Tiêu đề 2. P1C2. Vector (ভেক্টর) C+Merged Ok_Sha 17.4.24 (Mahee) Ok.pdf ✅

†f±i  Engineering Practice Content 1 wØZxq Aa ̈vq †f±i Vector weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| C B A wÎfz‡Ri †KvY ̧wj KZ? [BUET 22-23] A  = 2i  + j  + 3k  , B  = 3i  – 3j  – k  mgvavb: C  = B  – A  = i  – 4j  – 4k  A B = cos–1      A   . B  AB = cos –1 0 = 90 B C = cos–1      B   . C  BC = cos –1     3 + 12 + 4 19 × 33 = 40.64 C A = 90 – 40.64 = 49.35 2| B C 37 36 m 37 vb = 10 ms–1 vr = 3 ms–1 A K. BC Gi •`N© ̈ wbY©q Ki| (L) mivmwi B †Z †h‡Z KZ †Kv‡Y iIbv w`‡Z n‡e? [BUET 21-22] mgvavb: (K) BC = (vr + vb cos ) t = (vr + vb cos ). AB vb sin  = (3 + 10 cos 37). 36 10 sin 37 = 65.72 m (Ans) (L) mivmwi B †Z †h‡Z, vr + vb cos  = 0   = cos –1     – 3 10 = 107.46 (Ans.) 3| i  + j  †f±‡ii w`‡K A  = 2i  + 3j  †f±‡ii Dcvsk wbY©q Ki| [BUET 19-20] mgvavb: awi, B  = i  + j   B  Gi w`‡K A  Gi Dcvsk = A  . B  |B|  . B  |B|  = A  . B  |B|  2 B  = 2 + 3 1 2 + 12 . (i )  + j  = 5 2 (i )  + j  (Ans.) 4| †Kvb GKw`b 30 ms–1 MwZ‡Z Dj¤^fv‡e e„wó cowQj| hw` evqy 10 ms–1 MwZ‡Z DËi †_‡K `wÿ‡Y eB‡Z ïiæ K‡i Zvn‡j e„wó †_‡K iÿv †c‡Z †Zvgvi QvZv †Kvb w`‡K †g‡j ai‡Z n‡e †ei Ki| [BUET 06-07] mgvavb: wPÎ n‡Z,  = tan–1     10 30 = 18.43 (c~e© w`‡Ki mv‡_ `wÿY eivei)  10 ms–1 30 ms–1 `: D: c~: 5| cÖwZ NÈvq 1800 m †e‡M 240 m cÖk ̄Í GKwU b`x wb‡Pi w`‡K cÖevwnZ n‡”Q Ges cÖwZ NÈvq 3600 m †e‡M muvZv‡i mÿg GKRb muvZviæ GKwU wecixZ we›`y‡Z †h‡Z B”QzK| †m †Kvb w`K eivei muvZvi †`‡e Ges †mB we›`y‡Z †h‡Z KZ mgq †b‡e? [BUET 03-04] mgvavb: †b.Kvi †eM, v = 3.6 km/h † ̄av‡Zi †eM, u = 1.8 km/h  wecixZ cÖv‡šÍ †h‡Z n‡j † ̄av‡Zi mv‡_ †b.Kv Pvjbv Ki‡Z n‡e,  = cos –1     – u v   = cos –1     – 1.8 3.6 = 120 (Ans.) cÖ‡qvRbxq mgq, t = d vsin = 0.240 3.6 sin120  t = 0.077 hr (Ans.) 6| ci ̄ú‡ii mv‡_ j¤^fv‡e wμqvkxj `yBwU e‡ji jwä 80 N| hw` jwä GKwU e‡ji m‡1⁄2 60 †Kv‡Y AvbZ _v‡K, Z‡e ej `yBwUi gvb wbY©q Ki| [RUET 17-18] mgvavb: P = Rcos60  P = 80 × 1 2 = 40 N Q = Rsin60  Q = 80 × 3 2 = 40 3 N (Ans.) R = 80N 60 Q P
2  Physics 1st Paper Chapter-2 7| p Gi gvb KZ n‡j †f±i v  = (5x + 2y)i  + (2py – z)j  + (x – 2z)k  mwjbqWvj n‡e? [RUET 15-16] mgvavb: v  = (5x + 2y)i  + (2py – z)j  + (x – 2z)k    .v  = (5x + 2y) x + (2py – z) y + (x – 2z) z = 0  5 + 2p – 2 = 0  p = – 3 2 (Ans.) 8| †Kvb b`x‡Z GKwU †b.Kvi †eM † ̄av‡Zi AbyK~‡j I cÖwZK~‡j h_vμ‡g 18 Ges 6 kmh–1 | †b.KvwU KZ †e‡M †Kvb w`‡K Pvjbv Ki‡j †mvRv Aci cv‡o †cu.Qv‡e? [RUET 10-11] mgvavb: †b.Kvi †eM, u Ges † ̄av‡Zi †eM v  u + v = 18 u – v = 6  u = 12 km/h Ges v = 6 km/h v  u GLb,  = cos –1     – u v = cos –1     – 12 6   = 120 (Ans.) 9| 10 wK‡jvwgUvi/NÈv †e‡M e„wó co‡Q Ges 60 wK‡jvwgUvi/NÈv †e‡M c~e© n‡Z cwð‡g evZvm eB‡Q| c~e© n‡Z cwðg AwfgyLx PjšÍ Mvwoi MwZ‡eM wbY©q Ki hv‡ZÑ (a) Mvoxi mvg‡bi I wcQ‡bi KuvP wf‡R, (b) ïaygvÎ wcQ‡bi KuvP wf‡R| [RUET 04-05] mgvavb: – vc  va  = 60 km/h vr  = 10 km/h c~: c: (a) mvg‡bi I wcQ‡bi Dfq KuvP wfR‡Z n‡j Mvwoi mv‡c‡ÿ e„wói †eM n‡e Lvov wb¤œgyLx|  –vc  = va   vc = 60 km/h (Ans.) (b) – vc  va  vr  vrc  ïaygvÎ wcQ‡bi KuvP wfR‡j, wPÎvbymv‡i jwä n‡e vrc  A_©vr G‡ÿ‡Î, va > vc  vc < 60 km/h n‡e| (Ans.) 10| (a) †Kvb we›`y P Gi ̄’vbv1⁄4 P(2, –3, 4) n‡j we›`ywUi Ae ̄’vb †f±i wbY©q Ki| (b) A(2, –1, 3) Ges B(–1, 2, –3) we›`y؇qi ms‡hvMKvix w`K ivwkwU wbY©q Ki| [CUET 05-06] mgvavb: (a) OP  = 2i  – 3j  + 4k  (Ans.) (b) OA  = 2i  – j  + 3k  , OB  = – i  + 2j  – 3k   AB  = OB  – OA  = –3i  + 3j  – 6k  (Ans.) 11| GKwU BwÄb PvwjZ †b.Kvi †eM NÈvq 14 wK‡jvwgUvi| GKwU b`x AvovAvwo cvi n‡Z n‡j †b.KvwU‡K †Kvb w`‡K Pvjv‡Z n‡e? b`xi cÖ ̄’ 12.125 km n‡j Zv cvwo w`‡Z KZ mgq jvM‡e? † ̄av‡Zi †eM NÈvq 7 km| [CUET 04-05] mgvavb: AvovAvwo cvi n‡Z n‡j †b.Kv‡K † ̄av‡Zi mv‡_,  = cos –1     – u v = cos –1     – 7 14 = cos –1     – 1 2   = 120 †Kv‡Y Pvjbv Ki‡Z n‡e| (Ans.) Avevi, t = d vsin = 12.125 14 sin(120)  t = 1 hr (Ans.) 12| GKwU b`xi † ̄av‡Zi †eM 5 ms–1 | 10 ms–1 †e‡Mi GKwU †b.Kvi †mvRvmywRfv‡e b`x cvwo w`‡Z 1 min 40 second mgq jv‡M| b`xi cÖ ̄’ KZ? [CUET 03-04] mgvavb: †mvRvmywRfv‡e cvwo †`qvi †ÿ‡Î, t = d v 2 – u 2  d = 100 × 102 – 5 2 = 866.025 m (Ans.) u v 2 – u v 2 d 13| GKwU Mvwo 20 kmh–1 †e‡M c~e©w`‡K Pjgvb| evZvmI 4 kmh–1 †e‡M GKB w`‡K Pjgvb| G mgq e„wó Lvov wb‡Pi w`‡K 6 kmh–1 †e‡M co‡Z ïiæ K‡i| e„wó Mvwo‡Z Djø‡¤^i mv‡_ KZ †Kv‡Y AvNvZ Ki‡e? [BUTex 23-24] mgvavb: vw vc vr DËi (j) `wÿY (– j) cwðg (– i) c~e© (i) e„wói †eM, vr = – 6j  km/h; evZv‡mi †eM, vw = 4i  km/h Mvwoi †eM, vc = 20i  km/h evZvm I e„wói jwä †eM, vrw  = 4i  – 6j  km/h Mvwoi mv‡c‡ÿ, vrc  = vrw  – vc  = 4i  – 6j  – 20i  = – 16i  – 6j  Dj‡¤^i mv‡_ †KvY,  = tan–1     16 6