Tiêu đề 5. P1C5 Kaj-Shokti-Khamota With Solve_Sha 13.11.23 PDF.pdf ✅

KvR, kw3 I ÿgZv  Engineering Practice Content 1 cÂg Aa ̈vq KvR, kw3 I ÿgZv Work, Energy and Power ACS Physics Department Gi g‡bvbxZ cÖkœmg~n 1| 200N Gi ej cÖ‡qvM K‡i †Kv‡bv e ̄‘‡K e‡ji Awfgy‡L 300 m miv‡bv n‡j KZ KvR m¤úbœ n‡e †ei K‡iv| [Easy] mgvavb: W = Fs = 200  300 J = 6  104 J (Ans.) Tricks: ej Ges mi‡Yi AwfgyL fv‡jv K‡i †Lqvj Ki‡e| 2| 3kg f‡ii GKwU e ̄‘ 30ms–1 †e‡M MwZkxj Av‡Q| †_‡g hvevi c~e© gyn~Z© ch©šÍ Nl©Y e‡ji weiæ‡× e ̄‘wU‡K Kx cwigvY KvR Ki‡Z n‡e? [Easy] mgvavb: W = 1 2 mv2 = 1 2  3  302 J = 1350 J (Ans.) 3| GKRb †ivjvi PvjK Abyf~wg‡Ki mv‡_ 45 †KvY K‡i 20N e‡j †ivjvi Pvjvq| †ivjviwU mvg‡bi w`‡K 75m Pj‡j †m KZ KvR Ki‡e? [Easy] mgvavb: W = Fs cos = 20  75  cos45 J = 1060.66 J (Ans.) 4| 250 N IR‡bi GKwU evjK Lvov gB †e‡q kx‡l© DV‡Z 2000 J KvR m¤úbœ K‡i| gBwUi •`N© ̈ wbY©q K‡iv| [Easy] mgvavb: W = mgh  h = W mg  h = 2000 250 m  h = 8 m (Ans.) 5| GKwU BwÄb 20 m Mfx‡i GKwU K~c n‡Z cÖwZ wgwb‡U 100 kg cvwb‡Z †Zv‡j| BwÄb KZ...K cÖhy3 ej Ges Bwćbi ÿgZv wbY©q K‡iv| [Easy] mgvavb: BwÄb KZ...©K cÖhy3 ej, F = mg = 100 × 9.8 N = 980 N (Ans.) P = W t = mgh t = 9.8  100  20 60 W = 326.66 W (Ans.) 6| 50 kg f‡ii ce©Z Av‡ivnYKvix 15 wgwb‡U 150 m DuPz GKwU P~ovq Av‡ivnY K‡ib| Av‡ivnxi K...ZKvR I ÿgZv KZ? mgvavb: W = mgh = 50  9.8  150 J = 7.35  104 J (Ans.)  P = W t = 7.35  104 15  60 W = 81.67 W (Ans.) 7| 900 kg f‡ii GKwU wjdU 350 kg f‡ii †evSv mn 100 s G wbPZjv n‡Z 18 Zjvq 75 m Dc‡i D‡V| K...ZKvR I cÖhy3 ÿgZv wbY©q Ki| [Easy] mgvavb: W = mgh = 1250  9.8  75 J = 9.18  105 J (Ans.)  P = W t = 9.18  105 100 W = 9.18 kW (Ans.) 8| GKwU †gvUi Øviv 100 m Mfxi GKwU Kzqv †_‡K cÖwZ wgwb‡U 1200 kg cvwb DVv‡bv nq| †gvU‡ii ÿgZv IqvU, wK‡jvIqvU I Ak^ÿgZv GK‡K wbY©q K‡iv| [Easy] mgvavb: P = W t = 1200  9.8  100 60 W = 19600 W (Ans.) = 19600 103 kW = 19.6 kW (Ans.) = 19600 746 HP = 26.27 HP (Ans.) 9| f~wg †_‡K 20 m DuPz Qv‡` BU †Zvjvi Rb ̈ 10 kW Gi GKwU BwÄb e ̈envi Kiv n‡jv| 1 NÈvq BwÄbwU Kx cwigvY BU Qv‡` Zzj‡Z cvi‡e? [Easy] mgvavb: P = mgh t  m = 10000  3600 9.8  20 kg  m = 1.8  105 kg (Ans.)
2  Physics 1st Paper Chapter-5 10| 746 W ÿgZvi GKwU cv¤ú cÖwZ wgwb‡U Kx cwigvY cvwb 10 m D”PZvq Ic‡i DVv‡Z cvi‡e? [Easy] mgvavb: P = W t  m = Pt gh  m = 746  60 9.8  10 kg  m = 456.7 kg (Ans.) 11| 150 kg f‡ii GK e ̈w3 50 kg f‡ii GKwU †evSv wb‡q 4 m `xN© GKwU wmwo †e‡q 20 s G wb‡P bvg‡jv| hw` wmwowU †`qv‡ji mv‡_ 60 †Kv‡Y _v‡K Z‡e †jvKwUi ÿgZv wbY©q K‡iv| [Medium] mgvavb: P = W t = Fs cos t = 1960  4  cos60 20 W  P = 196 W (Ans.) 12| Abyf~wg‡Ki mv‡_ 60 †Kv‡Y 5 m j¤^v GKwU †njv‡bv Z‡ji cv`‡`k †_‡K kxl©‡`‡k 10 kg f‡ii GKwU eøK Zzj‡Z n‡e| Zj‡K Nl©Ynxb a‡i eøKwU‡K aaæe MwZ‡Z Zzj‡Z KZ KvR Ki‡Z n‡e wbY©q K‡iv| [Medium] mgvavb: W = mgh = 10  9.8  5 sin60 J = 424.35 J (Ans.) 13| GKwU cvwbc~Y© f~-Mf© ̄’ Rjvav‡ii MfxiZv 7.5 m Ges †PvOvK...wZ Rjvav‡ii e ̈vm 4 m| †h cv¤ú 30 wgwb‡U Rjvavi‡K m¤ú~Y© Lvwj Ki‡Z cv‡i Zvi ÿgZv KZ HP? [Medium] mgvavb: P = r 2 h    g  h 2 t W =  × 2 2  7.5  103  9.8  3.75 1800 W = 1925 746 HP = 2.58 HP (Ans.) 14| GKwU cv¤ú 4.9 wgwb‡U Kzqv †_‡K 10000 wjUvi cvwb 6 m Mo D”PZvq Zzj‡Z cv‡i| cv‡¤úi ÿgZvi 80% Kvh©Ki n‡j Gi ÿgZv wbY©q K‡iv| [Easy] mgvavb: P = mgh t   = 104  9.8  6 4.9  60  0.8 W = 2.5 kW (Ans.) 15| †Kv‡bv GKwU ̄’vb n‡Z GK wgwb‡U GKwU BwÄb 100 kg f‡ii GKwU e ̄‘ 20 m Ic‡i Zzj‡Z cv‡i| hw` BwÄbwUi ÿgZv 30% bó nq, Z‡e BwÄbwUi ÿgZv wbY©q K‡iv| [Easy] mgvavb: P = 100  9.8  20 60  0.7 W = 466.66 W (Ans.) 16| GKwU †μb 3.73 kW ÿgZv cÖ‡qv‡M 746 N IR‡bi GKwU †jŠn LЇK KZ Mo †e‡M Lvov Dc‡i Zzj‡Z cvi‡e? [Easy] mgvavb: P = Fv  v = 3.73  103 746 ms–1 = 5 ms–1 (Ans.) 17| GKwU •e`y ̈wZK cv¤ú f~-Mf© ̄’ Rjvavi †_‡K 9.1 m3 cvwb 1 NÈvq f~wg †_‡K 32 m D”PZvq Aew ̄’Z UvswK‡Z Zz‡j| f~Mf© ̄’ Rjvav‡ii cvwbi Zj f~-c„ô n‡Z 4 m MfxiZvq Aew ̄’Z| cv‡¤úi ÿgZv 70% Kvh©Kix n‡j cv‡¤úi ÿgZv wbY©q K‡iv| [Medium] mgvavb: P = mgh t = 9100  9.8  (32 + 4) 3600 W = 891.8 W  P = P  = 1247 W (Ans.) 18| 80% `ÿZv m¤úbœ GKwU †gvUi GKwU †μb wbqš¿Y K‡i hvi `ÿZv 50%| †gvUiwU 3.73 kW ÿgZv cÖ‡qvM Ki‡j †μ‡b 746 N IR‡bi GKwU e ̄‘i Ea©gyLx Mo‡eM KZ n‡e? [Medium] mgvavb: P = Fv  v = 3.73  80%  50%  103 746 ms–1  v = 2 ms–1 (Ans.) 19| 10 wU BU GKwUi Dci Av‡iKwU †i‡L †gvU 0.50 m DuPz GKwU ̄ͤ¢‡K mvRv‡Z KZ KvR Ki‡Z n‡e wbY©q K‡iv| (†`Iqv Av‡Q, cÖwZwU B‡Ui D”PZv 5  10–2 m Ges fi 1 kg) [Medium] mgvavb: W = n(n – 1) 2  mgh = 10 × 10 – 1 2  1  9.8 × 5 × 10–2 J = 22.05 J (Ans.)
KvR, kw3 I ÿgZv  Engineering Practice Content 3 20| 100 m Mfxi GKwU Kzqv †_‡K Bwćbi mvnv‡h ̈ cÖwZ wgwb‡U 1000 kg cvwb DVv‡bv nq| hw` BwÄbwUi ÿgZv 42% bó nq, Zvn‡j Gi Ak^ ÿgZv wbY©q K‡iv| [Easy] mgvavb:  P = P  = mgh t = 1000  9.8  100 60  0.58  746 HP = 37.75 HP (Ans.) 21| GKwU KYvi Ici F  = (6i )  – 3j  + 2k  N ej cÖ‡qv‡M KYvwUi r  = (2i )  + 2j  – k  m miY nq| ej KZ...K m¤úvw`Z Kv‡Ri cwigvY wbY©q K‡iv| mgvavb: W = F  .r  = (6i )  – 3j  + 2k  . (2i )  + 2j  – k  = 12 – 6 – 2 J = 4 J (Ans.) 22| GKwU †gvUi wgwb‡U 5.5  105 kg cvwb 100 m Ic‡i Zzj‡Z cv‡i| †gvUiwUi `ÿZv 70% n‡j Gi ÿgZv wbY©q K‡iv| [Easy] mgvavb: ÿgZv = mgh t   = 5.5  105  9.8  100 60  70% W = 1.28  107 W (Ans.) 23| GKwU cvwbc~Y© Kzqvi MfxiZv 12 m Ges e ̈vm 1.8 m| GKwU cv¤ú KzqvwU‡K 24 min G cvwbk~b ̈ Ki‡Z cv‡i| cv¤úwUi Ak^ ÿgZv KZ? [Medium] mgvavb: P = W t = mgh t =      1.8 2 2  12  103  9.8  12 2 24  60  746 HP  P = 1.67 HP (Ans.) 24| GKRb Qz‡Zvi †g‡Si Ici w`‡q KvV †evSvB GKwU Uawj `wo w`‡q †e‡a 10 m †U‡b wb‡q †Mj| `woi Uvb 200 N Ges Zv Abyf~wg‡Ki mv‡_ 37 †Kv‡Y Ic‡ii w`‡K| Qz‡Zv KZ...K K...ZKvR wbY©q Ki| †m hw` Abyf~wgKfv‡e ej cÖ‡qvM Ki‡Zv Z‡e KZ KvR n‡Zv? [Easy] mgvavb: 1g †ÿ‡Î KvR, W = Fs cos = 200  cos37  10 J = 1597.277 J (Ans.) 2q †ÿ‡Î KvR, W = Fs = 200  10 J = 2000 J (Ans.) 25| GKwU e ̄‘ mijc‡_ (3, 2, – 1) †_‡K (2, 1, – 4) we›`y‡Z †Mj| Gi Dci wμqvkxj ej 4 i  – 3 j  + 2k  | KvR wbY©q K‡iv| [Medium] mgvavb: r  = (2 – 3) i  + (1 – 2)j  – (4 – 1)k  = – i  – j  – 3k   KvR, W = F  . r  = – 4 + 3 – 6 J = – 7 J (Ans.) 26| 10 kg f‡ii GKwU w ̄’i e ̄‘i Ici GKwU w ̄’i gv‡bi ej cÖ‡qvM Kivi miY 5 s G 50 m n‡jv| K...ZKvR wbY©q K‡iv| [Easy] mgvavb: S = ut + 1 2 at2  a = 4 ms–2  W = Fs = mas = 10  4  50 J = 2000 J (Ans.) 27| GKwU Kv‡Vi eøK‡K Avbyf~wg‡Ki mv‡_ 30 †Kv‡Y 40 N ej cÖ‡qv‡M †g‡Si Dci w`‡q Uvbv n‡”Q| Nl©YRwbZ ej 8 N| eøKwUi miY 5 m n‡jÑ [Easy] K) cÖhy3 ej Øviv K...ZKvR L) Nl©Y ej Øviv K...ZKvR M) wbU KvR wbY©q K‡iv| mgvavb: K) W = Fs cos = 40  5  cos30 J = 173.21 J (Ans.) L) W = – Fs = – 8  5 J = – 40 J M) W = (173.21 – 40) J = 133.21 J (Ans.) 28| GKwU cv¤ú Øviv 1000 wjUvi cvwb 40 m Dc‡i GKwU U ̈vs‡K Zzj‡Z KZ kw3 e ̈q n‡e? GK Nb †mw›UwgUvi cvwbi fi 0.9985 g Ges 1000 Nb †mw›UwgUvi GK wjUvi| [Medium] mgvavb: W = mgh = vghJ = 106  0.9985  10–3  9.8  40 J = 391412 J (Ans.) 29| `vjv‡bi Qv‡`i mv‡_ jvMv‡bv 15 m `xN© GKwU gB Abyf~wg‡Ki mv‡_ 45 †Kv‡Y Av‡Q| 72 kg f‡ii GK e ̈w3 25 kg f‡ii GKwU †evSv wb‡q gB †e‡q 10 s mg‡q Qv‡` D‡V| †jvKwU KZ...©K K...ZKvR wbY©q K‡iv| H mg‡q Mo ÿgZv KZ wQj? [Medium] mgvavb: W = mgh = (72 + 25)  9.8  15 sin45 J = 10082.64 J (Ans.)
4  Physics 1st Paper Chapter-5  P = W t = 10082.64 10 W = 1008.264 W (Ans.) 30| GKwU B‡Ui D”PZv 7.5 cm Ges fi 2.5 kg| 10 wU BU‡K cici mvwR‡q GKwU ̄ͤ¢ •Zwi Ki‡Z KZ kw3 e ̈q n‡e? [Medium] mgvavb: W = n(n – 1) 2 mgh = 10(10 – 1) 2  2.5  9.8  7.5 100 J = 82.69 J (Ans.) Tricks: m~ÎwU wKfv‡e derive n‡q‡Q Zv †R‡b wb‡e| 31| GKwU Bwćbi cÖwZ NÈvq 37300 kg cvwb 18 m Dc‡i DV‡Z cv‡i| Bwćbi ÿgZv wbY©q K‡iv| [Easy] mgvavb: P = W t = mgh t = 37300  9.8  18 3600 W = 1827.7 W (Ans.) 32| `yBwU aaæe ej (i  + 2 j  + 3k  ) Ges (4 i  – 5 j  – 2k  ) GKwU KYvi Ici GKB mv‡_ KvR K‡i Ges KYvwUi 7k  Ae ̄’vb n‡Z 20 i  + 15 j  cm Ae ̄’vb miY nq| KYvi Ici †gvU K...ZKvR wbY©q Ki| [Medium] mgvavb: jwä ej, F  = F1  + F2  = 5 i  – 3 j  + k  N r  = 0.2 i  + 0.15 j  – 0.07k  m  W = F  .r  = 1 – 0.45 – 0.07 J = 0.48 J (Ans.) 33| ivqnvb 950 kg f‡ii GKwU Mvwo w`‡q 1 50 Xvj wewkó cvnvo 40 kmh–1 †e‡M DV‡Q| cvnv‡oi c„‡ôi mv‡_ Mvwoi PvKvi Nl©Y ̧Yv1⁄4 0.3, ivqnv‡bi fi 50 kg n‡j, ivqnv‡bi Dci cÖhy3 Awfj¤^ cÖwZwμqv ej Ges MvwowUi ÿgZv HP GK‡K cwigvc Ki| [Medium] mgvavb: mgcos  mg mgsin  R = mg cos = 50  9.8  cos     tan–1 1 50 N = 489.9 N (Ans.) P = Fv = (mgsin + kmgcos)  v = 1000  9.8(sin + 0.3cos) 40 3.6 W = 34837.9 W = 46.7 HP (Ans.) 34| GKwU cvwbc~Y© Kzqvi MfxiZv 12 m Ges e ̈vm 1.8 m| GKwU cv¤ú KzqvwU‡K 24 min G cvwbk~b ̈ Ki‡Z cv‡i| D3 Kv‡R 1 HP Gi AviI GKwU cv¤ú hy3 Kiv nj| wØZxq cv¤ú hy3 Kivq D3 Kv‡R KZ mgq mvkÖq n‡e? [Medium] mgvavb: P = W t = mgh t =      1.8 2 2  12  103  9.8  6 24  60  746 HP = 1.67 HP t2 = 1.67  24 (1.67 + 1) min = 15 min  mvkÖq n‡e = (24 – 15) min = 9 min (Ans.) 35| 10 wgUvi •`N© ̈, 5 wgUvi cÖ ̄’ I 3 wgUvi MfxiZv wewkó cvwb fwZ© †Kv‡bv cyKz‡ii 1 3 Ask cvwb k~b ̈ Ki‡Z 1 wU cv¤ú 2 NÈv mgq †bq| KZ ÿgZvi cv¤ú hy3 Ki‡j Aewkó Ask 2 NÈvq Lvwj Kiv hv‡e? [Medium] mgvavb: P = W t = Vgh t = 10  5  3  103  9.8  1 6  3  1 3 2  3600 W = 34.0278 W t = W P1 + P2  2  3600 = 10  5  2  3 3  103  9.8  2 34.0278 + P2  P2 = 238.19 W (Ans.)