VECTORS PREVIOUS VECTORS VECTORS PREVIOUS OBJECTIVE TYPE QUESTIONS
VECTORS - PREVIOUS VECTORS
VECTORS - PREVIOUS OBJECTIVE TYPE QUESTIONS PREVIOUS QUESTIONS
VECTORS - PREVIOUS 1) 0 Solution: ∴cos∆�� =1 (∴sin∆�� =∆�� )
VECTORS - PREVIOUS 2. A uniform sphere of weight W and radius 5 cm is being held by a string as shown in the figure. The tension in the string will be : [ J. M. O. L – 2013] 8 cm Solution:
VECTORS - PREVIOUS 1) 10 units Solution: V = u + at V = (3 i + 4 j) + (0. 4 i + 0. 3 j) 10 v = 3 i + 4 j + 4 i + 3 j v = 7 i + 7 j 3) 7 units 4) 8. 5 units
VECTORS - PREVIOUS
VECTORS - PREVIOUS Solution: 450 Fl sin = mg (1 -cos ) T F sin 450 = mg (1 -cos 450) mg KEY: 1
VECTORS - PREVIOUS z Solution: Torque = r x F x O y
VECTORS - PREVIOUS 6. An particle is moving east wards with a velocity of 5 ms-1. In 10 sec, the velocity changes to 5 ms-1 north wards. The average acceleration in this time is [AIEEE – 2005] 3) zero
VECTORS - PREVIOUS Solution: N NW S 5 KEY: 4 E
VECTORS - PREVIOUS 1) 3) /2 2) /4 4) /3 Solution: B A 1800 2 ABsin�� =0
VECTORS - PREVIOUS Solution: Conceptual
VECTORS - PREVIOUS 9. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N and perpendicular to the smaller force. Then the magnitude of the forces are [AIEEE-2002] 1) 12 N, 6 N 2) 13 N, 5 N 3) 10 N, 9 N 4) 16 N, 2 N Solution: Let P, Q are two forces R is resultant, P < Q Given P + Q = 18 . . . (1) Solving (1) & (2) P = 13 N, Q = 5 N R
VECTORS - PREVIOUS 1) –F 1/m 2) –F 1 F 3/m. F 1 3) (F 2 – F 3) /m 4) F 2/m
VECTORS - PREVIOUS Solution: F 3 F 1 + F 2 + F 3 = 0 F 2 + F 3 = - F 1 F 2 F 1 or KEY: 1
VECTORS - PREVIOUS COMPETITIVE QUESTIONS EAMCET
VECTORS - PREVIOUS 1. Sum of magnitudes of two forces is 25 N. The resultant of these forces is normal to the smaller force and has a magnitude of 10 N. Then the force are [TS E – 2015] 1) 14. 5 N, 10. 5 N Solution: 2) 16 N, 9 N 3) 13 N, 12 N 4) 20 N, 5 N
VECTORS - PREVIOUS Solution: The horizontal velocity of rain = velocity of running man = 5 kmph vertical velocity of rain = V kmph
VECTORS - PREVIOUS KEY: 4
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS 4) Mg Solution: F = Mg tan��
VECTORS - PREVIOUS 1) 1 unit 2) 9 units Solution: �� = (2 i – j - k). (3 i + 2 j – 5 k) �� =6– 2+5 �� =9 3) 13 units 4) 60 units
VECTORS - PREVIOUS 7. Sum of magnitudes of two forces acting at a point is 16 N. If their resultant is normal to smaller force, and has a magnitude 8 N, then the forces are [E – 2012] 1) 6 N, 10 N 2) 8 N, 8 N 3) 4 N, 12 N Solution: Given P + Q = 16 Q–P=4 ∴ P = 6 N, Q = 10 N Q R P 4) 2 N, 14 N
VECTORS - PREVIOUS 8. Displacement of a body (5 i + 3 j – 4 k)m when a force (6 i + 6 j + 4 k) n acts for 5 sec the power in watt is [M – 2012] 2) 9. 6 3) 6. 4 1) 16 4) 3. 2 Solution: P = 6. 4 watt
VECTORS - PREVIOUS 9. A certain vector in the xy plane has an x- component of 4 m and a y – component of 10 m. It is then rotated in the xy plane so that its xcomponent is doubled. Then its new y- component is (approximately): [E – 2011] 3) 5. 0 m 4) 4. 5 m 1) 20 m 2) 7. 2 m Solution: = 52 y = 7. 2 m
VECTORS - PREVIOUS 1) Cos 4) 1
VECTORS - PREVIOUS Solution: T 2 900 1 KEY: 3
VECTORS - PREVIOUS 4) Zero Solution:
VECTORS - PREVIOUS 1) /4 2) /2 3) 2 Solution: A=B /2 4) 0
VECTORS - PREVIOUS Solution: To keep the body in equilibrium
VECTORS - PREVIOUS KEY: 1
VECTORS - PREVIOUS 1) 3 2) 1. 5 3) -1. 5 4) -3 Force = ma Solution: From FLR vet So a and v must be perpendicular = 900 B
VECTORS - PREVIOUS C. s = 900 (2 i + cj). (3 i + 4 j) = 0 6+4 C = 0 4 C = 6 C = 3/2 KEY: 2
VECTORS - PREVIOUS Solution: = i – 2 j – 3 k + 4 i – 2 j + 6 k = 5 i – 4 j – 3 k
VECTORS - PREVIOUS 1) 3 Solution: 2)4 3) 5 4)6
VECTORS - PREVIOUS 17. Of the vectors given below, the parallel vectors are [M-2006] Solution: Conditions for parallel vectors Hence, A & C are correct
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS Solution: Q=8 P
VECTORS - PREVIOUS
VECTORS - PREVIOUS Solution: r 1 = 4 i – 4 j + 7 k, t = 10 sec r 2 = 2 i + 2 j + 5 k, v 1 = 0. 4 i v 2 = ? Here distance travel is same from given position r 1 + v 1 t = r 2 + v 2 t (4 i – 4 j + 7 k) + (0. 41)10 = (2 i + 2 j + 5 k) + v 2(10)
VECTORS - PREVIOUS = 6 i - 4 j + 7 k = 2 i + 2 j + 5 k – v v 2 = 0. 6 i – 0. 6 j + 0. 2 k KEY: 2
VECTORS - PREVIOUS
VECTORS - PREVIOUS Solution: Resultant of two perpendiculars is opposite to third particle 3 m/s 1 kg Res 1 kg KEY: 2 2 m/s 3 kg (rest)
VECTORS - PREVIOUS 1) 2 Solution: 2) 4 3) 6 4) 8
VECTORS - PREVIOUS 23. A boat which has a speed of 13 kmph in still water crosses a river of width 1 km along the shortest possible path in 12 minute. The velocity of the river water in kmph is [M-2002] 1) 12 2) 10 3) 8 Solution: Vb = boat velocity Vr = river velocity 4) 6
VECTORS - PREVIOUS Vr = 12 kmph KEY: 1
VECTORS - PREVIOUS −
VECTORS - PREVIOUS Solution: KEY: 2
VECTORS - PREVIOUS 25. An electron moves with speed 2 x 105 ms-1 along the positive X – direction in the presence of a magnetic induction (in tesla). The magnitude of the force experienced by the electron in newton is (charge of the electron = 1. 6 x 10 -19 C) [E-2001] 1) 1. 18 x 10 -13 2) 1. 28 x 10 -13 3) 1. 6 x 10 -13 4) 1. 72 x 10 -13 Solution:
VECTORS - PREVIOUS KEY: 3
VECTORS - PREVIOUS Solution: = 4 i +3 j + 6 k + (-j + 3 j- 8 k) = 4 i +3 j + 6 k – i + 3 j – 8 k
VECTORS - PREVIOUS Unit vector parallel to it is KEY: 1
VECTORS - PREVIOUS 1) Q(ae 1 + be 2)
VECTORS - PREVIOUS Solution: W = Q (e 1 i + e 2 j + e 3 k). (ai + bj) W = Q (e 1 a + e 2 b) KEY: 1
VECTORS - PREVIOUS Solution: C 2 = C 2 + 2 C 2 cos C 2 = 2 C 2[1 + cos ] = 1200
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS 3 2 KEY: 4
VECTORS - PREVIOUS 30. A particle is moving eastwards with a velocity 15 ms-1. Suddenly it moves towards north and moves with the same speed in time 10 sec. The average acceleration during this time is N Solution: 15 NW W N 15 E
VECTORS - PREVIOUS 31. Force is 6 i+Cj-2 k and displacement i+2 j+6 k. If the work done is 6 J, the value of ‘C’ is [E-1997] 2) 4 1) 6 3) 8 4) 9 Solution: W = F. S 6 = (6 i + Cj – 2 k). (i + 2 j + 6 k) 6 = 6 + 2 C - 12 12 = 2 C C=6
VECTORS - PREVIOUS 1) 2 Solution: Length in x – y plane means
VECTORS - PREVIOUS 1) Equal to (a+b) 2) Less than (a+b) 3) Greater than (a+b) 4) Not greater than (a+b) Solution: or So, not greater than (a + b)
VECTORS - PREVIOUS 34. A boat is moving with a velocity 3 i+4 j w. r. t. the ground. The water in the river is moving with a velocity 3 i-4 j. The relative velocity of boat w. r. t. water is [E-1991] 1) 6 i+8 j 2) -6 i-8 j Solution: = (3 i + 4 j) – (3 i – 4 j) = 0 i + 8 j V = 8 j 3) 8 j 4) 6 j
VECTORS - PREVIOUS 1) 3 ms-1 3) 5 ms-1 2) 4 ms-1 4) 6 ms-1 Solution: VR = 530 Vm
VECTORS - PREVIOUS KEY: 2
VECTORS - PREVIOUS 36. Wind is blowing to east along two parallel railway tracks. Two train moving with the same speed in opposite direction have the steam track of one double that of the other. The speed of each train is [1987] 1) Equal to that of the wind 2) Three times that of the wind 3) Double that of the wind 4) Half of that of the wind
VECTORS - PREVIOUS Solution: Vtrain Vwind Vtrain V + W = 2(V - W) V + W = 2 V - 2 W 3 W = V KEY: 2
VECTORS - PREVIOUS Solution: From Snell’s law
VECTORS - PREVIOUS 1) 3 2) 4 Solution: Perpendicular means (ai + aj + 3 k). (ai – 2 j – k) = 0 a 2 – 2 a -3 = 0 (a – 3) (a+1) = 0 a = 3 or a = -1 3) 9 4) 13
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS 40. The resultant of two forces, on double the other in magnitude is perpendicular to the smaller of the two forces. The angle between the two forces is [KET – 2002] Solution: R 2 = 600 180 - = 180 – 60 = 1200 1 180 -
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS 1) = 00 2) = 300 3) = 900 4) = 1800 Solution: Px = 2 cost P = P xi + P y j Py = 2 sin t
VECTORS - PREVIOUS F = F xi + F y j F. P = 0 F. P = F. P cos = 0 = 900 KEY: 3
VECTORS - PREVIOUS 43. Person aiming to reach the exactly opposite point on the bank of a stream is swimming with a speed of 0. 5 ms-1 at an angle of 1200 with the direction of flow of water. The speed of water in the stream is [DPMT – 2000] 2) 0. 67 m/s 1) 1 m/s 3) 0. 433 m/s 4) 0. 25 m/s Solution: 300 900
VECTORS - PREVIOUS 44. The angle which the velocity vector of projectile thrown with a velocity ‘v’ at an angle ‘ ’ to the horizontal will make with the horizontal after time ‘t’ of its being thrown up is [CPMT 98] 1) Solution: 2) tan-1( /t)
VECTORS - PREVIOUS Solution:
VECTORS - PREVIOUS = i(6 - 8) – j(-3) + k(4) = -2 i + 3 j + 4 k KEY: 1
VECTORS - PREVIOUS Thank you…
- Slides: 79