Tag: physics

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Multiple choice physics work and energy commercial unit of energy power work and power

A lamp rated 20w and an electric iron rated 50w are used for 2 hour everyday. Calculate the total energy consumed in 20 days.

  1. 14kwh

  2. 2.8kwh

  3. 40kwh

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

Energy consumed by lamp in 2 hour     $E _l = 0.02\times 2 = 0.04$ kWh per day

Energy consumed by iron in 2 hour     $E _i = 0.05\times 2 = 0.1$ kWh per day
$\therefore$ Total energy consumed by both appliance     $E _T = (E _l+E _i)\times 20 = (0.04+0.1)\times 20 = 2.8$ kWh

Multiple choice physics work and energy commercial unit of energy power work and power

One kilowatt hour is equal to

  1. $\displaystyle 36\times { 10 }^{ 5 }$ joules
  2. $\displaystyle 36\times { 10 }^{ 3 }$ joules
  3. $\displaystyle { 10 }^{ 3 }$ joules
  4. $\displaystyle { 10 }^{ 5 }$ joules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1$ kW  $ = 1000$ $\dfrac{J}{s}$            

$1$ h $ = 3600$ s
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =36\times 10^5$  $J$

Multiple choice physics work and energy commercial unit of energy power work and power

Number of KWh in 1Joule.

  1. $\displaystyle 3.6\times { 10 }^{ 6 }KWh$
  2. $\displaystyle 2.77\times { 10 }^{ -7 }KWh$
  3. $\displaystyle 600KWh$
  4. $\displaystyle 1.6\times { 10 }^{ -19 }KWh$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know  $1$ kWh $ = 3.6\times 10^6$ $J$

$\therefore$  $1$ $J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$ kWh

Multiple choice physics work and energy commercial unit of energy power work and power

Calculate the number of Joules in 1KWh.

  1. $\displaystyle 6\times { 10 }^{ -19 }J$
  2. $\displaystyle 3.6\times { 10 }^{ 6 }J$
  3. $\displaystyle 60J$
  4. $\displaystyle 59J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1$ kW $ = 1000$ $\dfrac{J}{s}$

$1$ h $=3600$ s 
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =3.6\times 10^6$  $J$

Multiple choice physics work and energy commercial unit of energy power work and power

When we pay for our electricity bill, we are paying for the ____________.

  1. charge used

  2. current used

  3. power used

  4. energy used
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A unit is defined as $kWh$, means a $1000 joule/ sec$ is used for $1 hour$ ,

$1KwH=1000\times 3600=3.6\times10^6joules$
we pay for unit and that is energy 

Multiple choice physics work and energy commercial unit of energy power work and power

$1kWh= $ _________?

  1. $3600000\ J$
  2. $10000\ J$
  3. $4.2\ J$
  4. $25000\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Kilowatt hour is the energy consumed by a body of power $1\ kW$ in $1\ hr$. 
Hence, 
$1\ kWh = 1 kW \times 1\ hr$
              $=  10^3 W \times 3600\ s$
              $= 3600000\ J$
Multiple choice physics work and energy commercial unit of energy power work and power

Kilowatt-hour is the unit of :

  1. potential difference.

  2. electric power.

  3. electrical energy.

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

The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour. The kilowatt-hour is commercially used as a billing unit for energy delivered to consumers by electric utilities.

Multiple choice physics work and power commercial unit of energy power work and energy

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $t$ is proportional to  :

  1. ${t^{3/4}}$
  2. ${t^{3/2}}$
  3. ${t^{1/4}}$
  4. ${t^{1/2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let's consider a body is moved along a straight line by a machine delivering a constant power $P$. The distance moved by the body is $S$. 


Power, $P=F.v$.. . . . . . (1)


Force, $F=ma$ . . . . . . . .(2)

where, $v=\dfrac{S}{t}$ 

acceleration, $a=\dfrac{S}{t^2}$

$m=$ mass

from  equation (1) and equation (2), we get

$P=\dfrac{mS}{t^2}\times \dfrac{S}{t}$

$S^2=\dfrac{Pt^3}{m}$

From the above equation, we get

$S^2\propto t^3$

$S\propto t^{3/2}$

The correct option is B.