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Vectors Practice Questions - DPP for NEET
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Page 1 1. (d) 90 2 sin 2 vv ° æö D= ç÷ èø 1 2sin4522 2 v vv = °= ´= 22 2 21 60 30 pp =´w=´´ = r cm/s 2. (b) 2 sin 2 vv q æö D= ç÷ èø 10 2 5 sin 45 2 = ´ ´ °= \ v a t D = D 10/21 10 2 == m/s 2 3. (c) For motion of the particle from (0, 0) to (a, 0) $ ˆˆ (0) FKi aj FKaj =- + Þ =- ur ur Displacement ˆˆ ˆˆ ( 0) (0 0) r ai j i j ai = + - += r $ So work done from (0, 0) to (a, 0) is given by $ . .0 W F r Ka j ai = =-= urr $ For motion (a, 0) to (a, a) $ () F K ai aj =-+ ur $ and displacement $ ˆ ˆ ˆˆ ( ) ( 0) r ai aj ai j a j = + - += r So work done from (a, 0) to (a, a) $ $ 2 . ( ). urr $ W F r K ai a j a j Ka = =- + =- So total work done = 2 Ka - 4. (d) q W P T sin T q cos T q As the metal sphere is in equilibrium under the effect of the three forces therefore 0 T PW + += ururuur From the figure cos TW q= ................... (i) sin TP q= ................... (ii) From equation (i) and (ii) we get tan PW =q and 2 22 T PW =+ 5. (b) Time taken to cross the river along shortest possible path is given by 22 = - d t vu v = velocity of boat in still water u = velocity of river water d = width of river 22 151 60 5 \= -u Þ u = 3 km/h 6. (d) Here d = 320 m = 320 km 1000 t = 4 min v = 5 3 u Putting values in 22 = - d t vu , u = 60 m/min 7. (c) 12 sin sin sin150 P QR == q q° Þ 1 1.93 sin sin150 R = q° Þ 1 1.93 sin150 1.93 0.5 1 sin 0.9659 R ´ °´ = == q P Q R 150° 2 q 1 q 8. (b) T T cos 30° T sin 30° 30 N W 30° Page 2 1. (d) 90 2 sin 2 vv ° æö D= ç÷ èø 1 2sin4522 2 v vv = °= ´= 22 2 21 60 30 pp =´w=´´ = r cm/s 2. (b) 2 sin 2 vv q æö D= ç÷ èø 10 2 5 sin 45 2 = ´ ´ °= \ v a t D = D 10/21 10 2 == m/s 2 3. (c) For motion of the particle from (0, 0) to (a, 0) $ ˆˆ (0) FKi aj FKaj =- + Þ =- ur ur Displacement ˆˆ ˆˆ ( 0) (0 0) r ai j i j ai = + - += r $ So work done from (0, 0) to (a, 0) is given by $ . .0 W F r Ka j ai = =-= urr $ For motion (a, 0) to (a, a) $ () F K ai aj =-+ ur $ and displacement $ ˆ ˆ ˆˆ ( ) ( 0) r ai aj ai j a j = + - += r So work done from (a, 0) to (a, a) $ $ 2 . ( ). urr $ W F r K ai a j a j Ka = =- + =- So total work done = 2 Ka - 4. (d) q W P T sin T q cos T q As the metal sphere is in equilibrium under the effect of the three forces therefore 0 T PW + += ururuur From the figure cos TW q= ................... (i) sin TP q= ................... (ii) From equation (i) and (ii) we get tan PW =q and 2 22 T PW =+ 5. (b) Time taken to cross the river along shortest possible path is given by 22 = - d t vu v = velocity of boat in still water u = velocity of river water d = width of river 22 151 60 5 \= -u Þ u = 3 km/h 6. (d) Here d = 320 m = 320 km 1000 t = 4 min v = 5 3 u Putting values in 22 = - d t vu , u = 60 m/min 7. (c) 12 sin sin sin150 P QR == q q° Þ 1 1.93 sin sin150 R = q° Þ 1 1.93 sin150 1.93 0.5 1 sin 0.9659 R ´ °´ = == q P Q R 150° 2 q 1 q 8. (b) T T cos 30° T sin 30° 30 N W 30° 12 DPP/ P 05 From the figure T sin 30° = 30 ...(i) T cos 30° = W ...(ii) By solving equation (i) and (ii) we get 303 WN = and T = 60 N 9. (c) Relative velocity = ˆˆ ˆˆ ˆ ˆ (3 4) (34) 68 i j i j ij + -- - =+ 10. (c) 30° 90° m v r v sin 30° = 1 2 r m v v = Þ 0.5 0.25 22 m r v v= == m/s 11. (a) To cross the river in minimum time, the shift is given by du v . 12. (d) Relative velocity = 10 + 5 = 15 m/s. Time taken by the bird to cross the train = 120 8 15 = sec 13. (d) vr = w´ r urr $ $ $ $ 3 4 1 18 13 2 5 66 i jk i jk = - =- -+ - $ $ 14. (d) | | 3( .) ur u r ur ur A B AB ´= sin 3 cos AB AB q=q Þtan3 q= Þ 60 q=° Now |||| R AB =+ ur ur ur 22 2 cos A B AB = ++q 22 1 2 2 A B AB æö = ++ ç÷ èø = 2 2 1/2 () A B AB ++ 15. (a) $ $ $ $ (7 3 )( 3 5) r F i j k i jk t= ´ = + + - ++ r r ur $$ $ $ $ $ 7 3 1 14 38 16 3 15 i jk i jk t= = -+ - $ r $ 16. (b)| |. ur ur ur ur B AB A´= Þ sin cos AB AB q=q Þ tan1 q= \ 45 q=° 17. (a) .0 PQ = u r ur Þ 2 2 30 aa - -= Þ a = 3 18. (a) 21 u u r ur r S rr =- . W FS = ur ur = $ $ $ $ (4 3).(11 11 15) i j k i jk + + ++ $$ = (4 × 11 + 1 × 11 + 3 × 15) = 100 J 19. (c) ˆˆ ˆ ˆ ˆˆ 3 2 , 35 A i j kB i jk = - + = -+ u r ur , ˆ ˆˆ 24 C i jk = +- ur 2 22 3 ( 2) 1 9 4 1 14 A = + - + = + += r 2 22 1 ( 3) 5 1 9 25 35 B= +- += + += r 222 2 1 ( 4) 4 1 16 21 r C= + +- = ++= As 22 B AC =+ therefore ABC will be right angled triangle. 20. (b) 12 12 . cos FF FF q= rr = ˆˆ ˆ ˆ ˆˆ (5 10 20 ).(10 5 15 ) 25 100 400. 100 25 225 i j k i jk + - -- + + ++ 50 50 300 525. 350 -+ = Þ 1 cos 2 q= 45 \q=° 21. (a) ˆˆ ˆˆ ˆ ˆ ˆˆ 4 32 r abc i j i jk i jk = + + = -- + - = +- r rrr 222 ˆ ˆˆ ˆ || 1 1 ( 1) r i jk r r r +- == + +- ˆ ˆˆ 3 i jk +- = 22. (a) D C A B 300 m 400 m Displacement AC AB BC =+ uuu r uuu r u uu r Page 3 1. (d) 90 2 sin 2 vv ° æö D= ç÷ èø 1 2sin4522 2 v vv = °= ´= 22 2 21 60 30 pp =´w=´´ = r cm/s 2. (b) 2 sin 2 vv q æö D= ç÷ èø 10 2 5 sin 45 2 = ´ ´ °= \ v a t D = D 10/21 10 2 == m/s 2 3. (c) For motion of the particle from (0, 0) to (a, 0) $ ˆˆ (0) FKi aj FKaj =- + Þ =- ur ur Displacement ˆˆ ˆˆ ( 0) (0 0) r ai j i j ai = + - += r $ So work done from (0, 0) to (a, 0) is given by $ . .0 W F r Ka j ai = =-= urr $ For motion (a, 0) to (a, a) $ () F K ai aj =-+ ur $ and displacement $ ˆ ˆ ˆˆ ( ) ( 0) r ai aj ai j a j = + - += r So work done from (a, 0) to (a, a) $ $ 2 . ( ). urr $ W F r K ai a j a j Ka = =- + =- So total work done = 2 Ka - 4. (d) q W P T sin T q cos T q As the metal sphere is in equilibrium under the effect of the three forces therefore 0 T PW + += ururuur From the figure cos TW q= ................... (i) sin TP q= ................... (ii) From equation (i) and (ii) we get tan PW =q and 2 22 T PW =+ 5. (b) Time taken to cross the river along shortest possible path is given by 22 = - d t vu v = velocity of boat in still water u = velocity of river water d = width of river 22 151 60 5 \= -u Þ u = 3 km/h 6. (d) Here d = 320 m = 320 km 1000 t = 4 min v = 5 3 u Putting values in 22 = - d t vu , u = 60 m/min 7. (c) 12 sin sin sin150 P QR == q q° Þ 1 1.93 sin sin150 R = q° Þ 1 1.93 sin150 1.93 0.5 1 sin 0.9659 R ´ °´ = == q P Q R 150° 2 q 1 q 8. (b) T T cos 30° T sin 30° 30 N W 30° 12 DPP/ P 05 From the figure T sin 30° = 30 ...(i) T cos 30° = W ...(ii) By solving equation (i) and (ii) we get 303 WN = and T = 60 N 9. (c) Relative velocity = ˆˆ ˆˆ ˆ ˆ (3 4) (34) 68 i j i j ij + -- - =+ 10. (c) 30° 90° m v r v sin 30° = 1 2 r m v v = Þ 0.5 0.25 22 m r v v= == m/s 11. (a) To cross the river in minimum time, the shift is given by du v . 12. (d) Relative velocity = 10 + 5 = 15 m/s. Time taken by the bird to cross the train = 120 8 15 = sec 13. (d) vr = w´ r urr $ $ $ $ 3 4 1 18 13 2 5 66 i jk i jk = - =- -+ - $ $ 14. (d) | | 3( .) ur u r ur ur A B AB ´= sin 3 cos AB AB q=q Þtan3 q= Þ 60 q=° Now |||| R AB =+ ur ur ur 22 2 cos A B AB = ++q 22 1 2 2 A B AB æö = ++ ç÷ èø = 2 2 1/2 () A B AB ++ 15. (a) $ $ $ $ (7 3 )( 3 5) r F i j k i jk t= ´ = + + - ++ r r ur $$ $ $ $ $ 7 3 1 14 38 16 3 15 i jk i jk t= = -+ - $ r $ 16. (b)| |. ur ur ur ur B AB A´= Þ sin cos AB AB q=q Þ tan1 q= \ 45 q=° 17. (a) .0 PQ = u r ur Þ 2 2 30 aa - -= Þ a = 3 18. (a) 21 u u r ur r S rr =- . W FS = ur ur = $ $ $ $ (4 3).(11 11 15) i j k i jk + + ++ $$ = (4 × 11 + 1 × 11 + 3 × 15) = 100 J 19. (c) ˆˆ ˆ ˆ ˆˆ 3 2 , 35 A i j kB i jk = - + = -+ u r ur , ˆ ˆˆ 24 C i jk = +- ur 2 22 3 ( 2) 1 9 4 1 14 A = + - + = + += r 2 22 1 ( 3) 5 1 9 25 35 B= +- += + += r 222 2 1 ( 4) 4 1 16 21 r C= + +- = ++= As 22 B AC =+ therefore ABC will be right angled triangle. 20. (b) 12 12 . cos FF FF q= rr = ˆˆ ˆ ˆ ˆˆ (5 10 20 ).(10 5 15 ) 25 100 400. 100 25 225 i j k i jk + - -- + + ++ 50 50 300 525. 350 -+ = Þ 1 cos 2 q= 45 \q=° 21. (a) ˆˆ ˆˆ ˆ ˆ ˆˆ 4 32 r abc i j i jk i jk = + + = -- + - = +- r rrr 222 ˆ ˆˆ ˆ || 1 1 ( 1) r i jk r r r +- == + +- ˆ ˆˆ 3 i jk +- = 22. (a) D C A B 300 m 400 m Displacement AC AB BC =+ uuu r uuu r u uu r DPP/ P 05 13 AC = 2 2 22 (AB) (BC) (400) (300) 500m + = += Distance = AB+BC =400+300=700 m 23. (a) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ A 3i 2j k,B i 3j 5k,C 2i j 4k = - + = - + = -+ rr r 2 22 A 3 ( 2) 1 9 4 1 14 = + - + = + += r 2 22 B 1 ( 3) 5 1 9 25 35 = +-+= + + = r 222 C 2 1 ( 4) 4 1 16 21 = + +- = ++= uv As 22 B AC =+ therefore ABC will be right angled triangle. 24. (a)AB0sin 00 ´ = \ q= \q= ° r r Two vectors will be parallel to each other. 25. (a), 26 (b), 27. (b) ˆ ˆˆ 2 11 1 11 r r i jk AB ´= = ˆ ˆˆ ( 1 1) (2 1) (2 1) i jk -- -+ - = ˆ ˆ jk -+ Unit vector perpendicular to A r and B r is ˆ ˆ 2 jk æö -+ ç÷ èø . Any vector whose magnitude is k (constant) times ˆ ˆˆ (2) i jk ++ is parallel to A r so, unit vector ˆ ˆˆ 2 6 i jk ++ is parallel to A r . 28. (b) A B AB + =- rr rr Þ A 2 + B 2 + 2AB cosq = A 2 + B 2 + 2AB cosq Hence cos q = 0 which gives q = 90° Also vector addition is commutative. Hence A B BA +=+ rr rr 29. (a) Let P r and Q r are two vectors in opposite direction, then their sum () P Q PQ +- =- rr rr If PQ = r r then sum equal to zero. 30. (d) The resultant of two vectors of unequal magnitude given by 22 2 cos R A B AB = ++q cannot be zero for any value of q.Read More
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FAQs on Vectors Practice Questions - DPP for NEET
| 1. What are vectors in physics? |  |
Ans. In physics, vectors are mathematical quantities that have both magnitude and direction. They are used to represent physical quantities such as displacement, velocity, and force.
| 2. How are vectors represented? | |
Ans. Vectors are represented using arrows. The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector.
| 3. What is the difference between scalars and vectors? | |
Ans. Scalars are physical quantities that have only magnitude, such as time, temperature, and mass. Vectors, on the other hand, have both magnitude and direction. For example, speed is a scalar quantity, while velocity (speed with direction) is a vector quantity.
| 4. What is the importance of vectors in physics? | |
Ans. Vectors are essential in physics as they allow us to accurately describe and analyze physical phenomena. They provide a mathematical framework to understand the motion, forces, and other quantities in the physical world.
| 5. How are vectors added or subtracted? | |
Ans. Vectors can be added or subtracted using the head-to-tail method. To add two vectors, place the tail of the second vector at the head of the first vector, and the resultant vector is drawn from the tail of the first vector to the head of the second vector. To subtract vectors, add the negative of the vector to be subtracted.
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