This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Effect of Gravity”.
1. A body of mass 0.25kg moving with a velocity 12m/s is stopped by applying a force of 0.6N. Calculate the time taken to stop the body.
Explanation: m = .25kg, u = 12m/s, v = 0, F = -0.6N
a = F/m = 0.6/0.25 = -24m/s2
t = (v-u)/a = (0-12)/(-24) = 5s.
2. Action and reaction take place on the same bodies.
Explanation: If the action and reaction were on the same body, the resultant force would be zero and there could never be accelerated motion. Therefore, action and reaction always act on different bodies.
3. Which of the following is the property of Newton’s third law?
a) A body moving with a uniform speed in a straight line cannot change the direction of motion by itself
b) It is applicable only to a point particle
c) It is a local relation
d) The action and reaction cannot cancel each other
Explanation: The forces of action and reaction cannot cancel each other. This is because of action and reaction, although equal and opposite, always act on different bodies and so cannot balance each other.
4. Book kept on a table is an example for ___________
b) First law of motion
c) Third law of motion
d) Second law of motion
Explanation: The book exerts a downward force on the table equal to its own weight. According to Newton’s third law, the table also exerts an equal and upward force on the book. As the book is under the action of two equal and opposite forces, it remains in equilibrium.
5. Which of the following is called a real law of motion?
a) Second law
b) First law
c) Third law
d) Law of inertia
Explanation: First law is contained in the second law: In the absence of external force a body at rest will remain at rest and a body in motion will continue to move. Third law is contained in the second law: In the absence of external force, the rate of change of momentum must be zero. This is the third law of motion. As both first and second law are confined in the second law, it the real law of motion.
6. For a man standing inside a life, when will his apparent weight be equal to real weigh?
a) When the life is moving upwards with a uniform acceleration
b) When the lift is moving downwards with a uniform velocity
c) At rest or moving with uniform velocity
d) Falling freely
Explanation: When the life is in rest or moving with a uniform velocity, the acceleration = 0. Net force on the man is zero. Therefore apparent weight will be equal to real weight of the man.
7. Conservation of linear momentum is a combination of ___________
a) Newton’s second law and third law
b) Newton’s second law and first law
c) Newton’s first law and third law
d) Newton’s second law and impulse force
Explanation: The second and third law of motion leads to one of the most important and fundamental principles of physics called law of conservation of momentum. When no external force acts on a body, the total linear momentum is conserved.
8. While firing a bullet, the gun should be ___________
a) Held away from the shoulder
b) Held tight to the shoulder
c) Held straight
d) Held with one hand stretched
Explanation: While firing a bullet, the gun should be held tight to the shoulder. The recoiling can hurt the shoulder. If the gun is held tightly against the shoulder, then the body and the gun together will constitute one system. Total mass becomes larger and the recoil velocity becomes small.
9. When a man jumps out of a boat to the shore, the boat slightly moves away from the shore.
Explanation: Initially, the total momentum of the boat and the man is zero. As the man jumps from the boat to the shore, he gains a momentum in the forward direction. To conserve momentum, the boat also gains an equal momentum in the opposite direction. So the boat slightly moves backwards.
10. A man weighing 60kg runs along the rails with a velocity of 18km/h and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start travelling along the rails.
Explanation: m1 = 60kg, u1 = 18km/h = 5m/s
m2 = 1quintal = 100kg, u2 = 0, v = ?
By conservation of linear momentum, (m1+m2)v = m1 u1+m2 u2
v = (60×5)/160 = 1.88m/s.
11. Fuel is consumes at the rate of 100kg/s in a rocket. The exhaust gases are ejected at speed of 4.5×104m/s. What is the thrust experienced by the rocket?
Explanation: dm/dt = 100kg/s, u = 4.5×104m/s
Thrust, F = u*(dm/st) = 4.5×104×100 = 4.5×106N.
Sanfoundry Global Education & Learning Series – Engineering Physics.
To practice all areas of Engineering Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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