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OBJECTIVE-I
1. A body of weight w1 is suspended from the ceiling of a room through a chain of weight w2 . The ceiling pulls the chain by a force
(a) w1 (b) w2 (c) w1+w2 (d) (w1+w2)/2
Answer: (c)
Reason: The ceiling is pulled by a force = w1+w2 and as per third law of motion the ceiling too pulls the chain by a force = w1+w2.
2. When a horse pulls a cart, the force that helps the horse to move forward is the force exerted by
(a) the cart on the horse (b) the ground on the horse
(c) the ground on the cart (d) the horse on the ground
Answer: (b)
Reason: The horse pushes the ground backwards and as per Newton's third law of motion the ground too pushes the horse forward equally.
3. A car accelerates on a horizontal road due to the force exerted by
(a) the engine of the car (b) the driver of the car
(c) the earth (d) the road
Answer: (d)
Reason: The wheels of the car push the ground backwards and as per Newton's third law of motion the road pushes the wheels with equal and opposite force forwards that accelerates the car.
4. A block of mass 10 kg is suspended through two light spring balances as shown in figure (5-Q2).
Figure for question number 4
(a) both the scales will read 10 kg (b) both the scales will read 5 kg (c) the upper scale will read 10 kg and the lower zero.
(d) the readings may be anything but their sum will be 10 kg
Answer: (a)
Reason: At any point between the load and the ceiling the tension force will be equal to the load as per the third law of motion.
5. A block of mass m is placed on a smooth inclined plane of inclination θ with the horizontal. The force exerted by the plane on the block has a magnitude
(a) mg (b) mg/cosθ (c) mg cosθ (d) mg tanθ
Answer: (c)
Force exerted by plane
Reason: See the figure on the right. Weight of the block is mg downwards. We resolve it in two components-along the plane and perpendicular to the plane. Former component does not exert force on the plane while the later component mg.cosθ is the force which the block exerts on the plane. If we take it as action then equal and opposite force mg.cosθ is exerted by the plane on the block as reaction satisfying the "Newton's Third law of Motion".
6. A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude
(a) mg (b) mg/cosθ (c) mg cosθ (d) mg tanθ
Answer: (b)
Accelerated Plane
Reason: When the plane is accelerated to balance the block, a horizontal force F acts on the block such that its component along the plane Fcosθ is equal and opposite to component of the weight along the plane mg.sinθ, see the figure on the right. So, Fcosθ = mg.sinθ → F= mg.tanθ.
The components of F and weight mg perpendicular to plane (Fsinθ and mgcosθ) together push the plane. Equal and opposite force is exerted by the plane on the block as per "Newton's Third Law of Motion" and its value is = Fsinθ + mgcosθ
= mg.tanθ.sinθ + mgcosθ =mg(sin2θ +cos2θ)/cosθ = mg/cosθ
7. Neglecting the effect of rotation of the earth. Suppose the earth suddenly stops attracting objects placed near its surface. A person standing on the surface of the earth will
(a) fly up (b) slip along the surface
(c) fly along a tangent to the earth's surface (d) remain standing
Answer: (d)
Reason: The person will loose weight but remain standing. As no force acts on him, so according to "Newton's First Law of Motion" he will remain in his state of rest as before.
8. Three rigid rods are joined to form an equilateral triangle ABC of side 1 m. Three particle carrying charges 20 µC each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame. The resultant force on the charged particle at A has the magnitude
(a) zero (b) 3.6 N (c) 3.6/3 N (d) 7.2 N
Answer: (a)
Reason: Since the charged particles are attached to rigid triangle ABC which is in rest in an inertial frame, so the particle at A is also at rest. It means the magnitude of resultant of forces acting on it is zero.
9. A force F1 acts on a particle so as to accelerate it from rest to a velocity v. The force F1 is then replaced by F2 which decelerates it to rest.
(a) F1 must be equal to F2 (b) F1 may be equal to F2
(c) F1 must be unequal to F2 (d) None of these
Answer: (b)
Reason: Neither the time nor the distance for accelerating and decelerating journey is mentioned. So both forces may or may not be equal.
10. Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than the mass of B. Suppose the air exerts a constant and equal force of resistance on the two bodies.
(a) The two bodies will reach the same height
(b) A will go higher than B
(c) B will go higher than A (d) Any of the above three may happen depending on the speed with which the objects are thrown.
Answer: (b)
Reason: The force of resistance of air will add to retardation of both the body but its magnitude will depend upon mass of the objects as per "Newton's Second Law of Motion" and it will be equal to Force/mass. Since the force is equal on both the objects, the object having greater mass (A) will be retarded less, so it will go higher than B.
11. A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes a time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will
(a) take a time longer than T to slide down the wedge
(b) take a time shorter than T to slide down the wedge
(c) remain at the top of the wedge
(d) jump off the wedge.
Answer: (c)
Reason: Both the wedge and block will be in free fall motion and have equal velocities downwards at any instant of time so there will be no relative motion.
12. In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle's motion. A particle of mass m projected upward takes a time t1 in reaching the maximum height and t2 in the return journey to the original point. Then
(a) t1 < t2 (b) t1 > t2 (c) t1 = t2 (d) The relation between t1 and t2 depends on the mass of the particle.
Answer: (b)
Reason: Since F is in the direction of particle's motion it will add an acceleration in the direction of motion. Let it be a. Now magnitude of upward acceleration =a-g and downward acceleration =a+g. Clearly a+g>a-g and the particle travels same distance/height. So the upward journey will take a longer time.
13. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then
(a) t1 = t2 (b) t1 < t2 (c) t1 > t2 (d) t1 < t2 or t1 > t2 depending on whether the lift is going up or down.
Answer: (a)
Reason: The stationary and uniformly moving elevator are both inertial frame so value of acceleration due to gravity will remain the same in both conditions. The distance traveled by the coin with respect to the frame of elevator is equal in both cases. So the time taken will also be equal.
14. A free 238U nucleus kept in a train emits an α - particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay as measured by the passenger is
(a) x+vt (b) x-vt (c) x (d) depends on the direction of the train.
Answer: (c)
Reason: In both of the cases the frame of reference is inertial so with reference to the frame of train the separation too will be equal.
===<<<O>>>===
Links for the chapter -
CHAPTER-8 - Work and Energy
Click here for → Newton's Law's of Motion - Exercises(1 to 12)
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CHAPTER-7 - Circular Motion
CHAPTER-8 - Work and Energy
HC Verma's Concepts of Physics, Chapter-6, Friction,
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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HC Verma's Concepts of Physics, Chapter-5, Newton's Law's of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -IIClick here for → Newton's Laws of Motion - Exercises(13 to 27)
Click here for→ Newton's Laws of Motion-Exercises(Q.No.28 to 42)-------------------------------------------------------------------------------
HC Verma's Concepts of Physics, Chapter-4, The Forces
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - Exercises
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In ques 8(obj1) whole system is at rest bcoz net force due to 3 particle addupto zero but on individual particle A net force is not zero
ReplyDeleteI think that answer given in book is itself wrong
Neither the book nor the explanation is wrong. When net force on a particle is not zero it must accelerate.
DeleteSo basically the net repulsive force on the charge A is balanced by the force exerted by the rods of triangle
DeleteKindly give explanation for d option of ques 13 obj1
ReplyDeleteWhen the lift is going up or down the speed of the lift will be the initial velocity of the coin. You can calculate the time by assuming the data like height of fall and speed of the lift for your satisfaction.
DeleteThanku so much sir
ReplyDeleteI am satisfied by your answer
Hi Sir,
ReplyDeleteCould you explain Q9 answer with an example please.
Dear student
DeleteSince F2 brings the object to rest let us assume that it acts till it comes to rest. Since F2 decelerates, its direction is opposite to F1. Let T = time till F1 acts and T' = time till F2 acts. If F2 > F1 then T' < T and If F2 < F1, then T' > T. But if F2 = F1, then T = T'. So we are can not be sure about the magnitude of F2. Since there is a possibility of F2 = F1 hence option (b) is true. Other options are not true.
Hi Sir,
ReplyDeleteAs force is a vector quantity, two forces opposite in direction are always unequal even if they have same magnitude.
Is the question considering only the magnitude of the two forces?
True, they are considering only magnitudes.
DeleteOk, Thank you Sir.
ReplyDeleteSir in the question 12 objective 1 it is told that the F is acting on the particle in its direction of motion. When the particle is in upward motion the force should add in upwrup and when it is downward the force should add in the downward direction so net Acceleration would be equal and hence the time would be equal . Yes or No . Plz explain sir ji .
ReplyDeleteDear student, answer to your question is no.
DeleteThe force of weight on the body mg is always downward while the force F changes its direction. So when F is upward, net force on the body is the difference of F and mg. But when F is downward the net force is the sum of F and mg. So the resultant forces in two cases are not the same. Hence the accelerations are also not the same.
Sir in Q 6 in which direction should the block be accelerated ? According to me it should in the left direction so that the pseudo force will act on the right direction and in this way it will balance the block . And sir is the force 'F' labelled in the diagram a pseudo force ?
ReplyDeleteObvious. Yes, F is the pseudo force.
DeleteSir in Q 7 what is the significance of the line ' neglect the effect of rotation of the earth '
ReplyDeleteWould the ans have been different if the above line was not given in the question ?
Your explanation is the best .
Yes, the answer would have been different. Due to earth's rotation, a person has an instantaneous tangential velocity on the surface of the earth. Option (c) would be the answer then.
DeleteSir in Q 7 why not option (a)?
ReplyDeleteThough there Is no gravitational pull ...but there is normal force which will push it away as it is a contact force and contact force would only disappear if the electrostatic force stopped, but that would be catastrophic because that is the force which holds atoms and molecules together, and also keeps them apart.
As soon as the gravitational pull is gone, the person loses his weight. He can not push the ground below his feet with his weight, so no normal force. In a standing position the normal force is equal and opposite to the weight. Hence the option (a) is not true.
DeleteSir in Q 8 what would have be the ans if it was non inertial frame?
ReplyDeleteSir in Q 10 u have mentioned that :
ReplyDelete' its magnitude will depend upon mass of the '
But I have studied that air resistance does not depend on the mass of the object
Dear student, the sentence is not like that. It has been said that the retardation (negative acceleration) will depend upon the mass of the body.
DeleteThank u so much sir
ReplyDeleteSir I think in Q 12 it should be :
ReplyDelete'Upward acceleration should be g-a instead of a-g' because if it is a-g then particle be move up forever.please correct me if wrong.🙏🙏🙏
What prompts you to assume that the numerical value of 'a' is more than 'g'? It is not so. And when I say upward acceleration is 'a-g', it doesn't mean particle is really accelerated upward. If the value of a-g is negative, it means the object is retarded, and it must be the case here because the maximum height is given.
DeleteThank u sir
ReplyDelete