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OBJECTIVE-II
1. An object follows a curved path. The following quantities may remain constant during the motion
(a) speed (b) velocity
(c) acceleration (d) magnitude of acceleration
Answer: (a), (d).
Explanation: On a curved path, instantaneous direction of movement changes continuously. So, a vector can not remain constant on a curved path because it has both magnitude and direction. In the above options, velocity and acceleration are vectors, therefore they will change. Speed and magnitude of acceleration do not have directions, so they may remain constant on a curved path, Hence options (a) and (d) are correct.
2. Assume that the earth goes round the Sun in a circular orbit with a constant speed of 30 km/s.
(a) the average velocity of the earth from 1st Jan, 90 to 30th June, 90 is zero.
(b) the average acceleration during the above period is 60 km/s².
(c) the average speed from 1st Jan, 90 to 31st Dec, 90 is zero.
(d) the instantaneous acceleration of the earth points towards the sun.
Answer:(d).
Explanation: Since during the mentioned period in (a) and (b) the earth covers a semi circle and there is a displacement between original point to final point, so average velocity can not be zero.
Acceleration is the rate of change of velocity, here the change in velocity may be 30-(-30)=60 km/s but the time taken is six months i.e. 181*24*60*60 s. Dividing the change in velocity by this time will give the quotient much much less than 60 km/s².
Since the earth moves with a constant speed of 30 km/s, so average speed will also be 30 km/s not zero.
Since it is a case of uniform circular motion, the tangential component of the instantaneous acceleration will be zero and only radial component will be there which is directed towards the center (Sun in this case). Only option (d) is correct.
3. The position vector of a particle in a circular motion about the origin sweeps out equal area in equal time. Its
(a) velocity remains constant.
(b) speed remains constant.
(c) acceleration remains constant.
(d) tangential acceleration remains constant.
Answer: (b), (d).
Explanation: Velocity and acceleration will not be constant because direction changes continuously. Hence option (a) and (c) are not correct.
With a constant speed the position vector will sweep equal area in equal time. Thus option (b) is correct.
Since speed is constant, it is a case of uniform circular motion in which tangential acceleration is always zero i.e, always constant. Thus option (d) is also correct.
4. A particle is going in a spiral path as shown in figure (7-Q3) with constant speed.
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| Figure for Problem no-4 |
(a) the velocity of the particle is constant.
(b) the acceleration of the particle is constant.
(c) the magnitude of acceleration is constant.
(d) the magnitude of acceleration is decreasing continuously.
Answer: (c).
Explanation: Option (a) and (b) are not correct for the same reason mentioned in explanation in problem no-1.
Since the speed is constant the particle will have only radial acceleration with constant magnitude. Option (c) is correct.
Option (d) is incorrect because option (c) is correct and both options are contradictory.
5. A car of mass 'm' is moving on a horizontal circular path of radius r. At an instant its speed is 'v' and is increasing at a rate 'a'.
(a) the acceleration of the car is towards the center of the path.
(b) the magnitude of the frictional force on the car is greater than mv²/r.
(c) the friction coefficient between the ground and the car is not less than a/g.
(d) the friction coefficient between the ground and the car is µ=tan-1(v²/rg).
Answer: (b), (c).
Explanation: Since the circular motion is not uniform it will have both the radial and tangential components of acceleration and the resultant acceleration will be other than towards the center of path. Option (a) is wrong.
Since the car will experience a force equal to mv²/r radially outward so the friction force must be more than it to keep it on the circular path. Option (b) is correct.
Normal force on the car by the ground =mg . So frictional force on the car =µmg. It must not be less than outward force ma i.e. µmg≮ ma, →µ≮ a/g. Option (c) is correct.
In limiting case µ=v²/rg, so option (d) is incorrect.
6. A circular road of radius r is banked for a speed v=40 km/h. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.
(a) the car cannot make a turn without skidding.
(b) if a car turns at a speed less than 40 km/h it will slip down.
(c) if the car turns at the correct speed of 40 km/h , the force by the road on the car is equal to mv²/r.
(d) if the car turns at the correct speed of 40 km/h , the force by the road on the car is greater than mg as well as greater than mv²/r.
Answer: (b), (d).
Explanation: The car can turn without skidding at 40 km/h, thus option (a) is wrong.
Since µ=0, so only force on the car by the road is the Normal force and no frictional force. So for speed less than 40 km/h it will slip down. Option (b) is correct.
As stated above only force by the road on the car is normal force (N) horizontal component of which (Nsinθ) produces a centripetal acceleration =v²/r . So Nsinθ=mv²/r, →N=mv²/rsinθ Where θ is angle of banking. Thus option (c) is wrong. See figure below,
Forces on a car on a banked Curve
As above, N=mv²/rsinθ , so N is always greater than mv²/r because sinθ is always <1. Similarly Ncosθ=mg, → N=mg/cosθ. So N is always >mg because cosθ is always <1. Thus option (d) is correct.
7. A person applies a constant force F on a particle of mass m and finds that the particle moves in a circle of radius r with a uniform speed v as seen from an inertial frame of reference.
(a) this is not possible.
(b) there are other forces on the particle.
(c) the resultant of the other forces is mv²/r towards the center.
(d) the resultant of the other forces varies in magnitude as well as direction.
Answer: (b), (d).
Explanation: Above situation is possible when resultant of external forces (R) is perpendicular to direction of motion and always directed towards center with a constant magnitude. So the direction of this resultant force is changing continuously. Since the force applied by the person (F) is constant hence there are other forces acting on the particle and the resultant of these other forces (P) is varying in magnitude and direction so that resultant (R) of the force applied by the person (F) and resultant of other forces (P) has a constant magnitude and always directed towards the center. So options (b) and (d) are correct and (a) & (c) are incorrect.
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Links to the chapter -
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CHAPTER- 10 - Rotational Mechanics
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q-1 TO Q-15
EXERCISES - Q-16 TO Q-30
CHAPTER- 10 - Rotational Mechanics
Questions for Short Answers
OBJECTIVE - I
OBJECTIVE - II
EXERCISES - Q-1 TO Q-15
EXERCISES - Q-16 TO Q-30
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
HC Verma's Concepts of Physics, Chapter-7, Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
HC Verma's Concepts of Physics, Chapter-6, Friction
Click here for → Friction OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → Friction - OBJECTIVE-II
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)
For more practice on problems on friction solve these "New Questions on Friction" .
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these "New Questions on Friction" .
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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 -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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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" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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in q.6 the free body diagram is wrong sir Nsin(theta) will be in direction of mv²/r you can read the same on page 104 in paragraph above equation 7.13
ReplyDeleteThank! you so much for the effort that you have put in to help others
Thanks Manraj Singh ! I've corrected it.
DeleteIn question no 5 why we consider the outward force
ReplyDeleteIn question no 5 why we consider the outward force
ReplyDeleteSince the body is on a circular path, a centrifugal force or you may say it a pseudo force will have to be considered to apply the Newton's Laws of motion.
DeleteIn Q 1st of objective 2 (c) should also be answer because in case of a parabolic motion which is also a curve acceleration I.e g is constant
ReplyDeleteReally grateful for your kind efforts
True.
DeleteKk sir you are a legend
Delete