12th Grade > Physics
WORK POWER AND ENERGY MCQs
Work And Energy, Sources Of Energy, Electric Potential Energy Potential And Dipoles
Total Questions : 109
| Page 10 of 11 pages
Answer: Option A. -> Solar energy - Chemical energy - Kinetic energy
:
A
Solar panels containphotovoltaic cells, which converts the sunlight to electricity.The electric cell stores this electrical energy in the form of chemical energy. When a toy is connected to this electric cell, the stored energy is used to move the toy. Therefore the toy possesses kinetic energy due to its motion.Thus,the energy changes occurring over here are solar energy to chemical energy and then this chemical energy is convertedto kinetic energy.
:
A
Solar panels containphotovoltaic cells, which converts the sunlight to electricity.The electric cell stores this electrical energy in the form of chemical energy. When a toy is connected to this electric cell, the stored energy is used to move the toy. Therefore the toy possesses kinetic energy due to its motion.Thus,the energy changes occurring over here are solar energy to chemical energy and then this chemical energy is convertedto kinetic energy.
Answer: Option B. -> 25 kW
:
B
Given, mass of the body m=50kg, initial velocity u=0ms−1, velocity v=100ms−1 and time t=10s.
The change in kinetic energy of the car would be equal to the work done by the engine.
Since, power is work done per unit time, so change in kinetic energy of the car and the time taken for that change would give us the power developed by the engine.
i.e.Work done =12×m×v2−12×m×u2=12×50×1002−12×50×02=250000J.
Power =WorkTime=25000010=25000W=25kW
:
B
Given, mass of the body m=50kg, initial velocity u=0ms−1, velocity v=100ms−1 and time t=10s.
The change in kinetic energy of the car would be equal to the work done by the engine.
Since, power is work done per unit time, so change in kinetic energy of the car and the time taken for that change would give us the power developed by the engine.
i.e.Work done =12×m×v2−12×m×u2=12×50×1002−12×50×02=250000J.
Power =WorkTime=25000010=25000W=25kW
Answer: Option B. -> False
:
B
We know that power is the energy consumed per unit time.
⇒ PowerConsumed=EnergyConsumedTime
The heart beats 72 times in one minute i.e. 60 seconds.
Energy consumed in one heart beat = 5 J
Energy consumed in 72heart beats = 5 ×72 J
On substituting all these values, we get
PowerConsumed=5×7260=6W
∴ The power consumed by the heart is6W.
:
B
We know that power is the energy consumed per unit time.
⇒ PowerConsumed=EnergyConsumedTime
The heart beats 72 times in one minute i.e. 60 seconds.
Energy consumed in one heart beat = 5 J
Energy consumed in 72heart beats = 5 ×72 J
On substituting all these values, we get
PowerConsumed=5×7260=6W
∴ The power consumed by the heart is6W.
Answer: Option B. -> 100 J
:
B
The change in gravitational potential energy of the baggage isequal to the work done against gravity in shifting the baggage from the first floor to the second floor.
Hence, it will be equal to 100 J.
:
B
The change in gravitational potential energy of the baggage isequal to the work done against gravity in shifting the baggage from the first floor to the second floor.
Hence, it will be equal to 100 J.
Answer: Option B. -> PE1 > PE2 < PE3
:
B
Potential Energy = Force × Displacement =mg×h where, m is the mass, g is acceleration due to gravity and h is the height.
Since the objects are of the same mass and acceleration due to gravity has a fixed value on earth, the force mg in all three cases aresame.
⇒Potential Energy =constant×h
⇒Potential energy is directly proportional to its height.
We know that,h1 > h2 < h3.
Therefore, PE1 > PE2 < PE3
:
B
Potential Energy = Force × Displacement =mg×h where, m is the mass, g is acceleration due to gravity and h is the height.
Since the objects are of the same mass and acceleration due to gravity has a fixed value on earth, the force mg in all three cases aresame.
⇒Potential Energy =constant×h
⇒Potential energy is directly proportional to its height.
We know that,h1 > h2 < h3.
Therefore, PE1 > PE2 < PE3
Answer: Option B. -> 50 J
:
B
Let initial velocity of the child be v1 and final velocity be v2
v1=2ms−1
v2=3ms−1
Since the child is in motion, he will have a kinetic energy ofmv22, which will be increasedwhen an external force acts on him.
Work done by the external force on the child =mv222–mv212
=20×(9–4)2
=10×5=50J.
:
B
Let initial velocity of the child be v1 and final velocity be v2
v1=2ms−1
v2=3ms−1
Since the child is in motion, he will have a kinetic energy ofmv22, which will be increasedwhen an external force acts on him.
Work done by the external force on the child =mv222–mv212
=20×(9–4)2
=10×5=50J.
Answer: Option A. -> Both the statements are true and statement 2 is the correct explanation of statement 1.
:
A
Any force is said to have done a positive work on an object if that force or any component of it displaces the object in the direction in which it acts. It can be seen that the kinetic energy of the body increases due to the application of force, therefore the velocity of the body increases which implies that the displacement is positive.
:
A
Any force is said to have done a positive work on an object if that force or any component of it displaces the object in the direction in which it acts. It can be seen that the kinetic energy of the body increases due to the application of force, therefore the velocity of the body increases which implies that the displacement is positive.
Answer: Option A. -> Positive, Negative
:
A
At the time of compressing, the work done by the boy on the ball is positive as it causes the ball to squeeze which is in direction of force applied.
But the ball resists to change its shape and always tries to regain its original shape by opposing the force applied, therefore, the work done by the ball on the boy is negative.
:
A
At the time of compressing, the work done by the boy on the ball is positive as it causes the ball to squeeze which is in direction of force applied.
But the ball resists to change its shape and always tries to regain its original shape by opposing the force applied, therefore, the work done by the ball on the boy is negative.
Question 99. Two boys were playing with a ball on a level ground. One of the boys throws the ball on the third floor of the building which is at a height of 5 m and the other boy manages to throw on the fourth floor which is at a height of 7 m from the level ground. Find the ratio of the work done by both the boys?
Answer: Option C. -> 5 : 7
:
C
The ratio of work done by both the boys is the respective change in gravitational potential energies of the ball. Taking the ground level as the reference, the change ingravitational potential energy is m×g×h where m- mass, g- acceleration due to gravity and h- height from the earth surface. Thechange for one ball would be m×g×5 and for the other it would be m×g×7 after starting from a zero gravitational potential energy from the ground.
∴The ratio of the work done would be equal to
m×g×5m×g×7=5:7
:
C
The ratio of work done by both the boys is the respective change in gravitational potential energies of the ball. Taking the ground level as the reference, the change ingravitational potential energy is m×g×h where m- mass, g- acceleration due to gravity and h- height from the earth surface. Thechange for one ball would be m×g×5 and for the other it would be m×g×7 after starting from a zero gravitational potential energy from the ground.
∴The ratio of the work done would be equal to
m×g×5m×g×7=5:7
Answer: Option C. -> 5ms−1
:
C
Given, the weight of a man, W=600N, g=10ms−2, kinetic energy, KE=750J
Let the mass of the man be m and the velocity be v.
We know, W=mg
So, 600=m×10
On solving we get, m=60kg
As we know, KE=12mv2
So, 750=12×60×v2
or,v=5ms−1
:
C
Given, the weight of a man, W=600N, g=10ms−2, kinetic energy, KE=750J
Let the mass of the man be m and the velocity be v.
We know, W=mg
So, 600=m×10
On solving we get, m=60kg
As we know, KE=12mv2
So, 750=12×60×v2
or,v=5ms−1
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