Tag: work, energy and power

Questions Related to work, energy and power

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

If the heart pushes 1 cc of blood in one second under pressure of 20000$Nm^{-2}$, the power of heart is

  1. 0.02 W

  2. 400 W

  3. $5\times 10^{-10}W$

  4. 0.2 W


Correct Option: A
Explanation:

Given,

$P=20000Nm^{-2}$
$\dfrac{dV}{dt}=1cc/s$
Power, $P=\dfrac{W}{t}=F.v=PA\dfrac{dx}{dt}$   (as, $F=PA$)
$P=P.\dfrac{dV}{dt}$
$P=20\times 10^3\times \dfrac{1}{(10^2)^3}$
$P=\dfrac{2}{100}=0.02W$
The correct option is A.

A pump can hoist 9000 kg of coal per hour from a mine of 120 m deep. calculate the power of the pump, in watts, assuming its efficiency is 75%, g=9.8 $m/s^{2}$

  1. 3.92 watts

  2. 39.2 watts

  3. 392 watts

  4. 3920 watts


Correct Option: D

A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him? 

  1. Zero

  2. Positive

  3. Negative

  4. Nothing can be said


Correct Option: A

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

  1. 100 litre

  2. 1000 litre

  3. 1200 litre

  4. 2000 litre


Correct Option: C

When a ball collides head-on and elastically with an identical ball on a horizontal frictionless surface,comes to rest while the second one moves with the same velocity as that of the first ball before coillision . 

  1. Can be derived by using momentum conservation alone.

  2. Can be derived by using energy conservation alone

  3. cannot be derived by using any to the two conservation principles.

  4. Can be derived by using both conservation of energy and momentum


Correct Option: A

A ball 'A' of mass 100 gm moving at 2m/s collides with another identical ball 'B' at 3m/s along the same line. There is no loss in energy collision. then, the speed of the balls 'A' and 'B' after the collision are ?

  1. 1m/s and 4m/s

  2. 2m/s and 3 m/s

  3. 3 m/s and 2 m/ s

  4. 2.5 m/s and 2.5 m/s


Correct Option: B

Two bodies A and B of masses 5 kg and 10 kg moving in free space in opposite directions with velocity form for second and 0.5 m per second respectively undergo a head on collision the force f of their mutual interaction varies with time T according to the given graph what can you conclude from the given information

  1. Period of Di formation is 0.2 second

  2. Coefficient of restitution is 0.5

  3. Body 0.5 m per second in the original direction

  4. Body be will 1.75 M per second in the reverse direction


Correct Option: A

A solid cylinder of mass 'M' and radius 'R' is rotating along its axis with angular velocity $\omega $ without friction. A particle of mass 'm' moving with velocity v collide against the cylinder and sticks to its rim. After the impact calculate angular velocity of cylinder.

  1. $\cfrac { I+R\omega }{ I+m{ R }^{ 2 } } $

  2. $\cfrac { mvR+IR }{ I+m{ R }^{ 2 } } $

  3. $\cfrac { I\omega +mvR }{ I+m{ R }^{ 2 } }$

  4. $\cfrac { I\omega +mR }{ I+mv{ R }^{ 2 } } $


Correct Option: C