Avinash is more powerful than Kapil.

(C) Power $(P)$ is defined as the rate of doing work, $P = \frac{W}{\Delta t}$.
Since work done to lift water is $W = mgh$, the formula becomes $P = \frac{mgh}{\Delta t}$.
Given: $P = 2 \, kW = 2000 \, W$, $h = 10 \, m$, $g = 10 \, m \, s^{-2}$, and $\Delta t = 1 \, \text{minute} = 60 \, s$.
Substituting the values: $2000 = \frac{m \times 10 \times 10}{60}$.
$2000 = \frac{100m}{60} \Rightarrow 2000 = \frac{5m}{3}$.
$m = \frac{2000 \times 3}{5} = 400 \times 3 = 1200 \, kg$.
Therefore, the pump can raise $1200 \, kg$ of water per minute.

(D) Power $(P)$ = $1200 \, W = 1.2 \, kW$.
Time $(t)$ per day = $30 \, \text{minutes} = 0.5 \, \text{hours}$.
The month of April has $30$ days.
Total energy consumed $(E)$ = Power $\times$ time per day $\times$ number of days.
$E = 1.2 \, kW \times 0.5 \, h \times 30 \, \text{days}$.
$E = 0.6 \times 30 = 18 \, kWh$.

(N/A) Power is defined as the rate of doing work or the rate of energy transfer.
Kilowatt $(kW)$ is a unit of power, representing $1000 \, J/s$. Kilowatt hour $(kWh)$ is a unit of energy, representing the energy consumed by a device of $1 \, kW$ power in $1 \, hour$.
Given: Height $(h) = 20 \, m$, Mass $(m) = 2000 \, \text{tonnes} = 2000 \times 1000 \, kg = 2 \times 10^6 \, kg$, Time $(t) = 1 \, \text{minute} = 60 \, s$, Acceleration due to gravity $(g) = 10 \, m \, s^{-2}$.
Potential Energy $(PE) = mgh = 2 \times 10^6 \times 10 \times 20 = 4 \times 10^8 \, J$.
Power = $\frac{\text{Energy}}{\text{Time}} = \frac{4 \times 10^8 \, J}{60 \, s} = \frac{40}{6} \times 10^7 \, W = 6.67 \times 10^6 \, W$ or $6.67 \, MW$.

(B) Power is defined as the rate of doing work. When lifting an object,the work done is equal to the change in potential energy,which is $mgh$.
Power $P = \frac{W}{t} = \frac{mgh}{t} = mg \left(\frac{h}{t}\right)$.
Since speed $v = \frac{h}{t}$,the relationship is $P = mgv$.
Given: Power $P = 100\, W$,speed $v = 1\, m\, s^{-1}$,and acceleration due to gravity $g = 10\, m\, s^{-2}$.
Rearranging the formula to find mass $m$: $m = \frac{P}{g \times v}$.
Substituting the values: $m = \frac{100}{10 \times 1} = 10\, kg$.

(C) One watt is defined as the power of an agent that performs work at the rate of $1 \, J \, s^{-1}$.
$1 \, \text{kilowatt} = 1000 \, J \, s^{-1}$.
Total power developed by the engine $= 150 \, kg \times 500 \, W \, kg^{-1} = 75,000 \, W = 7.5 \times 10^4 \, W$.
We know that $\text{Power} = \text{Force} \times \text{Velocity}$.
Therefore, $\text{Force} = \frac{\text{Power}}{\text{Velocity}} = \frac{7.5 \times 10^4 \, W}{20 \, m \, s^{-1}} = 3750 \, N$.

(D) Power $(P)$ is defined as the rate of doing work,which is $P = F \times v$. Since the objects are moving against gravity,the force required is $F = mg$.
$(i)$ For the butterfly:
Mass $m = 1.0 \, g = 1.0 \times 10^{-3} \, kg$,velocity $v = 0.5 \, m \, s^{-1}$,$g = 10 \, m \, s^{-2}$.
$P = (1.0 \times 10^{-3} \, kg) \times (10 \, m \, s^{-2}) \times (0.5 \, m \, s^{-1}) = 5 \times 10^{-3} \, W$.
$(ii)$ For the squirrel:
Mass $m = 250 \, g = 0.25 \, kg$,velocity $v = 0.5 \, m \, s^{-1}$,$g = 10 \, m \, s^{-2}$.
$P = (0.25 \, kg) \times (10 \, m \, s^{-2}) \times (0.5 \, m \, s^{-1}) = 1.25 \, W$.
Thus,the power for the butterfly is $5 \times 10^{-3} \, W$ and for the squirrel is $1.25 \, W$.

(N/A) $(i)$ Joule $(J)$
$(ii)$ Erg
$(iii)$ Kilowatt-hour $(kWh)$

(B) The practical unit of power used in engineering is the $Horsepower$ $(hp)$.
$1$ $hp$ is approximately equal to $746$ $W$ (Watts).

(B) The unit of power is the watt $(W)$.
$1$ horsepower $(hp)$ is a unit of power defined as the power exerted by a horse to pull a load.
By standard definition, $1$ horsepower is equal to $746$ watts.

(B) Horsepower $(hp)$ is a unit of measurement for power.
It is commonly used to express the rate at which a machine,such as an engine or motor,can perform work.
One mechanical horsepower is approximately equal to $746 \ W$ (watts).

(B) watt is the $SI$ unit of power.
It is defined as the rate of doing work or the rate of energy consumption.
One watt is equal to $1 \, J \, s^{-1}$,which means one joule of work done per second.

(20 W) The power $P$ is defined as the product of force $F$ and velocity $v$ when the force is applied in the direction of motion.
Given:
Force $F = 10\, N$
Velocity $v = 2\, m s^{-1}$
Formula:
$P = F \times v$
Calculation:
$P = 10\, N \times 2\, m s^{-1} = 20\, W$
Therefore,the power of the body is $20\, W$.

(N/A) The commercial unit of electrical energy is kilowatt-hour $(kWh)$.
The $SI$ unit of energy is Joule $(J)$.
Since $1 \text{ kW} = 1000 \text{ W}$ and $1 \text{ hour} = 3600 \text{ seconds}$,
$1 \text{ kWh} = 1000 \text{ J/s} \times 3600 \text{ s} = 3.6 \times 10^{6} \text{ J}$.
Therefore,the relation is $1 \text{ kWh} = 3.6 \times 10^{6} \text{ J}$.

(N/A) $1 \text{ kWh} = 1000 \text{ W} \times 1 \text{ h} = 1000 \text{ J/s} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$.
$1$ watt is defined as the power of an agent that performs work at the rate of $1$ joule per second.

(N/A) $(i)$ Power is defined as the rate of doing work or the rate of energy transfer. Its $SI$ unit is the watt $(W)$.
$(ii)$ Given: Mass $(m)$ = $50 \, kg$,Number of steps = $40$,Height of each step = $15 \, cm = 0.15 \, m$,Time $(t)$ = $8 \, s$,Acceleration due to gravity $(g)$ = $10 \, m \, s^{-2}$.
Total height $(h)$ = $40 \times 0.15 \, m = 6 \, m$.
Work done $(W)$ = $mgh = 50 \times 10 \times 6 = 3000 \, J$.
Power $(P)$ = $\frac{W}{t} = \frac{3000}{8} = 375 \, W$.

(112.5 W) Power is defined as the rate of doing work.
Given:
Mass of the boy $(m)$ = $45 \, kg$
Number of steps $(n)$ = $20$
Height of each step = $25 \, cm = 0.25 \, m$
Total height $(h)$ = $20 \times 0.25 \, m = 5 \, m$
Time taken $(t)$ = $20 \, s$
Acceleration due to gravity $(g)$ = $10 \, m s^{-2}$
Work done $(W)$ = Potential energy gained = $mgh$
$W = 45 \times 10 \times 5 = 2250 \, J$
Power $(P)$ = $\frac{\text{Work}}{\text{Time}} = \frac{W}{t}$
$P = \frac{2250}{20} = 112.5 \, W$
Thus,the power of the boy is $112.5 \, W$.

(N/A) The power of a device is said to be $1 \ W$ if it consumes energy at the rate of $1 \ J$ per second.
$(b)$ Given:
Energy $(E) = 1000 \ J$
Time $(t) = 10 \ s$
Power $(P) = E / t$
$P = 1000 \ J / 10 \ s = 100 \ W$
Thus,the power of the lamp is $100 \ W$.

(N/A) Power is defined as the rate of doing work or the rate of energy consumption. One watt is the power of an agent that does $1\, J$ of work in $1\, s$.
Given:
Energy consumed $(W)$ = $1000\, J$
Time taken $(t)$ = $10\, s$
Formula:
$P = \frac{W}{t}$
Calculation:
$P = \frac{1000\, J}{10\, s} = 100\, W$
Therefore,the power of the lamp is $100\, W$.

(N/A) The commercial unit of energy is kilowatt-hour $(kWh)$.
$1 \,kWh = 1 \,kW \times 1 \,h = 1000 \,W \times 3600 \,s = 3.6 \times 10^6 \,J$.
$(b)$ Given: Power $(P) = 60 \,W = 0.06 \,kW$,Time $(t) = 10 \,h$.
Energy consumed $(E) = P \times t = 0.06 \,kW \times 10 \,h = 0.6 \,kWh$.
Since $1 \,kWh = 1 \text{ unit}$,the bulb consumes $0.6 \text{ units}$ of electrical energy in one day.

(N/A) Power is defined as the rate of doing work or the rate of energy consumption.
The commercial unit of electrical energy is $kWh$ (kilowatt-hour), and the $SI$ unit is Joule $(J)$.
Given:
Power $(P)$ = $1.5 \, kW$
Time $(t)$ = $2 \, \text{hours}$
Electrical energy consumed $(E)$ = $P \times t$
$E = 1.5 \, kW \times 2 \, \text{h} = 3 \, kWh$.
Therefore, the electrical energy consumed in a day is $3 \, \text{units}$ (since $1 \, kWh = 1 \, \text{unit}$).

(A) Given: Mass $m = 800 \, kg$,height $h = 1500 \, cm = 15 \, m$,time $t = 20 \, s$,and acceleration due to gravity $g = 10 \, m s^{-2}$.
The work done $W$ to lift the water is equal to the potential energy gained: $W = mgh$.
Power $P$ is defined as the rate of doing work: $P = \frac{W}{t} = \frac{mgh}{t}$.
Substituting the values: $P = \frac{800 \times 10 \times 15}{20}$.
$P = \frac{120000}{20} = 6000 \, W$.
Since $1 \, kW = 1000 \, W$,we have $P = 6 \, kW$.

(A) Given: Mass of the man $m = 60 \, kg$,time taken $t = 40 \, s$,number of steps $= 30$,height of each step $= 20 \, cm = 0.2 \, m$.
Total height $h = 30 \times 0.2 \, m = 6 \, m$.
Acceleration due to gravity $g = 10 \, m/s^2$ (taking $g = 10 \, m/s^2$ for simplicity).
Work done $W = mgh = 60 \times 10 \times 6 = 3600 \, J$.
Power $P = \frac{W}{t} = \frac{3600 \, J}{40 \, s} = 90 \, W$.
Thus,the power of the man is $90 \, W$.

(A) The power $P$ expended is defined as the rate of doing work,given by the formula $P = W / t$.
Since the girl climbs at a constant speed,the work done $W$ is equal to the change in potential energy,which is $W = mgh$.
Given values are: mass $m = 40\, kg$,height $h = 6\, m$,and time $t = 15\, s$. Taking acceleration due to gravity $g = 9.8\, m/s^2$.
Substituting these values into the formula:
$P = \frac{mgh}{t} = \frac{40 \times 9.8 \times 6}{15} = \frac{2352}{15} = 156.8\, W$.
Thus,the power expended by the girl is $156.8\, W$.

(B) The work done $W$ in lifting a mass $m$ to a height $h$ against the gravitational force is given by $W = mgh$.
Power $P$ is defined as the rate of doing work,$P = \frac{W}{t}$.
Given values are: $m = 10^{5} \text{ kg}$,$g = 10 \text{ m s}^{-2}$,$h = 360 \text{ m}$,and $t = 1 \text{ hour} = 3600 \text{ s}$.
Substituting these values into the formula:
$P = \frac{10^{5} \times 10 \times 360}{3600}$
$P = \frac{10^{6} \times 360}{3600}$
$P = 10^{5} \text{ W}$.
Thus,the power required is $10^{5} \text{ W}$.

(C) Given: Power of the motor $P_{total} = 100\, W$, Time $t = 1\, \text{minute} = 60\, \text{seconds}$.
The power effectively used for stirring the water is $50\%$ of the total power supplied.
Power used for stirring $P_{stir} = 50\% \text{ of } 100\, W = \frac{50}{100} \times 100 = 50\, W$.
Work done on the water is calculated using the formula: $\text{Work} = \text{Power} \times \text{Time}$.
$\text{Work} = 50\, W \times 60\, \text{s} = 3000\, J$.
Therefore, the work done on the water in one minute is $3000\, J$.

(D) Energy consumed is given by $E = P \times t$.
Energy consumed by $4$ bulbs per day $= 4 \times 40 \, W \times 5 \, hrs = 800 \, Wh$.
Energy consumed by $4$ tube lights per day $= 4 \times 60 \, W \times 5 \, hrs = 1200 \, Wh$.
Energy consumed by $TV$ per day $= 100 \, W \times 6 \, hrs = 600 \, Wh$.
Energy consumed by washing machine per day $= 400 \, W \times 3 \, hrs = 1200 \, Wh$.
Total energy consumed per day $= 800 + 1200 + 600 + 1200 = 3800 \, Wh$.
Since April has $30$ days, total energy consumed in April $= 3800 \, Wh \times 30 = 114000 \, Wh$.
Converting to units $(kWh)$: $114000 \, Wh / 1000 = 114 \, kWh$ (or $114$ units).
Total electricity bill $= 114 \, \text{units} \times ₹ 1.80 / \text{unit} = ₹ 205.20$.

(B) The work done $(W)$ by the weightlifter is equal to the change in potential energy,given by $W = mgh$.
Here,mass $(m)$ = $240 \,kg$,height $(h)$ = $2.5 \,m$,and acceleration due to gravity $(g)$ = $9.8 \,m/s^2$.
$W = 240 \times 9.8 \times 2.5 = 5880 \,J$.
The time taken $(t)$ = $3 \,s$.
Average power $(P)$ is defined as the rate of doing work,$P = W / t$.
$P = 5880 / 3 = 1960 \,W$.
Therefore,the average power is $1960 \,W$.

(B) The statement is False.
Kilowatt hour $(kWh)$ is a unit of energy,not power.
Power is measured in Watts $(W)$ or Kilowatts $(kW)$,while energy is measured in Kilowatt hours $(kWh)$.

(A) The statement is True.
Power is defined as the rate at which work is done or the rate at which energy is transferred.
Mathematically, $Power (P) = \frac{Work (W)}{Time (t)}$.
Its $SI$ unit is the Watt $(W)$.

(FALSE) The statement is False.
The $SI$ unit of work is the joule $(J)$,which is defined as the work done when a force of $1$ newton displaces an object by $1$ meter.
Watt $(W)$ is the $SI$ unit of power,which is defined as the rate of doing work or the rate of energy transfer ($1$ watt = $1$ joule per second).