Title 5. P1C5. HSC PREP Papers 26_With Solve.pdf ✅

2  HSC Physics 1st Paper Chapter-5 3| GKwU KYvi Dci F  = (–5i  – 3j  – 6k  ) ej cÖ‡qvM Kivq KYvwU P(–6, 7, –1) we›`y †_‡K Q(–3, –8, 4) we›`y‡Z ̄’vbvšÍwiZ n‡jv| [iv. †ev. 23; g. †ev. 23] (K) Ae ̄’vb †f±i Kv‡K e‡j? (L) † ̄av‡Zi mv‡_ AvovAvwofv‡e b`x cvwo w`‡Z †b.Kv Kxfv‡e Pvjv‡Z n‡e? e ̈vL ̈v Ki| (M) DÏxc‡Ki KYvwUi miY wbY©q K‡iv| DËi: P  = – 6i  + 7j  – k  Q  = – 3i  – 8j  + 4k   DÏxc‡Ki KYvwUi miY = (– 3i )  – 8j  + 4k  –(– 6i )  + 7j  – k  = 3i  – 15j  + 5k  (Ans.) (N) GLv‡b DwjøwLZ ejwU †K›`agyLx ej wQj wKbv? MvwYwZKfv‡e we‡kølY K‡iv| DËi: †K›`agyLx ej Øviv K...ZKvR = 0  W = F  .S  = (– 5i )  – 3j  – 6k  .(3i )  – 15j  + 5k  = 0  DÏxc‡Ki ejwU †K›`agyLx ej| (Ans.) 4| GKwU KYvi Dci F  = (2i )  + 3j  – 2k  N ej cÖ‡qvM Kivq P(2, 1, –3) we›`y n‡Z Q (3, –2, 1) we›`y‡Z ̄’vbvšÍwiZ nq| Aci GKwU F  = (–i )  – j  + k  N, F  Gi mv‡_ cÖhy3 n‡q KYvwU‡K P we›`y †_‡K R(–2, 1, 3) we›`y‡Z ̄’vbvšÍwiZ K‡i| [Kz. †ev. 23] (K) hvwš¿K kw3 Kx? (L) Nl©Y ej AmsiÿYkxj ej, e ̈vL ̈v K‡iv| (M) PQ  mi‡Yi Rb ̈ K...ZKvR wbY©q K‡iv| DËi: P  = 2i  + j  – 3k  Q  = 3i  – 2j  + k   P we›`y n‡Z Q we›`y‡Z ̄’vbvšÍwiZ n‡j miY, S  1 = Q  – P  = (3i )  – 2j  + k  – (2i )  – j  + 3k  = i  – 3j  + 4k   KvR W1 = F  1.S  1 = (2i )  + 3j  – 2k  .(i )  + 3j  – 4k  = –15 GKK (Ans.) (N) PQ  Ges PR  mi‡Yi Dfq‡ÿ‡Î MwZkw3i cwieZ©b mgvb n‡e wKbvÑ MvwYwZK we‡kølYmn gšÍe ̈ K‡iv| DËi: P  = i  – 2j  + k  R  = – 2i  + j  + 3k   S  2 = (– 2i )  + j  + 3k  – (2i )  + j  – 3k  = – 4i  + 6k   F  2 = F  + F  = (2i )  + 3j  – 2k  + (– i )  – j  + k  = (i )  + 2j  – k  N  KvR, W2 = F  2.S  2 = (i )  + 2j  – k  .(– 4i )  + 6k  = –10 GKK (Ans.) †h‡nZz, W1  W2 ZvB PQ  Ges PR  mi‡Yi †ÿ‡Î MwZkw3i cwieZ©b mgvb n‡e bv| (Ans.) 5| GKwU cvwbc~Y© K‚‡ci MfxiZv I e ̈vm h_vμ‡g 20 m I 2 m| GKwU cv¤ú 30 min G K‚cwU‡K cvwbk~b ̈ Ki‡Z cv‡i| [g = 9.8 ms–2 ] [h. †ev. 23] (K) KvR-kw3 Dccv` ̈wU †jL| (L) MwZkw3i gvb me©`v abvZ¥K nqÑ e ̈vL ̈v Ki| (M) cv‡¤úi ÿgZv †ei K‡iv| DËi: P = mgh t = r 2 l  g  h 2 t = 3.1416  1 2  20  103  9.8  20 2 30  60 = 3420.85 W (Ans.) (N) K‚cwUi 1g PZzf©vM I 4_© PZz©fvM cvwbk~b ̈ Ki‡Z cÖ‡qvRbxq mg‡qi ZviZg ̈ n‡e wKbv? n‡j †Kgb n‡e wnmve K‡i †`LvI| DËi: 1g PZzf©v‡Mi †ÿ‡Î, t1 = mgh 1 p = r 2 l  g  h 1 p = 3.1416  1 2  20 4  103  9.8  0 + 20 4 2 3420.85 = 112.5 s 4_© PZzf©v‡Mi †ÿ‡Î, t2 = mgh 2 p = r 2 l  g  h 2 p = 3.1416  1 2  20 4  103  9.8  3 4 × 20 + 20 2 3420.85 = 787.5 s  t1 < t2 (Ans.)