Tag: work, energy and power

Quizzes (16) Questions (235)

Questions Related to work, energy and power

Multiple choice power work and power work, energy and power physics energy and its forms

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. $300\ KW$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice power work and power work, energy and power physics energy and its forms

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. $mg\, \left [ \displaystyle \frac{gt}{4}\, - u sin \theta \right ]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Average power of gravity = Work / time. Work done by gravity = -m * g * delta_y. Delta_y = u * sin(theta) * t - (1/2) * g * t^2. Work = -m * g * (u * sin(theta) * t - 0.5 * g * t^2). Power = Work / t = -m * g * (u * sin(theta) - 0.5 * g * t) = m * g * (0.5 * g * t - u * sin(theta)).

Multiple choice power work and power energy and its forms work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice power work and power energy and its forms work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work = mgh = 9000 kg * 9.8 m/s^2 * 120 m = 10,584,000 J. Power = Work / time = 10,584,000 J / 3600 s = 2940 W. Given 75% efficiency, the required power input is 2940 / 0.75 = 3920 W.

Multiple choice power work and power energy and its forms work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If we consider work done against gravity, walking from one point to the antipodal point involves moving away from Earth's center then descending back down. The net work against gravity is zero (positive climbing cancels negative descending). However, the question is poorly specified as the person does positive muscular work throughout.

Multiple choice power work and power energy and its forms work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Power = 4000 W. Time = 60 s. Work = Power * time = 240,000 J. Work = mgh -> 240,000 = m * 10 * 20 -> 240,000 = 200m -> m = 1200 kg. Since 1 kg of water = 1 liter, the volume is 1200 liters.

Multiple choice modelling collisions collisions momentum work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When two identical bodies undergo a one-dimensional elastic collision, energy conservation and momentum conservation together are necessary and sufficient to uniquely determine the final velocities, demonstrating that both principles are required.

Multiple choice modelling collisions collisions momentum work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an elastic collision between two identical bodies, the velocities are simply exchanged. Ball A initially moving at 2 m/s and Ball B moving at 3 m/s along the same line will swap velocities, meaning Ball A moves at 3 m/s and Ball B moves at 2 m/s after the collision.

Multiple choice modelling collisions collisions momentum work, energy and power physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice modelling collisions collisions momentum work, energy and power physics

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 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using conservation of angular momentum about the axis of the cylinder: L_initial = I*omega + m*v*R. L_final = (I + m*R^2)*omega_final. Equating them gives omega_final = (I*omega + m*v*R) / (I + m*R^2).