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OBJECTIVE-I
1. When a particle moves in a circle with a uniform speed
(a) Its velocity and acceleration are both constant
(b) Its velocity is constant but the acceleration changes
(c) Its acceleration is constant but the velocity changes
(d) Its velocity and acceleration both change.
ANSWER: (d)
Explanation to the Answer → Velocity and acceleration are both vectors and a vectors is fully defined with its magnitude and direction. Even when one entity changes, the vector changes. In the given case the particle moves in a circle with uniform speed. But movement in a circle makes its direction change every moment. So magnitude of the velocity is constant but its direction changes and hence the velocity changes.
The magnitude of acceleration of the particle is given by v²/r and its direction is towards the center ie towards radius, but as the particle moves the connected radius and hence direction of the acceleration changes. So acceleration also changes as the particle moves.
2. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circles in equal time, the ratio of their angular speeds ω1/ω2 is
(a) m1/m2
(b) r1/r2
(c) m1r1/m2r2
(d) 1
ANSWER: (d)
Explanation to the Answer → In completing a circle the angle covered is 2π what ever be the radius of the circle. The angular speed will be given by 2π/T where T is the time taken in covering the circle. In this case T is same for both cars, so both cars will have same angular speed ω = 2π/T . Hence ω1/ω2 =1
3. A car moves at a constant speed on a road as shown in the figure (7-Q2).The normal force by the road on the car is NA and NB when it is at the points A and B respectively.
(a) NA = NB (b) NA > NB (c) NA < NB (d) Insufficient information to decide the relation of NA and NB.
Figure for problem 3
ANSWER: (c)
Explanation to the Answer → Both points A and B are on the crest of the curves where the weight of the car 'mg' is perpendicular to the surface. But due to the movement of the car on the curve it will feel reduction in the weight by an amount mv²/r, where 'v' is the constant speed of the car and 'r' is the radius of the curve. Clearly this reduction in weight is inversely proportional to 'r' as the numerator is constant. Smaller the 'r', greater is the term mv²/r. Since 'r' at point A is smaller than at the point B, so reduction in weight will be more at A than at B. It means the apparent weight is less at A than B.
Normal forces at these points will be equal to the apparent weight at these points. So NA < NB .
4. A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius r with a uniform speed v. The centrifugal force on it is,
(a) mv²/r towards the center
(b) mv²/r away from the center
(c) mv²/r along the tangent through the particle.
(d) Zero
ANSWER: (d)
Explanation to the Answer → Centrifugal force is a pseudo force needed in an non-inertial frame of reference when we still want to use the Newton's Laws. In this case the frame of reference is inertial, so there is no centrifugal force,
5. A particle of mass 'm' rotates in a circle of radius 'a' with a uniform angular speed ω. It is viewed from a frame rotating about Z-axis with a uniform angular speed ω0 . The centrifugal force on the particle is
(a) mω²a (b) mω0²a (c) m{(ω+ω0)/2}²a (d) mωω0a
ANSWER: (b)
Explanation to the Answer → The frame of reference is rotating with a uniform angular speed ω0 , so it has an acceleration towards the center and it is a non-inertial plane. To apply Newton's Laws in this frame we apply a pseudo force equal to mω0²a to the particle but opposite to the direction of the acceleration of the frame. This pseudo force is the centrifugal force on the particle.
6. A particle is kept fixed on a turn table rotating uniformly. As seen from the ground, the particle goes in a circle, its speed is 20 cm/s and acceleration 20 cm/s². The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed and the acceleration will be
(a) 10 cm/s, 10 cm/s² (b) 10 cm/s, 80 cm/s²
(a) 40 cm/s, 10 cm/s² (a) 40 cm/s, 40 cm/s²
ANSWER: (a)
Explanation to the Answer → Velocity of the particle in a circular motion is equal to ωr , so if ω is constant the velocity is proportional to the radius. Here the particle is kept at half radius therefore velocity will be half ie = 10 cm/s.
Similarly acceleration is ω²r , here too, ω² is constant, so acceleration is also proportional to 'r'. In this case too, making radius half will make acceleration half ie = 10 cm/s².
7. Water in a bucket is whirled in a vertical circle with a string attached to to it. The water does not fall down even when the bucket is inverted at the top of its path. We conclude that in this position
(a) mg=mv²/r (b) mg is greater than mv²/r
(c) mg is not greater than mv²/r
(d) mg is not less than mv²/r
ANSWER: (c)
Explanation to the Answer →At the top position the weight of the water mg acts downward to bring it down while due to inertia water tries to escape upward with a force mv²/r on the bucket. Therefore if the water is not falling at the top position it means mg< mv²/r.
8. A stone of mass 'm' tied to a string of length 'l' is rotated in a circle with the other end of the string as the center. The speed of the stone is 'v'. If the string breaks, the stone will move
(a) towards the center
(b) away from the center
(c) along a tangent
(d) will stop
ANSWER: (c)
Explanation to the Answer →When a particle moves in a circle with a speed v, its instantaneous velocity has a magnitude v and direction along the tangent at that point. The direction of the instantaneous velocity is kept changing by a centripetal force which is tension in the string in this case. As soon as the string breaks, this force disappears and the direction of the instant velocity of the particle can no longer change. So it goes along the tangent to the instantaneous point on the circle.
9. A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the center. If the angular velocity of the turntable is doubled it will just slip at a distance of
(a) 1 cm
(b) 2 cm
(c) 4 cm
(d) 8 cm
ANSWER: (a)
Explanation to the Answer → The friction force to keep it rotating =mω²r. At 4 cm away from the center when it just slips, the static frictional force is at its limiting value. When the angular speed of the turntable is doubled the magnitude of the required centripetal force increases four times. So to keep it to limiting value of friction force, 'r' should be made one fourth. It will be clear as below
m(2ω)².r/4 =mω²r.
That is why it should be placed at 1 cm.
10. A motorcycle is going on an over-bridge of radius R. The driver maintains a constant speed. As the motor cycle is ascending on the over-bridge, the normal force on it
(a) increases
(b) decreases
(c) remains the same
(d) fluctuates
ANSWER: (a)
Explanation to the Answer →An over-bridge is not just an arch shape placed on a horizontal road because it will give a great jerk at the start. So to keep the driving smooth both ends of the over-bridge are kept concave upward. When a vehicle starts to ascend on this concave upward part of the over-bridge it puts extra force on the road in addition to the weight due to movement on the circular part. So the apparent weight and hence the normal force on the vehicle increases.
11. Three identical cars A, B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FB and FC be the normal forces exerted by the cars on the bridges when they are at the middle of the bridges.
(a) FA is maximum of the three forces.
(b) FB is maximum of the three forces.
(c) FC is maximum of the three forces.
(d) FA = FB = FC
ANSWER: (c)
Explanation to the Answer →Normal force on a plane bridge is equal to weight of the car. On a convex upward bridge due to the upward inertial force apparent weight reduces and so is the normal force on the vehicle. But at the middle of concave upward bridge this inertial force is downward making the apparent weight greater. Therefore the normal force is maximum in this case.
12. A train 'A' runs from east to west and another train 'B' of the same mass runs from west to east at the same speed along the equator. 'A' presses the track with a force F1 and 'B' presses the track with a force F2 .
(a) F1 > F2
(b) F1 < F2
(c) F1 = F2
(d) The information is insufficient to find the relation between F1 and F2 .
ANSWER: (a)
Explanation to the Answer → The earth spins from west to east at equator. To maintain the same speed the first train has to press the track more than normal as it goes against the speed of earth's surface while the second train gets help from the earth's movement and it has to press the track less than the normal which explains the answer.
13. If the earth stops rotating, the apparent value of g on its surface will
(a) increases everywhere
(b) decreases everywhere
(c) remains the same everywhere
(d) increases at some places and remain the same at some places.
ANSWER: (d)
Explanation to the Answer → If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear. But at poles it will not change because these places do not rotate in a circle on the earth's surface.
14. A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T1 and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends.
(a) T1 > T2
(b) T2 > T1
(c) T1 = T2
(d) The relation between T1 = T2 depends on whether the rod rotates clockwise or anticlockwise.
ANSWER: (a)
Explanation to the Answer → Taking the rod of uniform mass per unit length = m . Mass beyond a distance r from the pivot = m(L-r). And this mass will be assumed to be concentrated at middle of the (L-r) length ie at CG. So distance of the CG from the pivot = r+(L-r)/2=(L+r)/2. Force due to circular motion on this length = Tension at point r away from the pivot
=m(L-r).ω² .(L+r)/2
=mω²(L²-r²)/2
It is clear from this expression that as r increases Tension in the rod decreases. It explains the answer. See the figure below.
Diagram for tension in a rotating rod at 'r' distance away from the pivot
(Please Note:: Instead of asking this question as Objective Type or Multiple choice, it can be asked as "Prove that/Show that T1 > T2". In that case you can prove it with diagram as shown above.)
15. A simple pendulum having a bob of mass 'm' is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the car is in air is
(a) mg
(b) mg-mv²/r
(c) mg+mv²/r
(d) zero
ANSWER: (d)
Explanation to the Answer → During the jump of the car from the cliff its trajectory is like a projectile and from the maximum height it is freely falling under gravity with an acceleration g downward. Taking it as frame of reference, it is a non-inertial frame of reference. So a pseudo force equal to mg has to be applied to the bob in upward direction which neutralizes the weight and making it zero. So tension in the string is zero.
16. Let θ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is 'm', the tension in the string is mg.cosθ
(a) always
(b) never
(c) at the extreme positions
(d) at the mean position.
ANSWER: (c)
Explanation to the Answer → At any position tension in the string is = mg.cosθ+mv²/r
(Where r is length of pendulum and v is instantaneous velocity of bob)
v is zero only at extreme positions. So tension is mg.cosθ only at extreme positions.
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Links for the chapter -
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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