Tag: physics

Questions Related to physics

A body is moving along a straight line delivering power given as P = at, then work (W) done is given for time 0 to t is

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

  1. W = a

  2. W = $\frac{1}{2}a t^2$

  3. W = $2at^2$

  4. W = $a^2t^2$

Show answer
Correct Option: B
Explanation:

Given that ,

Power , $P= at$
We know that 
$P=\dfrac{dW}{dt}$

$\implies \dfrac{dW}{dt}= at$

$\implies dW= at dt$
$\implies W= \int _0^t atdt$
$\implies W=\dfrac12 at^2$

$\therefore $ Work done for time 0 to t ,   $W=\dfrac12 at^2$  

A force of $ 2\hat { i } +3\hat { j } +2kN $ acts on a body for 4 s and produces a displacement of $3\hat {i} +4\hat {j} +5 \hat {k} m $ calculate the power ?

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

  1. 5 w

  2. 6 w

  3. 7 W

  4. 9 w

Show answer
Correct Option: C
Explanation:

Given that,

Force  ,$F= 2 \hat i + 3\hat j + 2\hat k $  N
Displacement , $S= 3\hat i + 4\hat j +5 \hat k $ m 
Time Taken , $t= 4\ s$

Power, $P=\dfrac Pt = \dfrac{F\cdot S}{t}= \dfrac{6+12+10}{4}= \dfrac{28}4 = 7\ W$

A pump of $200W$ power is lifting $2kg$ water from an average depth of $10m$ in one second. Velocity of water delivered by the pump is :

$(g=10m/s^2)$

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

  1. $10m/s$

  2. $2m/s$

  3. $4 m/s$

  4. $1 m/s$

Show answer
Correct Option: A
Explanation:

acceleration due to gravity $g = 10m/sec^2$

height $H = 10 meters$
mass $= 2kg$
Potential energy $= mgh = 2\times 10 \times 10 = 200J$
as power $= \dfrac{work \ done}{time}$
when power of motor $= 200w$
$200 = 200/t$
$\Rightarrow t = 1sec$
here displacement of water= height $= 10m$
time = 1sec
Hence,
Velocity $V= \dfrac{Displacement}{Time}$
              $V= \dfrac{10}{1}$
              $V= 10m/sec$

A small diesel engine uses a volume of $1.5 \times 10^4\, cm^3$  of fuel per hour to produce a useful power
output of 40 kW. It may be assumed that 34 kJ of energy is transferred to the engine when it uses $1.0\, cm^3$  of fuel.
What is the rate of transfer from the engine of energy that is wasted?

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

  1. 850 kW

  2. 920 kW

  3. 840 kW

  4. 810 kW

Show answer
Correct Option: D
Explanation:

Energy produced by $1.0\ cm^3$ of fuel = $34\ kJ$

So, energy produced by  $1.5 \times 10^4\ cm^3$ of fuel in one hour = $1.5 \times 10^4 \times 34\ kJ$
                                                                                                   = $5.1 \times 10^5\ kJ$
Energy produced in one second = $\dfrac{5.1 \times 10^5}{60}\ kJ/s$
                                                         = $850\ kW$
So, rate of energy wasting = $(850-40)\ kW$
                                             = $810\ kW$

A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest :

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

  1. at the highest position of the body

  2. at the instant just before the body hits the earth

  3. it remains constant throughout

  4. at the instant just after the body is projected

Show answer
Correct Option: B
Explanation:

$Power,$ $P$ $=$$\overrightarrow{F}$.$\overrightarrow{v}$ $=$ $Fv$ $cos$$\theta$
$Just$ $before$ $hitting$ $the$ $earth$ $θ$ $=$ $0°.$ $Hence,$ $the$ $power$ $exerted$ $by$ $the$ $gravitational$ $force$ $is$ $greatest$ $at$ $the$ $instant$ $just$ $before$ $the$ $body$ $hits$ $the$ $earth.$

1 kWh is equal to

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

  1. $3.6 \times 10^6 MJ$

  2. $3.6 \times 10^5 MJ$

  3. $3.6 \times 10^2 MJ$

  4. $3.6 MJ$

Show answer
Correct Option: D
Explanation:
1 kilowatt hour is the energy produced by 1 kilowatt  power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

$\implies 1kWh=3.6\times 10^6J$

$\implies 1kWh=3.6MJ$

Answer-(D)

Number of kilowatt-hours =$\dfrac { volt\times ampere\times time }{ 1000 } $. Then:

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

  1. time in seconds

  2. time in minutes

  3. time in hours

  4. time in days

Show answer
Correct Option: C
Explanation:

Kilowatt-hours is the power generated in one hour=$\dfrac{volt\times current\times time( hour)}{1000}$


Answer-(C)

If 1 unit of electricity cost $0.20$, how much does it cost to switch on a heater marked $120 V$, $3 A$ for $90$ min?

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

  1. $ 0.11$

  2. $ 2.70$

  3. $ 64.80$

  4. $ 108.00$

Show answer
Correct Option: A
Explanation:
Voltage across the heater $V = .12$ kilo-volts 
Current flowing through the heater $I = 3 A$
Thus power of the heater $P = VI$
$\therefore$ $P = 0.12 \times 3 = 0.36 $ $kW$
Time of operation $t = 90$ min $ = 1.5 $ hr
Thus energy consumed $E = Pt$
$\implies$ $E = 0.36 \times 1.5 = 0.54$ $kWhr$
Cost to switch on heater =  $0.54 \times 0.2 = 0.11$

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude $P _{0}$. The instantaneous velocity of this car is proportional to:

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

  1. $t^{1/2}$

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

  3. $t/\sqrt{m}$

  4. $t^{2}P _{0}$

Show answer
Correct Option: A
Explanation:

Power =F.V


Power delivered as at 

Cont. magnitude $P _0$

$P _0=F.V$

$P _0=ma\times V$

$\dfrac{P _0}{m}=\dfrac{dv}{dt}\times V$

$\displaystyle \left(\dfrac{P _0}{m}\right)\int^t _0 dt=\int^v _0 vdv$

$\dfrac{P _0t}{m}=\dfrac{V^2}{2}$

$V^2=\left(\dfrac{2P _0}{m}\right)t$

$V=\sqrt{\dfrac{2P _0}{m}}\times t^{\dfrac{1}{2}}$

$\boxed{V\alpha\,t^{\dfrac{1}{2}}}$

$1$ kWh$=$ ______________J.

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

  1. $3.6\times 10^6$

  2. $36\times 10^6$

  3. $3.6\times 10^7$

  4. $3.6\times 10^5$

Show answer
Correct Option: A
Explanation:

$1$ kilowatt-hour(kWh) is a unit of energy. Normally, we want energy to be expressed in joules(J) and time in seconds(s).
Energy(kWh)$=$Power(kW)$\times$(h)$=1000$W$\times 3600$s$=1000$J/s$\times 3600$s$=3600000$J$=3.6\times 10^6$J.