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Work and power - class-X

Description: work and power
Number of Questions: 45
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Tags: work, energy and power work and power physics
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An energy of 4 kJ causes a displacement of 64 m in 2.5 s. The power delivered is

  1. 16 W

  2. 160 W

  3. 1600 W

  4. 16000 W


Correct Option: C
Explanation:

Given that


Energy, $E=4\ kJ=4000\ J$
Time taken , $t=2.5\ secs$

Power delivered , $P=\dfrac Et=\dfrac{4000}{2.5}= 1600\ W$

A constant power is delivered to a body moving along a straight line. the distance travelled by the body in time t is proportional to 

  1. $t ^ { 1 / 2 }$

  2. $t ^ { 3 / 2 }$

  3. $t ^ { 5 / 2 }$

  4. $t ^ { 7 / 2 }$


Correct Option: B
Explanation:

$\begin{array}{l} power\, P\, =Fv=mav \ acceleration\, a=\frac { S }{ { { t^{ 2 } } } } \, and\, velocity\, v=\frac { S }{ t } , \ where\, \, S\, \, is\, the\, dis\tan  ce\, moved \ Since\, \, the\, power\, P\, and\, mass\, m\, are\, cons\tan  t,\, we\, \, get \ P=\frac { { mS } }{ { { t^{ 2 } } } } \times \frac { S }{ t } \to { S^{ 2 } }={ t^{ 2 } }\to S\propto { t^{ \frac { 3 }{ 2 }  } } \end{array}$

Hence,
option $B$ is correct answer.

In the above question, if the work done on the system along the curved path $ba$ is $52\ J$, heat absorbed is 

  1. $-140\ J$

  2. $-172\ J$

  3. $140\ J$

  4. $172\ J$


Correct Option: A

A machine gun fires 360 bullets per minute. Each bullet moves with a  velocity of 600 ms$^{-1}$. If the power of the gun is 5.4 kw, the mass of each bullet is,

  1. 5 kg

  2. 0.5 kg

  3. 0.05 kg

  4. 5 gm


Correct Option: D
Explanation:

Given,

Number of bullets, $n=360\,nos.$

Mass of bullet, $m$

Power, $=5400watt$

Kinetic energy of 360 bullets, $K.E=n\times \dfrac{1}{2}m{{v}^{2}}=360\times \dfrac{1}{2}\times m\times {{600}^{2}}=648\times {{10}^{5}}m\,\,J$

$ Power=\dfrac{kineticenergy}{time} $

$ 5400=\dfrac{648\times {{10}^{5}}m}{60} $

$ m=5\times {{10}^{-3\,}}kg\,=\,5\,gram\, $

Hence, weight of each bullet is $5\,gram$

The driving side belt has a tension of $1600\ N$ and the slack side has $500\ N$ tension. The belt turns a pulley $40\ cm$ in radius at a rate of $300\ rpm$. The pulley drives a dynamic having $90\%$ efficiency. How many kilowatt are being delivered by the dynamo?

  1. $12.4\ kW$

  2. $6.2\ kW$

  3. $24.8\ kW$

  4. $13.77\ kW$


Correct Option: A

A heating unit on an electric stove is rated at $880  W$. It is connected to a power supply of $220  V$. The current it will consume is

  1. 2 amp

  2. 4 amp

  3. 6 amp

  4. 8 amp


Correct Option: B
Explanation:

Power $= V \times I$
or, $880 = 220 \times I$
or, $I = \displaystyle\frac{880}{220} = 4  amp$.

The average power required to lift a $100kg$ mass through a height of $50$ metres in approximately $50$ seconds would be ( in J/s)

  1. $50$

  2. $5000$

  3. $100$

  4. $980$


Correct Option: D
Explanation:

$Power= Work/Time= F.s/t=mgs/t=(100\times 9.8\times 50)/50= 980W $


Hence correct answer is option $D $ 

A tap supplies water at 22C. A man takes 11 litre water per minute at 37C from the geyser. The power of geyser is

  1. 525 W

  2. 1050 W

  3. 1575 W

  4. 2100 W


Correct Option: B

A rectangular block of dimensions $6m\times 4m\times 2m$ and of density $1.5\ gm/c.c$ is lying on horizontal ground with the face of large area in contact with the ground. The work done in arranging it which its smallest area in contact with a ground is, $(g=10ms^{-1})$

  1. $2880\ kJ$

  2. $1440\ kJ$

  3. $3800\ kJ$

  4. $720\ kJ$


Correct Option: B
Explanation:

m=v*density

m=$6\times 4\times 2\times 1.5\times 10^3=72\times 10^3$kg
when face with large area is in contact with the ground its height is 2m
Centre of mass is at a height of 1m

when face with small area is in contact with the ground its height is 6m
Centre of mass is at a height of 3m
W=$72\times 10^4(3-1)=1440 kJ$

A small body of mass $m$ is located on a horizontal plane. The body acquires a horizontal velocity ${v} _{0}$. Find mean power developed by the frictional force, during the whole time of its motion. Coefficient of friction is $\mu$

  1. $\cfrac { -\mu mg{ v } _{ 0 } }{ 3 } $

  2. $\cfrac { -\mu mg{ v } _{ 0 } }{ 2 } $

  3. $\cfrac { -\mu mg{ v } _{ 0 } }{ 5 } $

  4. $\cfrac { -\mu mg{ v } _{ 0 } }{ 6 } $


Correct Option: B

A body of mass $10kg$ is moving along positive $x-$axis with $5\ m/s$ at $t=0$ and is moving along negative $x-$axis with same speed at $t=10\ s$. Average power of the force acting on the body is:

  1. $Zero$

  2. $25\ W$

  3. $50\ W$

  4. $100\ W$


Correct Option: A
Explanation:

since there is no acceleration 

there is no power
P=0

A light bulb has the rating 100 W, 220 V. If the supply voltage is 110 V, then power consumed by the bulb is

  1. 50 W

  2. 75 W

  3. 25 W

  4. 20 W


Correct Option: C
Explanation:
$P =\dfrac{ V^2}R$  

If $V = 220$ V we have

$100 W =\dfrac{ 220^2}R$

$R = \dfrac{220^2}{100} Ω = 484 Ω$. This is the resistance of the bulb.

When $V = 110$ V, power consumed $=\dfrac{V^2}R= \dfrac{110^2}{484} = 25$ W.

So, 25 W power is consumed when it is operated on 110 V.


A force $\vec {F}=(3\hat {i}+4\hat {j})N$ acts on $2kg$ movable object that moves from an initial position $\vec {r} _{1}=(-3\hat {i}-2\hat {j})m$ to a final position $\vec {r} _{1}=(5\hat {i}+4\hat {j})m$ in $6s$. The average power delivered by the force during the interval of $6s$ is equal to :

  1. $8\ watt$

  2. $\dfrac{50}{6}\ watt$

  3. $15\ watt$

  4. $\dfrac{50}{3}\ watt$


Correct Option: A
Explanation:

 

Given,

Force, $\vec {F}=(3\hat {i}+4\hat {j})N$

Displacement, $\vec{d}={{\vec{r}} _{2}}-{{\vec{r}} _{1}}=(5\hat{i}+4\hat{j})-(-3\hat{i}-2\hat{j})=\left( 8\hat{i}+6\hat{j} \right)\,m$

Work, $W=\vec{F}.\vec{d}=\left( 3\hat{i}+4\hat{j} \right)\left( 8\hat{i}+6\hat{j} \right)=48\,J$

Average power $P=\dfrac{W}{t}=\dfrac{48}{6}=8\,W$

Average power is $8\,W$ 

A weight lifter lifts $300\ kg$ from the ground to a height of $2$ meter in $3$ seconds. The average power generated by him is:-

  1. $5880\ watt$

  2. $4410\ watt$

  3. $2205\ watt$

  4. $1960\ watt$


Correct Option: D

Human heart pumps $70\ cc$ of blood at each beat against a pressure of $125\ mm$ of $Hg$. If the pulse frequency is $72$ per minute, the power of the heat is nearly.

  1. $1.2\ W$

  2. $1.4\ W$

  3. $1.6\ W$

  4. $1.8\ W$


Correct Option: B

A force F acting on a body depends on its displacement $S$ as $F \propto S^{1/3}$. The power delivered by $F$ will depend on displacement as:

  1. $S^{2/3}$

  2. $S^{-5/3}$

  3. $S^{1/2}$

  4. $S^0$


Correct Option: A
Explanation:

We know Power=$\dfrac{work \ done}{time}$

Also work done = Force $\times$displacement

As given $F \ \alpha \ {S}^{1/3}$

$F=KS^{ 1/3 }$------(1) (where $K$= constant of proportionality)

Now Power= $\dfrac{F \times S}{time}$----(2)
Putting value of 1 in 2 we get 

$P= {\dfrac{KS^{\dfrac{2}{3}}}{t}}$

An engine of ine metric ton is going up an inclined plane, 1 in 2 at the rate of 36 kmph. If the coefficient of friction is $1/ \sqrt{3}$, the power of engine is 

  1. 9.8W

  2. 98W

  3. 980W

  4. 98kW


Correct Option: D

A train of mass $6 \times 10^2$ metric tones is pulled by a locomotive. The speed of the train will be $36 \,kmhr^{}-1$. The locomotive pulls the train on the train on the level track, whose mass is $125$ metric tones. The force of friction acts on the locomotive and the train is $1 \times 10^1$ newton per metric tonne. Calculate the power of the locomotive.

  1. 72500

  2. 6000

  3. 5000

  4. 4000


Correct Option: A
Explanation:

$m = 600$
$m\ell = 125$
$m \,net = 725$
$f = \dfrac{10}{tan}$
$t _{net} = 10 \times 125$

$V = 36 \,km/h = 36 \times \dfrac{5}{18} m/s$

$P = t _{net} \times V = 72500 \,W$

The power of water pump is The power of water pump is  $4kW.$  If  $\left( g=10{ m }{ { s }^{ -2 } } \right) ,$  the amount of water it can raise in $1$ minute to a height of $20 m$  is then

  1. $100$ litre

  2. $1000$ litre

  3. $1200$ litre

  4. $2000$ litre


Correct Option: C
Explanation:

Given that,

Power, $P=4\,kW=4000\,W$

Height, $h=20\,m$

Time, $t=60\,\sec $

$ power=\dfrac{work}{time} $

$ 4000=\dfrac{mgh}{t} $

$ 4000=\dfrac{m\times 10\times 20}{60} $

$ m=1200\,Kg $

The pump can raise 1200 litre in one minute

An object of mass accelerates uniformly from rest to a speed $v _f$ in time $t _f$ Then the instantaneous power delivered to the object,as a function of time $t$ is -

  1. $mt\left(\dfrac{{v _f}^2}{t _f}\right)$

  2. $mt\dfrac{v _f}{t _f}$

  3. $\dfrac{1}{2}mt^2\left(\dfrac{v _f}{t _f}\right)^2$

  4. $\dfrac{1}{2}mt^2\left(\dfrac{v _f}{t _f}\right)$


Correct Option: A

A force applied by the engine of a train of mass 2.05 x $10^{6} kg$ changes its velocity from 5 $ms^{-1}$ to 25  $ms^{-1}$ in 5 minutes. The power of the engine is then

  1. 1.025 MW

  2. 2.05 MW

  3. 5 MW

  4. 6 MW


Correct Option: B
Explanation:

Given that,

Mass of train, $m=2.05\times {{10}^{6}}\,Kg$

Time, $t=5\,\min utes=300\,s$

$ v=25\,m/s $

$ u=5\,m/s $

Acceleration,

$ a=\dfrac{v-u}{t} $

$ a=\dfrac{25-5}{300} $

$ a=\dfrac{2}{30}=\dfrac{1}{15}\,m/{{s}^{2}} $

Using equation of motion,

$ {{v}^{2}}-{{u}^{2}}=2as $

$ {{(25)}^{2}}-{{(5)}^{2}}=2\times \dfrac{1}{15}\times s $

$ s=4500\,m $

Power,

$ P=\dfrac{work\,\,done}{time} $

$ P=\dfrac{F\times d}{t} $

$ P=\dfrac{m\times a\times s}{t} $

$ P=\dfrac{2.05\times {{10}^{6}}\times 1\times 4500}{15\times 300} $

$ P=2.05\times {{10}^{6}}\,W $

$ P=2.05\,MW $

A motor lifts $100 kg$ of water in $2 min$ from a well of $60m$ depth then the electric power of the motor is$(Taken g=10 m/s^2)$

  1. $1000 W$

  2. $750 W$

  3. $1200 W$

  4. $500W$


Correct Option: D
Explanation:

A motor lifts $=100kg$ of water

Time $=2min=2\times60=120s$
Depth$=60m$ depth then,
electric power of the motor$=?$
Taking $=10m/s^2$
$P=Power=\cfrac{mgh}{t}\ \quad=\cfrac{100\times10\times60}{120}\ \quad=500W$

A force $'F'$ accelerates a block of mass $'m'$ along a straight line to velocity $'v'$ from rest and displace it through a distance $'s'$. What is the average power developed?

  1. $\dfrac {v^{2}}{2F}$

  2. $Fv$

  3. $\dfrac {mv^{2}}{2s}$

  4. $\dfrac {Fv}{2}$


Correct Option: D

The average power required to lift a 100 Kg mass trough a highest of 50 meters in approximatley 50 second would be

  1. 50 J/s

  2. 5000 J/s

  3. 100 J/s

  4. 980 J/s


Correct Option: D

A pump ejects $12000kg$ of water at speed of $4m/s$ in $40$ second. Find the average rate at which the pump is working

  1. $0.24KW$

  2. $2.4KW$

  3. $24KW$

  4. $24W$


Correct Option: B

A refrigerator transfer 180 joule of energy in one second from temperature $ { -3 }^{ 0 }C$ to $ { 27 }^{ 0 }C.$ Calculate the average power consumed, assuming no energy losses in the process. 

  1. 18 W

  2. 54 W

  3. 20 W

  4. 120 W


Correct Option: A

A man pulls a lawn roller with a force of $20\ kgf$. He applies the force at an angle $60^{\circ}$ with the ground through $10\ m$. If he takes $1$ minutes in doing so, calculate the power developed.

  1. $25\ J$

  2. $16.67\ J$

  3. $34\ J$

  4. $58\ J$


Correct Option: B

If the average power radiated by the star is $10 ^ { 16 } \mathrm { W }$ , the deuteron supply of the star is exhausted in a time of the order of 

  1. $10 ^ { 6 }$ seccond

  2. $10 ^ { 14 }$ second

  3. $10 ^ { 12 }$ second

  4. $10 ^ { 16 }$ second


Correct Option: C

A boat moving with constant speed v in still waters experiences a total frictional force F. The power developed by the boat is

  1. $\frac{1}{2}Fv$

  2. $Fv$

  3. $\frac{1}{2}Fv^2$

  4. $Fv^2$


Correct Option: B

A body of mass m is projected at an angle $\displaystyle \theta $ with the horizontal with an initial velocity $\displaystyle v _{0}.$ The average power of gravitational force over the whole time of flight is

  1. $\displaystyle mg\cos \theta $

  2. $\displaystyle \frac{1}{2}mg\sqrt{u\cos \theta }$

  3. $\displaystyle \frac{1}{2}mgu\sin \theta $

  4. zero


Correct Option: D
Explanation:

The projected body again comes back down. Hence the net displacement in the vertical direction will be 0.
$\therefore W=mgh=mg\times 0=0$

A time varying power $P=2t$ is applied on a particle of mass $m$. Find average power over a time interval from t=0 to t=t :

  1. $\displaystyle P _{av}= t$

  2. $\displaystyle P _{av}= 2t$

  3. $\displaystyle P _{av}= 4t$

  4. $\displaystyle P _{av}= 8t$


Correct Option: A
Explanation:
Average power over a time interval $t=0$ to $t=t$ is 
$\dfrac{\int _0^tPdt}{\int _0^tdt}$
$=\dfrac{\int _0^t 2tdt}{t}$
$=\dfrac{t^2}{t}=t$

A car of mass 1000 kg accelerates from rest to $100 : km : h^{-1}$ in 5 seconds. What is the average power of the car ?

  1. $7.71\times 10^5 W$

  2. $7.71\times 10^4 W$

  3. $15.42\times 10^4 W$

  4. $15.42\times 10^5 W$


Correct Option: B
Explanation:


Mass of the car m$=1000\ kg$
Initial velocity of car ${v} _{1}=0$
Final velocity of car ${v} _{2}$$=100km/h=100\times\dfrac{5}{18}m/s=27.78m/s$
Average power of car$=$change in kinetic energy per unit time$=\dfrac{\dfrac{1}{2}m{{v} _{2}}^{2}-\dfrac{1}{2}m{{v} _{1}}^{2}}{t}$,since ${v} _{1}=0$
So,average power of car$=\dfrac{\dfrac{1}{2}m{{v} _{2}}^{2}}{t}$
                                 $=\dfrac{\dfrac{1}{2}\times1000\times{27.78}^{2}}{5}$
                                 $=7.71\times{10}^{4}W$
                                

A particle of mass $m$ is lying on smooth horizontal table. A constant force $F$ tangential to the surface is applied on it. Find average power over a time interval from $t=0$ to $t=t$.

  1. $\displaystyle \frac{F^{2}t}{3m}$

  2. $\displaystyle \frac{F^{2}t}{2m}$

  3. $\displaystyle \frac{2F^{2}t}{3m}$

  4. $\displaystyle \frac{3F^{2}t}{2m}$


Correct Option: B
Explanation:

Average power of a time interval $t=0$ to $t=t$


is $\dfrac{\int^t _0 Fvdt}{ _0^t\int dt}$


$=\dfrac{F\int _0^t vdt}{t}$

$=\dfrac{F\int _0^t \dfrac{F}{m}tdt}{t}$

$=\dfrac{F^2t}{2m}$

A block of mass $1$ $kg$ starts moving with constant acceleration $\displaystyle a= 4m/s^{2}.$. Find the average power of the net force in a time interval from $t=0$ to $t=2s$.

  1. $16 W$

  2. $1.6 W$

  3. $15 W$

  4. $1.5 W$


Correct Option: A
Explanation:
Velocity of an object after time $t$ is given by $v=at$
Average power from time $t _1$ to $t _2$

$=\dfrac{\int _{t _1}^{t _2}Fvdt}{\int _{t _1}^{t _2}dt}$

$=\dfrac{\int _0^t(ma)(at)dt}{\int _0^tdt}$

$=ma^2\dfrac{t}{2}$

$=1\times 4^2\times\dfrac{2}{2}W$

$=16W$

A given $1700 kg$ car goes from $18 {m}/{s}$ to $0 {m}/{s}$. If this transition took $9 sec$, what was the average power supplied by the force causing this deceleration?
Take the car system to be otherwise isolated (i.e. the decelerating force was the only force acting on the car).

  1. $-1700 W$

  2. $-15300 W$

  3. $-30600 W$

  4. $-61200 W$

  5. Cannot be determined from the information given


Correct Option: C
Explanation:

Given :   $u =18$ m/s             $u = 0$ m/s                $m = 1700$ kg             $t = 9$ s

Using   $v = u+at$
$\therefore$  $0 = 18 + a \times 9$                 $\implies a = -2$  $m/s^2$
Using       $v^2  - u^2 = 2aS$
$\therefore$  $0 - (18)^2  =2(-2) S$                      $\implies S = 81$  m

Thus work done by decelerating force        $W  = -ma S $
$\therefore$  Power supplied     $P =\dfrac{-maS}{t}  = \dfrac{- 1700 \times 2 \times 81}{9}  =-30600$  W

Your uncle pushes a $60.0 kg$ crate along a floor with average speed $v=0.65 {m}/{s}$ for $5.0$ seconds as he moves furniture to clean up the garage.
If the coefficient of friction between the floor and the crate is $\mu=0.340$, what is the average power output of your uncle during this time?

  1. $26.0 W$

  2. $130 W$

  3. $383 W$

  4. $650 W$

  5. None of the above


Correct Option: B
Explanation:

Given :  $\mu = 0.340$             $v = 0.65$  m/s                $m =60.0$  kg

As the crate moves with constant speed, thus the force applied by uncles must be equal to the frictional force.
$\therefore$   $F  = \mu mg = 0.340 \times 60.0 \times 9.8  = 199.92$  N
Average power output        $P = F v = 199.92 \times 0.65  \approx 130$  W

A $0.25 kg$ book falls from a height of $1.00 m$, initially at rest. What is the AVERAGE power output of gravity on the book in this time period?

  1. $2.45 W$

  2. $5.43 W$

  3. $10.9 W$

  4. $24.1 W$

  5. None of the above


Correct Option: B
Explanation:

Given :   $m = 0.25$ kg               $S =1.00$ m                    $u =0$ m/s                $a = g = 9.8m/s^2$

Using   $S = ut+\dfrac{1}{2}at^2$
$\therefore$    $1.00 = 0 + \dfrac{1}{2} \times 9.8t^2$                                     $\implies t = 0.45$ s
 Work done by gravity         $W = mgh  =0.25 \times 9.8 \times 1.00 = 2.45$  J
$\therefore$Output power  of gravity        $P = \dfrac{W}{t} = \dfrac{2.45}{0.45} = 5.43$  $W$

Find the power of a pump which takes $10 s$ to draw $100 kg$ of water from a tank situated at a height of $20 m$.

  1. $2\times { 10 }^{ 4 }W$

  2. $2\times { 10 }^{ 3 }W$

  3. $200 W$

  4. $1 kW$


Correct Option: B
Explanation:

Power = $\dfrac{work done}{time}$


             =  $\dfrac{mgh}{time}$

            =  $\dfrac{100 \ast 10 \ast 20}{10}$

             =  $\dfrac{20000}{10}$

               = 2000W  i.e  2 $\times$ 10$^{3}$W

A $40 N$ force accelerates a $20 kg$ mass through a distance of $16 m$. If there were no other forces acting on the mass and the mass started at rest, what was the average power output of the force?

  1. $0 W$

  2. $160 W$

  3. $226 W$

  4. $640 W$

  5. Cannot be determined from the information given


Correct Option: B
Explanation:

Given :  $F = 40$ N            $m = 20$ kg              $S = 16$ m               $u = 0$ m/s

Thus acceleration of the mass        $a = \dfrac{F}{m} = \dfrac{40}{20} =2$  $m/s^2$
Using    $S = ut + \dfrac{1}{2}at^2$
$\therefore$  $16 = 0 + \dfrac{1}{2} \times 2 t^2$                      $\implies t = 4$ s
Work done by the force     $W = FS = 40 \times 16  =640$ J
$\therefore$ Power output        $P = \dfrac{W}{t} = \dfrac{640}{4} = 160$  W

An engine of $4.9 kW$ power is used to pump water from a well which is $20 m$ deep. What quantity of water in kiloliters can it pump out in $30$ minutes?

  1. $45 kl$

  2. $75 kl$

  3. $25 kl$

  4. $90 kl$


Correct Option: A
Explanation:

Power$=\dfrac{work}{time}=\dfrac{mgh}{t}$

$=\dfrac{4.9\times 10^3\times 30\times 60}{9.8\times 20}=m$
$m=45kl$

A girl of mass $40 kg$ climbs $50$ stairs each of average height $20 cm$ in $50 s$. Find the power of the girl $\left( g=10m{ s }^{ -2 } \right) $.

  1. $50 W$

  2. $50\times 20W$

  3. $80 W$

  4. $50\times 20\times 2W$


Correct Option: C
Explanation:

Power $=\dfrac{work \quad done}{time}$

$=\dfrac { change\quad in\quad PE }{ time } \ =\dfrac { 40\times 10\times 50\times 20\times { 10 }^{ -2 } }{ 50 } \ =80W$

In a factory, due to a sudden strike the work usually done in a day took a longer time. Which of the following happened?

  1. Power increased

  2. Power decreased

  3. Energy increased

  4. None of these


Correct Option: B
Explanation:

Power $=\dfrac{work \quad done}{time}$

Since time is increased,power decreased.

In each heart beat, a heart pumps $80\ ml$ of blood at an average pressure of $100\ mm$ of $Hg$. Assuming $60$ heart beats per second, the power output of the heart is $(\rho _{Hg} = 13.6\times 10^{3} kgm^{-3})(g = 9.8\ ms^{-2})$.

  1. $60.97\ W$

  2. $63.97\ W$

  3. $1.12\ W$

  4. $83.97\ W$


Correct Option: B
Explanation:

Power=$P\times\Delta V\times beat\ rate$

$\rho gh\Delta V\times 60$
$13.6\times10^3\times9.8\times 100\times 10^{-3}\times80\times10^{-6}\times60=63.97\ W$

An area of land is an average of $2\ m$ below sea level. To prevent flooding, pumps are used to lift rainwater up to sea level. What is the minimum pump output power required to deal with $1.3 \times 10^9\ Kg$ of rain per day?

  1. $15\ KW$

  2. $30\ KW$

  3. $100\ KW$

  4. $300\ KW$


Correct Option: D
Explanation:

Given,

$m=1.3\times 10^9 kg$
$g=10m/s^2$
$h=2m$
$t=1day$
Power, $P=\dfrac{mgh}{t}$
$P=\dfrac{1.3\times 10^9\times 10\times 2}{1\times 24\times 60\times 60}$
$P=300\times 10^3=300kW$
The correct option is D.

A stone is projected with velocity $u$ at an angle $\theta$ with horizontal. Find out average power of the gravity during time t.

  1. $mg\, \left [ \displaystyle \frac{gt^2}{2}\, - u sin \theta \right ]$

  2. $mg\, \left [ \displaystyle \frac{gt}{2}\, + u sin \theta \right ]$

  3. $mg\, \left [ \displaystyle \frac{gt}{2}\, - u sin \theta \right ]$

  4. $mg\, \left [ \displaystyle \frac{gt}{4}\, - u sin \theta \right ]$


Correct Option: C
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