Tag: physics

Questions Related to physics

$1  kwh$ is equal to

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

  1. $3.6 \times {10}^{6} J$

  2. $100 J$

  3. $1000 J$

  4. $3.6 \times {10}^{3} J$

Show answer
Correct Option: A
Explanation:

$1  kwh = 1  kw \times 1  h$
     $= 1000  w \times 1 \times 60 \times 60$
     $= 3.6 \times {10}^{6}  J$

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

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

  1. 14kwh

  2. 2.8kwh

  3. 40kwh

  4. All

Show answer
Correct Option: B
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

One kilowatt hour is equal to

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

  1. $\displaystyle 36\times { 10 }^{ 5 }$ joules

  2. $\displaystyle 36\times { 10 }^{ 3 }$ joules

  3. $\displaystyle { 10 }^{ 3 }$ joules

  4. $\displaystyle { 10 }^{ 5 }$ joules

Show answer
Correct Option: A
Explanation:

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

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

Number of KWh in 1Joule.

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

  1. $\displaystyle 3.6\times { 10 }^{ 6 }KWh$

  2. $\displaystyle 2.77\times { 10 }^{ -7 }KWh$

  3. $\displaystyle 600KWh$

  4. $\displaystyle 1.6\times { 10 }^{ -19 }KWh$

Show answer
Correct Option: B
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

Calculate the number of Joules in 1KWh.

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

  1. $\displaystyle 6\times { 10 }^{ -19 }J$

  2. $\displaystyle 3.6\times { 10 }^{ 6 }J$

  3. $\displaystyle 60J$

  4. $\displaystyle 59J$

Show answer
Correct Option: B
Explanation:

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

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

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

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

  1. charge used

  2. current used

  3. power used

  4. energy used

Show answer
Correct Option: D
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 

$1kWh= $ _________?

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

  1. $3600000\ J$

  2. $10000\ J$

  3. $4.2\ J$

  4. $25000\ J$

Show answer
Correct Option: A
Explanation:
Kilowatt hour is the energy consumed by a body of power $1\ kW$ in $1\ hr$. 
Hence, 
$1\ kWh = 1 kW \times 1\ hr$
              $=  10^3 W \times 3600\ s$
              $= 3600000\ J$

Kilowatt-hour is the unit of :

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

  1. potential difference.

  2. electric power.

  3. electrical energy.

  4. charge.

Show answer
Correct Option: C
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.

Kilowatt hour is the unit of:

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

  1. Power

  2. Energy

  3. Impulse

  4. Force

Show answer
Correct Option: B
Explanation:

KiloWatt is the unit of power and hour is the unit of time. Product of power and time equals 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  :

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

  1. ${t^{3/4}}$

  2. ${t^{3/2}}$

  3. ${t^{1/4}}$

  4. ${t^{1/2}}$

Show answer
Correct Option: B
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.