Title Vector Varsity Daily-2 (Set-B) Solution.pdf ✅

1 Daily-02 [Set-B (Solve Sheet)] wm‡jevm: †f±i c~Y©gvb: 30 †b‡MwUf gvK©: 0.25 mgq: 20 wgwbU 1. P  = M   N  Ges R  = N   M  n‡j, P  I R  Gi ga ̈eZ©x †KvY KZ? [If P  = M   N  and R  = N   M  , what is the angle between P  and R  ?] 0 90 30 180 DËi: 180 e ̈vL ̈v: P  N  M  M  N  R  †h‡nZz P  , M  I N  Dci j¤^ I R  †f±i P  †f±‡ii wecixZ w`‡K Aew ̄’Z Ges M  I N  Gi Dci j¤^|   = 180 2. P   Q  n‡j, |(P ) |   Q   P  = ? [If P   Q  |(P ) |   Q   P  = ?] 0 P  |P|  P 2Q DËi: P 2Q e ̈vL ̈v: |P |   Q  = PQ sin90 = PQ |(P ) |   Q   P  = |P|  |Q|  |P|  sin90 = P2Q 3. `ywU †f±i A  = 2i  + 4j  – 2k  I B  = 4i  – aj  – 4k  | a Gi gvb KZ n‡j, †f±iØq mgvšÍivj n‡e? [Given vectors A  = 2i  + 4j  – 2k  and B  = 4i  – aj  – 4k  , for what value of a will the two vectors be parallel?] 4 –4 –8 8 DËi: –8 e ̈vL ̈v: A  I B  †f±iØq mgvšÍivj n‡j, Zv‡`i X, Y I Z A‡ÿi GKK †f±‡ii mn‡Mi AbycvZ mgvb n‡e| 2 4 = 4 –a = –2 –4  1 2 = 4 –a = 1 2  4 –a = 1 2  a = – 8 4. P   B  = C  , C  †f±iwU B  †f±‡ii mv‡_ KZ †KvY Drcbœ K‡i? [If P   B  = C  , what angle does the vector C  make with the vector B  ?] 90 45 0 30 DËi: 90 e ̈vL ̈v: C  B  P  C  †f±iwU P  I B  Df‡qi mv‡_ j¤^| C  †f±iwU P  I B  Gi cÖwZwUi mv‡_ j¤^ †Kbbv Zv Dfq †f±‡ii j¤^Z‡j Av‡Q| 5. NÈvq 40 km †e‡M DËi w`‡K Pjgvb GKwU Mvwoi PvjK NÈvq 30 km †e‡M GKwU UavK‡K cwðg w`‡K Pj‡Z †`Lj| UavKwUi cÖK...Z †eM KZ? [A car is moving north at a speed of 40 km/h. The driver sees a truck moving west at a speed of 30 km/h. What is the actual velocity of the truck?] 70 kmh–1 75 kmh–1 50 kmh–1 78 kmh–1 DËi: 50 kmh–1 e ̈vL ̈v: N W E S vT  vc  vT  v  TC
2 VT = V 2 TC + V2 C = 402 + 302 = 50 kmh–1 GLv‡b, Mvwoi cÖK...Z †eM, VC = 40 kmh–1 Mvwoi mv‡c‡ÿ Uav‡Ki †eM, VTC = 30 kmh–1 Uav‡Ki cÖK...Z †eM, VT = ? 6. GKwU b`x‡Z † ̄av‡Zi †eM 18ms–1 Ges †bŠKvi †eM 9 ms–1 n‡j, †mvRvmywR Aci cv‡o †cuŠQv‡Z n‡j † ̄av‡Zi mv‡_ KZ †Kv‡Y †bŠKv Pvjv‡Z n‡e? [A river flows at a speed of 18 m/s, and a boat can travel at a speed of 9 m/s in still water. At what angle should the boat be steered with respect to the current to reach the opposite bank directly?] 120 126.86 None of these 135.33 DËi: None of these e ̈vL ̈v: †h‡nZz, † ̄av‡Zi †eM  †bŠKvi †eM  KLbB †bŠKvwU Aci cv‡o †cuŠQv‡Z cvi‡e bv| KviY, cos–1     – † ̄av‡Zi †eM †bŠKvi †eM = cos–1     – 18 9 = cos–1 (– 2) = cos–1 (– 2) Gi gvb Awb‡Y©q| 7. `yBwU mggv‡bi †f±i GKwU we›`y‡Z wμqvkxj| G‡`i jwäi gvb †h‡Kv‡bv GKwU †f±‡ii gv‡bi mgvb n‡j, ga ̈eZ©x †KvY KZ? [Two vectors of equal magnitude act at a point. If the magnitude of their resultant is equal to the magnitude of one of the vectors, what is the angle between them?] 90 120 45 0 DËi: 120 e ̈vL ̈v: P = Q = R n‡j, R = P 2 + Q2 + 2PQ cos  P = P 2 + P2 + 2P2 cos  P 2 = 2P2 + 2P2 cos  –2P2 cos = P2  cos = P 2 – 2P2 = – 1 2 = cos 120   = 120 (Ans.) 8. `ywU †f±‡ii †hvMdj I we‡qvMd‡ji gvb ci ̄úi mgvb n‡j, †f±iØq ci ̄ú‡ii Kx n‡e? [If the magnitude of the sum and difference of two vectors are equal, what is the relationship between the vectors?] mgvšÍivj (Parallel) j¤^ (Perpendicular) wecixZ (Antiparallel) †Kv‡bvwU bq DËi: j¤^ (Perpendicular) e ̈vL ̈v: awi, `ywU †f±i A  , B  Gi ga ̈eZ©x †KvY  |A |  + B  = |A |  – B  A  B    A 2 + B2 + 2AB cos = A2 + B2 – 2ABcos [eM© K‡i|]  2AB cos + 2ABcos = 0  4ABcos = 0  cos = 0 4AB  cos = 0  cos = cos90   = 90   =  2 (j¤^) 9. V  = 3xi  – 8yj  + 3zk  †f±i †ÿÎwUi WvBfvi‡RÝ Gi Rb ̈ wb‡¤œi †KvbwU mwVK n‡e? [Given the vector field V  = 3xi  – 8yj  + 3zk  what is the divergence of V  ?]   .V  = 0 DËi: e ̈vL ̈v:   . V  =      x i  +  y j  +  z k  . (3xi )  – 8yj  + 3zk  = i  .i   x 3x – j  .j   y 8y + k  .k   z 3z = 3 – 8 + 3     i  . i  = 1  x x n = nxn–1 = –2 (i)   . V  positive n‡j, Flask wbM©Z n‡e| (ii)   . V  Negative n‡j, Flask AvMZ n‡e| (iii)   . V  = 0 n‡j, AvMZ I wbM©Z Flask mgvb n‡e|
3 10. A  = i  + 2j  + 5k  I B  = i  – 5j  – 2k  n‡j, †KvbwU mwVK? [If A  = i  + 2j  + 5k  and B  = i  – 5j  – 2k  Which one is right?] (A )  + B   (A )  – B  = 0i  – 28j  + 14k  (A )  + B  . (A )  – B  = 0 A  + B  = 4i  + 3k  , A  – B  = 2i  + 2j  †Kv‡bvwUB bq DËi: (A )  + B  . (A )  – B  = 0 e ̈vL ̈v: (A )  + B  (A )  – B  = (2i )  – 3j  + 3k  (7j )  + 7k  = – 21 + 21 = 0 11. 3N I 4N gv‡bi `ywU e‡ji jwä wb‡Pi †KvbwU n‡Z cv‡i bv? [If two forces of magnitude 3N and 4N act on a point, which of the following cannot be the magnitude of their resultant force?] 7N 1N 2N None of these DËi: None of these e ̈vL ̈v: P  = 3N Q  = 4N Rmax = = P + Q = 7N Rmin = P  Q = 1N Avgiv Rvwb, Rmin  R  Rmax  jwäi gvb 1N n‡Z 7N Gi g‡a ̈ †h‡Kvb wKQz n‡Z cv‡i| 12. `yBwU KYv 3 ms–1 †e‡M I 5 ms–1 †e‡M 120 †KvY Drcbœ K‡i †Kv‡bv GKwU we›`y‡K AwZμg Kij| 2s ci Zv‡`i ga ̈Kvi `~iZ¡ KZ? [Two particles move with velocities of 3 m/s and 5 m/s at an angle of 120° to each other. If they pass through the same point at the same time, what is the distance between them after 2 seconds?] 24 m 34 m 14 m 44 m DËi: 14 m e ̈vL ̈v: 120 3ms–1 (3  2) = 6m 5ms–1 (5  2) = 10m S cos120 = 102 + 62 – S 2 2  10  6  – 1 2 = 100 + 36 – S 2 2  60  S 2 = 136 + 60  S = 14 m 13. hw` Mvwoi MwZi w`‡K evZvm eB‡Z _v‡K Zvn‡j †Kvb †ÿ‡Î mvg‡bi KuvP wfR‡e? [If a car is moving faster than the wind, which part of the car's windshield will get wet?] evZv‡mi †eM = Mvwoi †eM (Wind speed = car speed) evZv‡mi †eM  Mvwoi †eM (Wind speed < car speed) evZv‡mi †eM  Mvwoi †eM (Wind speed > car speed) †KvbwUB bq (None of these) DËi: evZv‡mi †eM  Mvwoi †eM (Wind speed < car speed) e ̈vL ̈v: (i) evZvm I Mvwo/ ch©‡eÿK wecixZ w`‡K _vK‡j  me©`vB mvg‡bi KuvP wfR‡e (ii) evZvm I Mvwo/ ch©‡eÿK GKB w`‡K _vK‡j  Vc > Va n‡j, mvg‡bi KuvP wfR‡e  Vc < Va n‡j, wcQ‡bi KuvP wfR‡e  Vc = Va n‡j, Dfq KuvP wfR‡e 14. A  = 2i  – 4j  I B  = 4i  – 2j  n‡j, A  eivei B  Gi Dcvsk KZ? [If A  = 2i  – 4j  and B  = 4i  – 2j  what is the angle between A  and B  ?] – 4 5 (2i )  – 4j  4 5 (2i )  + 4j  (2i )  + 4j  4 5 (2i )  – 4j  DËi: 4 5 (2i )  – 4j  e ̈vL ̈v: A  eivei B  Gi Dcvsk = B cos. a  = A  . B  A . A  A = 8 + 8 20 . 2i  – 4j  20 = 16 20 (2i )  – 4j  = 4 5 (2i )  – 4j  15. O we›`y n‡Z GKRb e ̈w3 C we›`y‡Z †M‡j Zvi `~iZ¡ I mi‡Yi gvb KZ? [If a person moves from point O to point C, what will be the magnitude of his displacement and distance?] 0 (0, 0) A (2, 0) B (2, 4) C (1, 5) 6, 26 26, 6 + 2 6 + 2, 26 26, 2 DËi: 6 + 2, 26
4 e ̈vL ̈v: 0 (0, 0) A (2, 0) B (2, 4) C (1, 5) miY, OC  = (1 – 0) i  + (5 – 0) j  = 1i  + 5j  mi‡Yi gvb, |OC|  = 5 2 + 12 = 26 `~iZ¡ = OA + AB + BC = 2 + 4 + (2 – 1) 2 + (5 – 4) 2 = 2 + 4 + 1 2 + 12 = 6 + 2 16. M  . N  = 0 Ges M  I P  †f±i ci ̄úi j¤^ n‡j, M  Gi mgvšÍivj †f±i †KvbwU? [M  . N  = 0 and M  and P  are parallel vectors, what is the vector parallel to M  ?] N  0 P  N   P  DËi: N   P  e ̈vL ̈v: M  . N  = 0 gv‡bI M  I N  †f±i `ywU ci ̄úi j¤^| Avevi N  I P  †f±i Df‡qB †h Z‡j Av‡Q Zvi Dci j¤^ n‡jv N   P  | ZvB N   P  Ges M  †f±iØq ci ̄úi mgvšÍivj| 17. P  = 4i  + 8j  + 6k  Q  = 2i  + 4j  + 3k  Zvn‡j, P   Q  = ? [P  = 4i  + 8j  + 6k  Q  = 2i  + 4j  + 3k  Then, P   Q  = ?] 0 8 9 Impossible to find out DËi: 0 e ̈vL ̈v:        i   4 2 j  8 4 k  6 3 = (24 – 24) i  + (12 – 12) j  + (16 – 16) k  = 0 18. 6 ms–1 †e‡M †`Š‡o hvevi mgq GKRb †jvK 8 ms–1 †e‡M j¤^fv‡e cwZZ e„wói m¤§yLxb n‡jv| e„wói †_‡K iÿv †c‡Z n‡j Zv‡K wb‡Ri w`‡Ki mv‡_ KZ †Kv‡Y QvZv ai‡Z n‡e? [A person is running at a speed of 6 m/s in the rain, which is falling vertically at a speed of 8 m/s. At what angle should he hold his umbrella to protect himself from the rain?] 36.86 59.5 92.36 42.72 DËi: 36.86 e ̈vL ̈v: tan = e„wói †eM †jv‡Ki †eM = 6 8  = tan–1 6 8   = 36.86; Avbyf‚wgKfv‡e Z_v †jv‡Ki w`‡K _v‡K| VM = 8 VR = 6 19. i   j  + j   i  = ? 0 i  k  j  DËi: 0 e ̈vL ̈v: i   j  = k  j   i  = – k  i   j  + j   i  = k  – k  = 0 20. GKwU b`x‡Z † ̄av‡Zi †eM 6 ms–1 Ges †bŠKvi †eM 7ms–1 Ges b`xi cÖ ̄’ 70 3 m. hw` † ̄av‡Zi mv‡_ 60 †Kv‡Y Pvjv‡bv nq Z‡e b`x cvi n‡Z KZ mgq jvM‡e? [A river flows at a speed of 6 m/s, and a boat can travel at a speed of 7 m/s in still water. If the width of the river is √ m, how long will it take the boat to cross the river if it is steered at an angle of 60° to the current?] 20 h 15 sec 20 sec None of these DËi: 20 sec e ̈vL ̈v: t = d vsin = 70 3 7 sin60 = 70 3  2 7  3 = 20 sec GLv‡b, t = mgq d = b`xi cÖ ̄’ v = †bŠKvi †eM  = † ̄av‡Zi mv‡_ †bŠKvi †KvY