Tiêu đề 03.MOTION IN A PLANE.pdf ✅

Objective Physics Volume-I 345 YCT 03. Motion in Plane (a) Scalar and Vector Quantity 1. If A, B are perpendicular vectors ˆ ˆ ˆ A = 5i + 7j- 3k ˆ ˆ ˆ B = 2i + 2j- ck . The value of c is (a) –2 (b) 8 (c) –7 (d) –8 EAMCET-1991 Ans. (d) : A 5i 7j 3k = + − ˆ ˆ ˆ B 2i 2j ck = + − ˆ ˆ ˆ For perpendicular vectors A B 0 ⋅ = ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ 5i 7j 3k 2i 2 j ck 0 + − ⋅ + − = 10 14 3c 0 + + = 24 = – 3c c = –8 2. The resultant of the vectors A and B depends also on the angle θ between them. The magnitude of the resultant is always given by (a) A B 2ABcos + + θ (b) ( ) A B 2ABcos + + θ (c) 2 2 A B 2ABcos + + θ (d) ( ) 2 2 2 A B 2ABcos + + θ EAMCET-1992 Ans. (c) : Resultant vector, R From ∆DOF, (OD)2 =(OF)2 + (DF)2 (OD)2 = (OE + EF)2 + (DF)2 2 R = ( A + B cosθ) 2 + ( B sinθ) 2 = A2 + B2 cos2 θ + 2AB cosθ + B2 sin2 θ = A2 + B2 (cos2 θ + sin2 θ) + 2AB cosθ 2 R = A2 + B2 + 2AB cosθ 2 2 R A B 2ABcos = + + θ 3. A and B are vectors such that A + B = . A - B Then, the angle between them is (a) 90o (b) 60o (c) 45o (d) 0o EAMCET-1993 Ans. (a) : Given, | A B | | A B | + = − Squaring both sides, 2 2 | A B | | A B | + = − 2 2 2 2 A B 2A B A B 2A B + + ⋅ = + − ⋅ 4A B 0 ⋅ = A B 0 ⋅ = | A || B | cos 0 θ = cos 0 θ = θ = ° 90 4. When two vectors A and B of magnitude a and b are added, the magnitude of the resultant vector is always (a) equal to (a + b) (b) less than (a + b) (c) greater than (a + b) (d) not greater than (a + b) EAMCET-1993 Ans. (d) : Given, | A | a, B b = = 2 2 | A B | a b 2abcos + = + + θ 2 2 max | A B | a b 2ab + = + + [For max, θ = 0] | A B | a b + = + max ( ) Hence, magnitude of resultant vector is not greater than (a + b) 5. If a unit vector is represented by ˆ ˆ ˆ 0.5i + 0.8j+ ck , the value of c is (a) 1 (b) 0.11 (c) 0.011 (d) 0.39 TS-EAMCET-10.09.2020, Shift-1 EAMCET-1994 Ans. (b) : A 0.5i 0.8j ck ˆ = + + ˆ ˆ ˆ ˆ ∴ = | A | 1 2 2 2 0.5 0.8 c 1 + + = 2 c 0.11 = c 0.11 =
Objective Physics Volume-I 346 YCT 6. The angle made by the vector ˆ ˆ A = i + j with x- axis is (a) 90o (b) 45o (c) 22.5o (d) 30o EAMCET-1996 Ans. (b) : Given that, A i j = +ˆ ˆ 2 2 | A | 1 1 2 = + = Ax = 1, Ay = 1 If θ is the angle made by the vector with x–axis than, Ax 1 cos cos | A | 2 θ θ = ⇒ = θ = ° 45 7. The angle between two vectors ˆ ˆ ˆ 6i + 6j- 3k and ˆ ˆ ˆ 7i + 4j+ 4k is given by (a) 1 1 cos 3 −       (b) 1 5 cos 3 −       (c) 1 2 sin 3 −       (d) 1 5 sin 3 −         EAMCET-1999 Ans. (d) : Given that, A 6i 6j 3k = + − ˆ ˆ ˆ B 7i 4j 4k = + + ˆ ˆ ˆ A B | A || B | cos ⋅ = θ ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ 6i 6j 3k 7i 4 j 4k + − ⋅ + + 36 36 9 49 16 16 cosθ = + + ⋅ + +       42 24 12 81 81.cos 9 9cos + − = = × θ θ 54 81cos = θ 54 cos 81 θ = 6 2 cos 9 3 θ = = 2 2 2 2 sin 1 cos 1 3 θ θ   = − = −        2 4 5 sin 1 9 9 θ = − = 5 sin 3 θ = 1 5 sin 3 θ − = 8. The component of vector ˆ ˆ ˆ A = a i + a j + a k x y z along the direction of ˆ ˆ i - j is (a) ax – ay + az (b) ax – ay (c) (a a / 2 x y − ) (d) (ax + ay + az) EAMCET-2008 Ans. (c) : Given, x y z A a i a j a k ˆ ˆ ˆ = + + Let B i j ˆ ˆ = − Component of vector A along any vector B A.B B = Component of vector A = x y z ˆ ˆ ˆ a i a j a k + + along ( ) B i j ˆ ˆ = − ( ) ( ) ( ) ( ) x y z 2 2 ˆ ˆ ˆ ˆ ˆ a i a j a k . i j 1 1 + + − = + − x y a a 2 − = 9. For vectors A and B making an angle θ which one of the following relations is correct? (a) A B B A × = × (b) A B ABsin × = θ (c) A B ABcos × = θ (d) A B B A × = − × DCE-2009 Ans. (d) : We know that, Cross product of vectors A and B A B ABsin × = θ Cross product of vectors B and A B A BAsin × = − θ So, A B B A × = − × 10. Given two vectors A = i + 2j 3k − − ˆ ˆ ˆ and B = 4i 2j + 6k ˆ ˆ − ˆ . The angle made by (A + B) with x-axis is (a) 30° (b) 45° (c) 60° (d) 90° AP EAMCET(Medical)-2007 Ans. (b) : A i 2j 3k = − + − ˆ ˆ ˆ B 4i 2j 6k ˆ ˆ ˆ = − + A B ( i 2j 3k) (4i 2j 6k) ˆ ˆ ˆ ˆ ˆ ˆ + = − + − + − + A B 3i 0j 3k + = + + ˆ ˆ ˆ α is angle with x–axis cosα = x component of A B ( ) | A B | − + + cosα = 3 3 9 0 9 3 2 = + + cosα = 1 2 α = 45o 11. Of the vectors given below, the parallel vectors are, ɵ ɵ → → → → ɵ ɵ ɵ ɵ ɵ ɵ ɵ A = 6i + 8j B = 210i + 280k C = 5.1i + 6.8j D = 3.6i + 6j+ 48k
Objective Physics Volume-I 347 YCT (a) A → and B → (b) A → and C → (c) A → and D → (d) C → and D → AP EAMCET(Medical)-2006 Ans. (b) : If component of vector is same then vectors will be same. A 6i 8j ˆ ˆ = + B 210i 280k ˆ ˆ = + C 5.1i 6.8j ˆ ˆ = + D 3.6i 6j 48k ˆ ˆ ˆ = + + ( ) 1.7 C 5.1i 6.8j 6i 8j ˆ ˆ ˆ ˆ 2 = + = + ∵ Hence, it is clear that A and C are parallel and we can write as, 1.7 C A 2 = This implies that A is parallel to C. 12. A vector Q which has a magnitude of 8 is added to the vector P , which lies along the X- axis. The resultant of these two vectors is a third vector R , which lies along the Y-axis and has a magnitude twice that of P . The magnitude of P is: (a) 6 5 (b) 8 5 (c) 12 5 (d) 16 5 AP EAMCET(Medical)-2004 Ans. (b) : Given, Q = 8 units R = 2P 2 2 2 Q R P = + 2 2 2 (8) (2P) P = + 2 2 = + 4P P 2 64 5P = 2 64 8 P P 5 5 = ⇒ = 13. Angle (in rad) made by the vector ˆ ˆ 3 i j + with the X-axis: (a) 6 π (b) 4 π (c) 3 π (d) 2 π AP EAMCET(Medical)-2005 Ans. (a): A 3 i j ˆ ˆ = + Angle with x–axis : cosθ = Ax | A | cosθ = ( ) ( ) 2 2 3 3 2 3 1 = + θ = 30° 6 π θ = 14. The unit vector parallel to resultant of the vectors A 4i 3j 6k ˆ ˆ ˆ → = + + and B i 3j 8k ˆ ˆ ˆ = − + − is: (a) ( ) 1 ˆ ˆ ˆ 3i 3j 2k 7 + − (b) ( ) 1 ˆ ˆ ˆ 3i 6j 2k 7 + − (c) ( ) 1 ˆ ˆ ˆ 3i 6j 2k 49 + − (d) ( ) 1 ˆ ˆ ˆ 3i 6j 2k 49 − + AP EAMCET(Medical)-2000 Ans. (b) : A 4i 3j 6k = + + ˆ ˆ ˆ B i 3j 8k = − + − ˆ ˆ ˆ R A B (4i 3j 6k) ( i 3j 8k) = + = + + + − + − ˆ ˆ ˆ ˆ ˆ ˆ R 3i 6j 2k = + − ˆ ˆ ˆ unit vector, R Rˆ | R | = 2 2 2 | R | 3 6 2 = + + = + + 9 36 4 = 49 | R | 7 = ˆ ˆ ˆ 3i 6j 2k Rˆ 7 + − = ( ) 1 R 3i 6j 2k ˆ ˆ ˆ ˆ 7 = + − 15. Pressure is a scalar quantity because (a) it is the ratio of force to area and both force and area are vector quantities (b) it is the ratio of magnitude of force to area (c) it is the ratio of component of the force normal to the area (d) it depends on the size of the area chosen SCRA-2015 Ans. (c) : Pressure is a scalar quantity because pressure is the ratio of normal force to the area and the direction of force is not required. 16. The component of a vector r along X-axis will have a maximum value, if : (a) r is along positive X-axis (b) r is along positive Y-axis (c) r is along negative Y-axis (d) r makes an angle of 450 with the X-axis Karnataka CET-2016
Objective Physics Volume-I 348 YCT Ans. (a) : rx = r cosθ rx is maximum when θ = 0 r will be along positive X-axis for maximum value. 17. Which of the following is not a vector quantity? (a) Weight (b) Nuclear spin (c) Momentum (d) Potential energy Karnataka CET-2014 Ans. (d) : Weight, Nuclear spin and Momentum are vector quantities because they have both magnitude as well as direction, whereas potential energy has magnitude only but no direction, thus it is a scalar quantity. 18. Two equal forces (P each) act at a point inclined to each other at an angle of 1200 . the magnitude of their resultant is : (a) P/2 (b) P/4 (c) P (d) 2P Karnataka CET-2004 Ans. (c) : Given, Q = P, θ = 120° ∵ 2 2 R P Q 2PQcos = + + θ = 2 2 P P 2P Pcos120 + + × ° = 2 2 1 P P 2P P P 2   + − × × =     19. The resultant of two forces 3P and 2P is R. If the first force is doubled then the resultant is also doubled. The angle between the two forces is : (a) 900 (b) 1800 (c) 600 (d) 1200 Karnataka CET-2001 Ans. (d) : Case – I ( ) ( ) 2 2 2 R 3P 2P 12P cos = + + θ R P 13 12cos = + θ Case -II ( ) ( ) 2 2 2 R 6P 2P 24P cos 1 = + + θ = + θ P 40 24cos From case (I) and case (II), R1 40 24cos 2 R 13 12cos + θ = = + θ [∵R 2R 1 = ] 40 + 24 cosθ = 52 + 48 cosθ 24 cosθ = –12 cosθ = 1 2 − ⇒ 1 1 cos 2 −   θ = −    θ= 1200 20. Three vectors satisfy the relation A.B = 0 and A.C = 0 then A is parallel to : (a) C (b) B (c) B C× (d) B.C JCECE-2013 COMEDK 2013 Karnataka CET-2003 Ans. (c) : A.B 0 A B = ⇒ ⊥ A.C 0 A C = ⇒ ⊥ A is perpendicular to both B and C and B C× is also perpendicular to both B and C . Therefore, A is parallel to B C× 21. Three forces F1, F2 and F3 together keep a body in equilibrium. If F1 = 3 N along the positive x- axis, F2 = 4 N along the positive y-axis, then the third force F3 is (a) 5N making an angle 1 3 tan 4 −   θ =     with negative y-axis (b) 5N making an angle 1 4 tan 3 −   θ =     with negative y-axis (c) 7N making an angle 1 3 tan 4 −   θ =     with negative y-axis (d) 7N making an angle 1 4 tan 3 −   θ =     with negative y-axis J&K CET- 2010 Ans. (c) : F1, F2, F3 keep a body in equilibrium then resultant of force, ΣF = 0 F1 + F2 +F3 = 0 3 + 4 + F3 = 0 F3 = – 7N Magnitude of F3 = 7N θ is angle made with-Y axis tanθ = 3 4 θ = tan–1 3/4 F3 make angle with negative y-axis. 22. Magnitudes of four pairs of displacement vectors are given. Which pair of displacement vectors, under vector addition, fails to give a resultant vector of magnitude 3 cm ? (a) 2 cm, 7 cm (b) 1 cm, 4 cm (c) 2 cm, 3 cm (d) 2 cm, 4 cm J&K CET- 2009