Title 01A. NEWTON LAWS OF MOTION ( 1 - 14 ).pdf ✅

NISHITH Multimedia India (Pvt.) Ltd., 1 JEE - ADVANCED - VOL - II CURRENT ELECTRICITY NISHITH Multimedia India (Pvt.) Ltd., NEWTON LAWS OF MOTION NEWTON’S LAWS OF MOTION LEVEL – V SINGLE ANSWER QUESTIONS 1. At the instant t = 0 a force F = kt2 (k is a constant) acts on a small body of mass m resting on a smooth horizontal plane. The time, when body leaves the surface is :  F m (A) mgsin k  (B) mg kcos (C) mg k sin  (D) mg ksin 2. At t = 0, force F = ct is applied to a small body of mass m resting on a smooth horizontal plane (c is a constant). The force is at an angle  with the horizontal .The velocity of the body at the moment of its breaking off the plane and the distance travelled by the body up to this moment are m F  A) 2 3 cos / , 2 2 3 2 sin 6 sin mg mg con m s m c c                 B) 2 2 3 cos cos / , 2 2 3 2 sin 6 sin mg m g m s m c c                 C) 2 3 cos sin / , 2 2 3 2 sin 6 cos mg m g m s m c c                 D) 2 2 3 cos sin / , 2 2 3 2 sin 6 cos mg m g m s m c c                 3. In the arrangements shown, the pulleys, strings and springs are weightless and the systems can move freely without friction. The extension of spring in figure 1 is x1 and that in figure 2 is x2 . Then k 5 kg 10 kg k 5 kg 5 kg Fig 1 Fig 2 A) x1 = x2 B) x2 > x1 > 0 C) x1 > x2 = 0 D) x1 > x2 > 0 4. Figure shows a system of four pulleys with two masses mA = 3 kg and mB = 4 kg. At an instant, force acting on block A, if block B is going up at an acceleration of 3 m/s2 and pulley Q is going down at an acceleration of 1 m/s2 is P Q R A B (A) 7 N acting upward (B) 7 N acting downward (C) 10.5 N acting upward (D) 10.5 N acting downward.
Jr Chemistry E/M NEWTON LAWS OF MOTION 2 NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED - VOL - II NISHITH Multimedia India (Pvt.) Ltd., 5. Two masses m and M are attached with strings as shown. For the system to be in equilibrium we have m M 45o  45o (A) 2M tan 1 m    (B) 2m tan 1 M    (C) M tan 1 2m    (D) m tan 1 2M    6. A thin rod AB is moving in a vertical plane. At a certain instant when the rod is inclined at 60o to the horizontal, point A is moving horizontally with 3 m/s while B is moving in the vertical direction. Then velocity of B is B A 60o 3 m/s vB (A) 1 3 m/s (B) 3 m/s (C) 2 3 m/s (D) 3 2 m/s 7. Two blocks of masses m1 and m2 are con- nected by a massless (light) inextensible cord that is passed over a light pulley slid- ing along the smooth sides of a righ-angled wedge of mass m, which rests on a smooth horizontal plane. The horizontal displace- ment of the wedge when m1 slides down by h along the wedge is  1 m m m2 A)  1 2  1 2 h m cos m sin m m m      towards right B)  1 2  1 2 h m cos m sin m m m      towards left C)  1 2  1 2 h m cot m m m m     towards right D)  1 2  1 2 h m m tan m m m     towards right 8. A block is dragged on a smooth plane with the help of a rope, which moves with a velocity v as shown in the figure. The horizontal velocity of the block is m  v (A) v (B) v /sin  (C) v sin  (D) vcos 9. Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a fricitonless horizontal surface with the string held tight so that each mass is at a distance a from the centre P (as shown in the figure). Now, the mid point of the string is pulled vertically upwards withy a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is [2007] P m m F a a (A) 2 2 F a 2m a x  (B) 2 2 F 2m a  x x (C) F 2m a x (D) 2 2 F a 2m  x x
NISHITH Multimedia India (Pvt.) Ltd., 3 JEE - ADVANCED - VOL - II CURRENT ELECTRICITY NISHITH Multimedia India (Pvt.) Ltd., NEWTON LAWS OF MOTION MULTIPLE ANSWER QUESTIONS 10. The string shown in the figure is passing over small smooth pulley rigidly attached to trolley A. If speed of trolley is constant and equal to VA and that of block B at the instant shown in figure is VB , then x=3cm h=4cm A B (A) vB = vA (B) VA = 0 (C) B A 3 v v 5  (D) VB = 0 11. A particle of mass m moving at t = 0 due to a force F = F0 sint where F0 and are con- stant. Then correct statement is/are (A) it will stop first time at   (B) It will travel distance   0 2 F S m during this time (C) During this distance maximum velocity of particle is 0 max F v m   (D) it will stop for first time at 2 /   12. In the figure,. the pulley P moves to the right with a constant speed u. The downward speed of A is vA , and velocity of B is vB . B A u (A) B A v v  (B) A B v v 2u      (C) B A v u v   (D) Both blocks have accelerations of the same magnitude. 13. A reference frame attached to the earth : (1986; 2M) (A) is an inertial frame by definition (B) cannot be an inertial frame because the earth is revolving round the sun (C) is an inertial frame because Newton’s laws are applicable in this frame (D) cannot be an inertial frame because the earth is rotating about its own axis 14. The system shown in figure is in equilibrium and at rest. The spring and string are massless. Now the string is cut. The acceleration of mass 2m and m just after the string is cut will be: A B 2m m (A) Acceleration of 2m is g/2 upwards (B) Acceleration of 2m is g upwards (C) Acceleration of m is g downwards (D) Acceleration of m is g/2 downwards PASSAGE TYPE QUESTIONS PASSAGE : 1 A body of mass m kg  1.8 is placed on an in- clined plane, the angle of inclination is 0   37 , and is attached to the top end of the slope with a thread which is parallel to the slope. Then the plane slope is moved with a horizontal accelera- tion of a.Friction is negligible. 15. The acceleration, if the body pushes the plane with a force of 3 4 mg is : A) 5 2 / 43 m s B) 2 0.5 / m s C) 2 0.75 / m s D) 5 2 / 6 m s
Jr Chemistry E/M NEWTON LAWS OF MOTION 4 NISHITH Multimedia India (Pvt.) Ltd., JEE - ADVANCED - VOL - II NISHITH Multimedia India (Pvt.) Ltd., 16. The tension in thread in the above question is : A) 12 N B) 10 N C) 8 N D) 4 N 17. At what acceleration will the body lose con- tact with plane : A) 40 2 / 3 m s B) 2 7.5 / m s C) 2 10 / m s D) 2 5 / m s Passage II A particle slides down a smooth inclined plane of elevation  fixed in an elevator going with an acceleration a The base of the incline has a length L. L 0 a  18. The acceleration of particle with respect to the incline : (A) gsin (B) a sin 0  (C) 0 (g a )sin   (D) 0 (gsin a cos )    19. The time taken by the particle to reach the bottom : (A) 1/2 2L t gsin         (B) 1 2 0 2L (g a )sin cos          (C) 1/2 0 2L a sin        (D) 1/2 0 2L (g a )sin         20. If the elevator going up with constant veloc- ity, the time taken by the particle to reach the bottom is (A) 1/2 2L gsin cos         (B) 1/2 2L gsin        (C) 1/2 2L gcos        (D) none of these. MATRIX MATCHING QUESTIONS 21. In the diagram strings, springs and the pul- ley are light and ideal. The system is in equi- librium with the strings taut T  0 , match the column. Masses are equal. M A MB C M W Z x y Column - 1 Column- 2 A) Just after string W breaks P) 0 A a  B) Just after spring X breaks Q) 0 B a  C) Just after string Y breaks R) 0 C a  D) Just after spring Z breaks S) B C a a  22. In the situation shown, all surfaces are fric- tionless and triangular wedge is free to move, In column-2, the direction of certain vectors are shown. Match the direction of quantities in Column-1 with possible vector in Column-2.  Column - 1 Column-2 A) acceleration of the block X relative to ground (P)  B) acceleration of block X relative to wedge (Q) C) normal force by block on wedge (R)  D) net force on the wedge (S)