12th Grade > Physics
WORK POWER AND ENERGY MCQs
Work And Energy, Sources Of Energy, Electric Potential Energy Potential And Dipoles
Total Questions : 109
| Page 11 of 11 pages
Answer: Option D. -> 9 m
:
D
Maximum energy depends on the work done in compressing the spring which is directly proportional to the height attained by the ball after the spring is released. The gravitational potential energy is expressed as "m x g x h” and it would be the maximum for the height of 9m. So, he has to domaximum work to compress the spring when the ball rises to a height of 9m.
:
D
Maximum energy depends on the work done in compressing the spring which is directly proportional to the height attained by the ball after the spring is released. The gravitational potential energy is expressed as "m x g x h” and it would be the maximum for the height of 9m. So, he has to domaximum work to compress the spring when the ball rises to a height of 9m.
Answer: Option B. -> 60 J
:
B
Given, mass of the box,m=2kg, vertical height, h=3m and acceleration due to gravity, g=10ms−2 .
We know that, gravitational potential energy =mgh=2×10×3=60Joule
:
B
Given, mass of the box,m=2kg, vertical height, h=3m and acceleration due to gravity, g=10ms−2 .
We know that, gravitational potential energy =mgh=2×10×3=60Joule
Answer: Option B. -> 170J
:
B
Given:
Mass, m=10kg,
Acceleration due to gravity, g=10ms−2,
Displacement, s=1.7m.
Force F is calculated as
F=m×g=10×10=100N
Work done by the man onluggage=Force×displacement
W=100×1.7=170J
Hence, the required work done is 170J.
It can be noted that the work is done by the man on the luggage because he lifts it against the gravitational force of the earth.
:
B
Given:
Mass, m=10kg,
Acceleration due to gravity, g=10ms−2,
Displacement, s=1.7m.
Force F is calculated as
F=m×g=10×10=100N
Work done by the man onluggage=Force×displacement
W=100×1.7=170J
Hence, the required work done is 170J.
It can be noted that the work is done by the man on the luggage because he lifts it against the gravitational force of the earth.
Answer: Option B. -> 50 J
:
B
Let initial velocity of the child be v1 and final velocity be v2
Given, v1=2ms−1, v2=3ms−1 and massof the child =20kg.
Since the child is in motion, he will have a kinetic energy ofmv22, which will be increased after an application of force by an external agent on him.
Now, Applying work energy theorem ,
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
Given, v1=2ms−1, v2=3ms−1 and massof the child =20kg.
Since the child is in motion, he will have a kinetic energy ofmv22, which will be increased after an application of force by an external agent on him.
Now, Applying work energy theorem ,
Work done by the external force on the child =mv222–mv212
=20×(9–4)2
=10×5=50J
Answer: Option B. -> False
:
B
Potential energy depends on the position of an object with respect to a reference point. In this case, the earth is the reference point. A man on a lift going up, gains height with respect to the earth.
Thus, the potential energy of the person inside the lift increases.
:
B
Potential energy depends on the position of an object with respect to a reference point. In this case, the earth is the reference point. A man on a lift going up, gains height with respect to the earth.
Thus, the potential energy of the person inside the lift increases.
Answer: Option B. -> False
:
B
Moving train has velocity and will have only kinetic energy. A moving train will not gain any height and hence has no potential energy. Therefore, the given statement is wrong.
:
B
Moving train has velocity and will have only kinetic energy. A moving train will not gain any height and hence has no potential energy. Therefore, the given statement is wrong.
Answer: Option D. -> Just before touching the ground floor
:
D
When the object isat a height of 20m above the ground, it has only potential energy.
Since the velocity of the object increases when it comes down, its kinetic energy will increase. When theobject is just about to touchthe ground, its potential energy will be zeroand kinetic energy will be maximum.
:
D
When the object isat a height of 20m above the ground, it has only potential energy.
Since the velocity of the object increases when it comes down, its kinetic energy will increase. When theobject is just about to touchthe ground, its potential energy will be zeroand kinetic energy will be maximum.
Answer: Option B. -> False
:
B
Work done by a force is defined as the product of force and displacement of the body along the direction of force. Since Hulk was unable to move the wall, the displacement is zero and irrespective of the large amount of force he applied, work done will be zero.
:
B
Work done by a force is defined as the product of force and displacement of the body along the direction of force. Since Hulk was unable to move the wall, the displacement is zero and irrespective of the large amount of force he applied, work done will be zero.
Answer: Option B. -> 60 W
:
B
Given, energy E=7.2kJ=7.2×103J and time t=2minutes=2×60=120s.
We know, power, P=EnergyorWorkt=7.2×103120=60W.
∴ The power of electric bulb is 60W.
:
B
Given, energy E=7.2kJ=7.2×103J and time t=2minutes=2×60=120s.
We know, power, P=EnergyorWorkt=7.2×103120=60W.
∴ The power of electric bulb is 60W.