Tiêu đề 2. P1C2. HSC PREP Papers 26_With Solve.pdf ✅

†f±i  HSC Prep Papers 1 †f±i Vector wØZxq Aa ̈vq Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU †f±‡ii †hvM-we‡qvM I jwä Ñ 1 1 2 2 2 Ñ 8 AvqZ GKK †f±i 1 1 1 Ñ Ñ 1 2 6 †f±i ivwki wefvRb (Dcvsk) 1 Ñ Ñ 1 Ñ 1 Ñ 3 e„wó I QvZv aiv msμvšÍ Ñ 1 Ñ 1 Ñ 2 2 6 †f±‡ii ̧Y, †ÿÎdj I j¤^ Awf‡ÿc, GKB mgZj, ga ̈eZ©x †KvY 5 4 Ñ 3 7 4 5 29 b`x cvivcvi Ñ 2 Ñ 1 4 4 9 20 †f±i Acv‡iUi Ñ Ñ Ñ 1 Ñ 4 Ñ 5 *we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| 1 km cÖ‡ ̄ i GKwU b`x cvi nIqvi Rb ̈ `yBRb muvZviæ, muvZvi cÖwZ‡hvwMZvq AskMÖnY K‡i| cÖ_g muvZviæ 6 kmh–1 †e‡M † ̄av‡Zi cÖwZK‚‡ji mv‡_ 60 †Kv‡Y Ges wØZxq muvZviæ 6 kmh–1 †e‡M AvovAvwofv‡e muvZvi KvUv ïiæ K‡i| b`x‡Z † ̄av‡Zi †eM 3 kmh–1 | [Xv. †ev. 23] (K) †K.wYK fi‡eM Kx? (L) Kx Kx k‡Z© Kv‡Ri gvb k~b ̈ n‡Z cv‡i? e ̈vL ̈v K‡iv| (M) cÖ_g muvZviæi jwä †eM wbY©q K‡iv| DËi: u 60  w v jwä †eM, w = u 2 + v2 + 2uvcos  w = 6 2 + 32 + 2  3  6  cos 120  w = 5.196 km/h (Ans.) (N) D3 cÖwZ‡hvwMZvq †Kvb muvZviæ Av‡M b`x cvi n‡Z cvi‡e? DËi: Avgiv Rvwb, t = d usin cÖ_g mvZviæi †ÿ‡Î, t1 = d u1sin1 = 1 6  sin 120  t1 = 0.1924 hr wØZxq mvZviæi †ÿ‡Î, t2 = d u2sin2 = 1 6  sin 90  t2 = 0.167 hr  t2 < t1 nIqvq wØZxq mvZviæ Av‡M b`x cvi n‡Z cvi‡e| (Ans.) 2| GKwU Mvwoi †cQ‡bi Møvm Qv‡`i mv‡_ 30 †Kv‡Y †njv‡bv| MvwowU v  = 18i  †e‡M GKwU iv ̄Ívq PjwQj| nvVvr e„wó u  = – 12j  †e‡M cov ïiæ n‡jv| [Xv. †ev. 23] (K) wkwkiv1⁄4 Kv‡K e‡j? (L) `iRvi nvZj cÖv‡šÍ †`qv nq †Kb? e ̈vL ̈v K‡iv| (M) Mvwoi mvg‡bi Møv‡m e„wó KZ †e‡M co‡e? DËi: Mvwoi mv‡c‡ÿ Av‡cwÿK †eM, v rc  = vr  – vc  = – 12j  – 18i   |v | rc  = 122 + 182 = 21.63 GKK (Ans.) (N) DÏxc‡Ki Mvwoi wcQ‡bi Møvm e„wó‡Z wfR‡e wK bvÑ MvwYwZK we‡kølYc~e©K gZvgZ `vI| DËi: 30 30 33.7 vrc 30 vrc ÔMÕ n‡Z cvB, vrc  = – 18j  – 12i   tan = 12 18   = tan–1     2 3 = 33.7   > 30 nIqvq e„wó Mvwoi wcQ‡bi KvP‡K wfRv‡e| (Ans.) 3| P  = 5i  + 3j  – mk  ; Q  = i  + j  + 4k  ; GLv‡b P  I Q  ci ̄úi j¤^| hw` P  Ges Q  Gi gvb h_vμ‡g †b.Kv Ges GKwU b`xi † ̄av‡Zi `aæwZ wb‡`©k K‡i Z‡e me©wb¤œ c‡_ b`x cvi n‡Z †b.KvwUi 2 wgwbU mgq jv‡M| [iv. †ev. 23; w`. †ev. 23; g. †ev. 23] (K) mgZjxq †f±‡ii msÁv `vI| (L) †Kv‡bv cÖevnxi AvqZ‡bi cwieZ©b wbY©‡q WvBfvi‡RÝ Gi f~wgKv Av‡Q wK-bv? e ̈vL ̈v Ki| (M) ‘m’ Gi gvb wnmve Ki| DËi: P  I Q  †f±iØq ci ̄úi j¤^ n‡j, G‡`i WU ̧Ydj k~b ̈ n‡e| A_©vr, P  .Q  = 0  (5i )  + 3j  – mk  .(i )  + j  + 4k  = 0

†f±i  HSC Prep Papers 3 b`x cvi n‡Z †b.Kvi cÖ‡qvRbxq mgq, t = d u 2 – v 2 = 500 4 2 – 2 2  t = 144.33 s  B we›`y‡Z MvwowU _vK‡e, t = 1200 15 + 40 = 120 s < t  †b.Kvi hvÎxiv Mvwo‡Z DV‡Z cvi‡e bv| (Ans.) 7| Y Q(–2, 2, 1) O(0, 0, 0) P(2, – 1, 3) Z X [wm. †ev. 23] (K) † ̄..jvi A‡cÿ‡Ki MÖ ̈vwW‡q‡›Ui msÁv †jL| (L) KvR I UK© Gi GKK AwfbœÑ e ̈vL ̈v K‡iv| (M) PQ  †f±‡ii mgvšÍivj GKwU GKK †f±i wbY©q K‡iv| DËi: OQ  = – 2i  + 2j  + k  OP  = 2i  – j  + 3k  PQ  = OQ  – OP  = (– 2i )  + 2j  + k  – (2i )  – j  + 3k  = – 4i  + 3j  – 2k   PQ  Gi mgvšÍiv‡j GKwU GKK †f±i = PQ  |PQ|  = – 4i  + 3j  – 2k  4 2 + 32 + 22 = 1 29 (– 4i )  + 3j  – 2k  Note: cÖ‡kœ GKwU †f±i PvIqvq,  e ̈envi Kiv nqwb| (N) OPQ mg‡KvYx wÎfzR wK bv hvPvB K‡iv| DËi: OP  = 2i  – j  + 3k   |OP|  = 2 2 + 12 + 32 = 14 GKK OQ  = – 2i  + 2j  + k   |OQ|  = 2 2 + 22 + 12 = 3 GKK PQ  = – 4i  + 2j  – 2k   |PQ|  = 4 2 + 32 + 22 = 29 GKK  OP2 + OQ2 = 14 + 9 = 23 PQ2 = 29  OP2 + OQ2  PQ2 myZivs OPQ mg‡KvYx wÎfzR bq| (Ans.) 8| Kv‡Z©mxq ̄ vbv1⁄4 e ̈e ̄ vq wZbwU we›`y O (0, 0, 0), P (2, 4, 2) Ges Q (2, –4, –4)| [Xv. †ev. 22] (K) mgvb †f±i Kx? (L) †f±i Acv‡iUi † ̄..jvi ivwk‡K †f±i ivwk‡Z iƒcvšÍi K‡iÑ e ̈vL ̈v Ki| (M) PQ  Gi gvb wbY©q Ki| DËi: P Gi Ae ̄ vb †f±i, OP  = 2i  + 4j  + 2k  Q Gi Ae ̄ vb †f±i, OQ  = 2i  – 4j  – 4k   PQ  = OQ  – OP  = – 8j  – 6k   |PQ|  = 8 2 + 62 = 10 GKK (Ans.) (N) P I Q Gi Ae ̄ vb †f±iØq ci ̄úi j¤^ n‡e wK-bv hvPvB Ki| DËi: OP  = 2i  + 4j  + 2k  OP  = 2i  – 4j  – 4k   OP  .OQ  = 2  2 + 4  (–4) + 2  (–4) = 4 – 16 – 8 = – 20  0  OP  .OQ   0 e‡j †f±i `ywU ci ̄úi j¤^ bq| 9| kvšÍ evZv‡m 6 kmh–1 †e‡M e„wó co‡Q| G mg‡q mvB‡K‡j P‡o Avwe` 8 kmh–1 †e‡M evwo wdi‡Q| nVvr Avwe‡`i Pjvi wecixZ w`‡K 2 kmh–1 †e‡M evZvm cÖevwnZ n‡Z jvMj| Dfq †ÿ‡Î e„wó †_‡K evuP‡Z Avwe` QvZv e ̈envi Kij| [Xv. †ev. 22] (K) gyw3 †eM Kx? [6ô Aa ̈vq] (L) ej I miY k~b ̈ bv n‡jI KvR k~b ̈ n‡Z cv‡i wK? e ̈vL ̈v Ki| [5g Aa ̈vq] (M) w ̄ i evZv‡m e„wói jwä †eM wbY©q Ki| vrc   vr  vc  –vc  DËi: mvB‡K‡j †eM, vc  = 8i  kmh–1 e„wói †eM, vr  = – 6j  kmh–1  mvB‡Kj Av‡ivnxi mv‡c‡ÿ e„wói †eM, vrc  = vr  – vc   vrc  = – 6j  – 8i  |v | rc  = 6 2 + 82 = 10 kmh–1 (Ans.)  = tan–1     8 6 = 53.13 (Dj‡¤^i mv‡_) (Ans.)
4  HSC Physics 1st Paper Chapter-2 (N) evZvm cÖevwnZ nIqvi Av‡M I c‡i GKBfv‡e QvZv ai‡j Avwe` e„wó †_‡K iÿv cv‡e wK-bv? MvwYwZKfv‡e hvPvB Ki| DËi: evZvm cÖevwnZ nIqvi Av‡M, vrc  = – 6j  – 8i    = tan–1     8 6 = 53.13 (Dj‡¤^i mv‡_) 8 6  wØZxq †ÿ‡Î, evZv‡mi †eM, va  = –2i  kmh–1  Avwe‡`i mv‡c‡ÿ e„wói †eM, vrc  = ( vr )  + va  – vc  = (– 6j )  – 2i  – 8i  = – 10i  – 6j   = tan–1     10 6 = 59.03 (Dj‡¤^i mv‡_) 10  6   evZvm cÖevwnZ nIqvi c~‡e© 53.19 †Kv‡Y Ges evZvm cÖevwnZ nIqvi c‡i Dj‡¤^i mv‡_ 59.03 †Kv‡Y QvZv ai‡Z n‡e| (Ans.) 10| 500 m cÖ‡ ̄ i GKwU b`x‡Z 6 kmh–1 †e‡M † ̄avZ cÖevwnZ n‡”Q| GB b`xwU gvnxi I wbwa cÖwZ‡hvwMZvi D‡Ï‡k ̈ mvuZvi †K‡U cvi nIqvi wm×všÍ wb‡jv| gvnxi 10 kmh–1 †e‡M † ̄av‡Zi mv‡_  †Kv‡Y Ges wbwa 9 kmh–1 †e‡M † ̄av‡Zi mv‡_ j¤^fv‡e muvZvi KvU‡Z ïiæ Kij| [P. †ev. 22] (K) miY †f±i Kx? (L) †b.Kvi ̧Y Uvbvi mgq `wo hZ j¤^v nq †b.Kv ZZ `aæZ P‡j †Kb? e ̈vL ̈v Ki| (M)  - †Kv‡Yi gvb KZ n‡j gvnxi †mvRvmywR b`xi Aci cv‡o †cu.Qv‡e? DËi: cÖkœg‡Z, ucos + v = 0   = cos–1     – v u = cos–1     – 6 10 = 126.87 (Ans.) u v   (N) DÏxcK Abymv‡i †K wRZ‡e? MvwYwZK we‡kølYmn gZvgZ `vI| DËi: awi, gvnxi I wbwa Gi b`x AwZμ‡g cÖ‡qvRbxq mgq h_vμ‡g t1 I t2  t1 = d u 2 1 – v 2  t1 = 0.5 102 – 6 2  t1 = 0.0625 hour  t2 = d u2 = 0.5 9 = 0.0556 hour u1 v u 2 1 – v 2  t2 < t1 nIqvq wbwa cÖwZ‡hvwMZvq wRZ‡e| (Ans.) 11| GKw`b GKwU A‡ji ZvcgvÎv I evZv‡mi †eM cvIqv †M‡jv h_vμ‡g, Q = 2xy2 z 3 – 4xy I v  = (y2 cosx + z3 )i ^ + (2ysinx – 4) j ^ + (3xz2 + 2) k ^ . [P. †ev. 22] (K) WU ̧Yb Kx? (L) N~Y©biZ c„w_ex m~h© n‡Z `~‡i m‡i †M‡j Gi †eM K‡g hvq †Kb? e ̈vL ̈v Ki| [6ô Aa ̈vq] (M) (1, –1, 2) we›`y‡Z H A‡ji ZvcgvÎvi †MÖwW‡q›U wbY©q Ki| DËi: Grad(Q) =      x i ^ +  y j ^ +  z k ^ .(2xy2 z 3 – 4xy) = (2y2 z 3 – 4y) i ^ + (4xyz3 – 4x) j ^ + (6xy2 z 2 ) k ^  (1, –1, 2) we›`y‡Z Grad(Q) Gi gvbÑ Grad(Q) = (16 + 4)i ^ + (– 32 – 4)j ^ + 24k ^ = 20i ^ – 36j ^ + 24k ^ (Ans.) (N) Hw`b H A‡ji evZv‡m †Kv‡bv N~Y©b wQ‡jv wKbv Zv MvwYwZK we‡køl‡Yi gva ̈‡g gZ `vI| DËi: Curl(v)  =       i ^  x y 2 cosx + z3 j ^  y 2ysinx – 4 k ^  z 3xz2 + 2 = i ^ (0 – 0) + j ^ (3z2 – 3z2 ) + k ^ (2ycosx – 2ycosx) = 0i ^ + 0j ^ + 0k ^ = 0  †h‡nZz Curl(V)  = 0, myZivs v  AN~©Ybkxj| A_©vr Hw`b H A‡ji evZv‡m †Kv‡bv N~Y©b wQj bv| (Ans.) 12| B 3kmh–1 AiæY gvwS 8 kmh–1 eiæY gvwS 2 km † ̄avZ AiæY gvwS 8 kmh–1 †e‡M †b.Kv Pvwj‡q b`xi cÖ ̄ eivei cvi nq| eiæY gvwS GKB †e‡M b`xi cÖ ̄ eivei †b.Kv Pvjvq| b`xi cÖ ̄ 2 km| [iv. †ev. 22] (K) †f±i †hv‡Mi wÎfzR m~ÎwU †jL| (L) Uawj †e‡Mi nvZj j¤^v ivLvi myweav Kx? (M) DÏxc‡K AiæY gvwS‡K †Kvb w`‡K †b.Kv Pvjv‡Z n‡qwQj? DËi: cÖkœg‡Z, v + ucos = 0   = cos–1     – v u   = cos–1     – 3 8   = 112.02 (Ans.) v  u (N) DÏxc‡Ki †Kvb gvwS Kg mg‡q b`x cvi n‡e? MvwYwZKfv‡e e ̈vL ̈v Ki| DËi: AiæY gvwSi b`x cvi n‡Z cÖ‡qvRbxq mgq t1 n‡j, t1 = d u 2 1 – v 2 = 2 8 2 – 3 2  t1 = 0.27 hr