Aspect Physics 1

ASPECT PHYSICS cÖ_g cÎ †f±i 95 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES A I B ‡f±i؇qi ci¯úi O we›`y‡Z ‡Kv‡Y wµqvkxj| G‡`i †¯‥jvi ¸Yb A.B Ges †f±i ¸Yb A × B Øviv cÖKvk Kiv nq| Z_¨ †_‡K 11 I 12 bs cÖ‡kœi DËi `vI: 11. A = – B n‡j A × B Gi gvb 0 (k~b¨) n‡e, hLb †f±i؇qi ga¨eZx© †KvY- [Bm&nvK, Avwgi] A. = 2 B. = C. > D. < 2 Ans B 12 wP‡Îi A = B ‡f±i `ywU- [Bm&nvK, Avwgi] A. WU ¸Yb wewbgq m~Î †g‡b P‡j B. µm ¸Yb wewbgq m~Î †g‡b P‡j C. WU I µm ¸Yb DfqB wewbgq m~Î †g‡b P‡j D. †KvbwUB bq Ans A 13. wP‡Î A I B ‡h Z‡j Av‡Q POQ ‡mB Zi‡ji Ici j¤^| A B Gi w`K-| [Bm&nvK, Avwgi] A. Gi w`‡K B. Gi w`‡K C. AGi mgvšÍivj D. B Gi mgvšÍivj Ans A 14. GKwU mvgvšÍwi‡Ki mwbœwnZ `ywU evû hw` `ywU †f±i Øviv wb‡`©wkZ nq Z‡e Gi †ÿÎdj- [Bm&nvK, Avwgi] A. †f±i `ywU †hvMd‡ji mgvb B. †f±i `ywUi we‡qvMd‡ji mgvb C. †f±i `ywUi †¯‥jvi ¸Yd‡ji mgvb D. †f±i `ywUi †f±i ¸Yd‡ji mgvb Ans D 15. †f±i Acv‡iUi¸‡jvi wb¤œiƒc •ewkó¨ i‡q‡Q- [Bm&nvK, Avwgi] i. †¯‥jvi ivwki †MÖwW‡q›U GKwU †f±i ivwk ii. †Kv‡bv †f±i †ÿ‡Îi WvBfvi‡RÝ k~b¨ n‡j †f±iwU AN~Y©bkxj iii. †Kv‡bv †f±‡ii Kvj© H †f±‡ii N~Y©b wb‡`©k K‡i wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans B 16. †Kv‡bv AšÍixKiY‡hvM¨ †f±i A‡cÿ‡Ki Kvj© n‡jv- [Bm&nvK, Avwgi] A. × V B. V C. .V D. V . Ans A 17. (j ) ˆ + k ˆ 2 †f±iwU GKwU- [Bm&nvK, Avwgi] A. k~b¨ †f±i B. GKK †f±i C. 2 gv‡bi †f±i D. †Kv‡bvwUB bq S B info A = j ˆ + k ˆ |A | = 1 2 + 12 = 2 A Gi GKK †f±i = A |A | = j ˆ + k ˆ 2 18. wZbwU †f±i A , B I C Ggb †h A × B + B × C + C × A = 0; †f±i wZbwU n‡jv- [Bm&nvK, Avwgi] A. GK‡iLxq B. GKZjxq C. GKZjxq bq D. †Kv‡bvwUB bq Ans B 19. A – B – C Ges A – B = C n‡j A I B †f±i `ywUi ga¨eZx© †KvY KZ? [Bm&nvK, Avwgi] A. 0° B. 30° C. 60° D. 90° Ans A 20. hw` †f±i Gi w`K eivei GKK †f±i nq, Zvn‡j- [Zcb, AvwRR] A. B. C. D. S A info GKK †f±i= †f±i †f±‡ii gvb 21. wb‡Pi †Kvb wµqvwU wewbgq m~Î †g‡b P‡j bv? [Zcb, AvwRR] A. †f±i ivwki WU ¸Yb B. †f±i ivwki µm ¸YY C. †f±i ivwki †hvM D. Dc‡ii †Kv‡bvwUB bq Ans B 22. †f±i Gi mgvšÍivj GKK †f±i [Zcb, AvwRR] A. B. C. D. S B info mgvšÍivj GKK †f±i= 23. n‡j Gi gvb n‡e [Zcb, AvwRR] A. –A 2 B. 0 C. –B 2 D. 1 S B info A = – B n‡j, A × B = B × B – B I B Gi ga¨ewZ© †KvY 180° – B × B = |B | . |B | × sin 180° = 0 24. †Kvb ej Øviv K…ZKvR, W = †Kvb GK †¶‡Î Ges k~b¨ bv n‡jI K…ZKvR k~b¨| G †_‡K Avgiv ej‡Z cvwi- [Zcb, AvwRR] A. Ges Gi w`K GKB B. Ges wecixZgyLx C. Ges ci¯ú‡ii Dci j¤^ D. Ges ci¯úi mgvšÍivj Ans C 25. †f±i A abvZ¥K X-Aÿ eivei Aew¯’Z| Ab¨ GKwU †f±i B Ggbfv‡e Aew¯’Z †hb A × B Gi gvb k~b¨ nq| Zvn‡j B n‡Z cv‡i- [Zcb, AvwRR] A. 4j ˆ B. –4j ˆ C. – (i ˆ + j ˆ ) D. (j ˆ + k ˆ ) S C info A × B Gi gvb k~b¨ n‡Z n‡j †f±i `yBwUi ga¨ewZ© †KvY 0° ev 180° n‡e| 26. [Zcb, AvwRR] A. B. C. D. 1 S C info (j ˆ + k ˆ ) × i ˆ = i ˆ × i ˆ = 0 OP OQ n ˆ A A A nˆ n ˆ A.|A| A A nˆ A A nˆ 2 A A n ˆ k ˆ j 2 ˆ i ˆ A 2 9 2 k ˆ 9 2 j ˆ 9 1 i - ˆ 3 2 k ˆ 3 2 j ˆ 3 1 i - ˆ 5 2 k ˆ 5 2 j ˆ 5 1 i - ˆ k ˆ j ˆ i - ˆ A 4 1 4 9 3 k ˆ 3 ˆ j 2 3 ˆ i 1 3 2 A A A B A B F.S F S F S F S F S F S i ? ˆ k ˆ j ˆ i ˆ 2 i ˆ 0

96 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 27. gvb k~b¨ bq G iKg GKwU †f±i‡K Zvi gvb w`‡q fvM Ki‡j Kx cvIqv hvq? [Zcb, AvwRR] A. GKK †f±i B. mgZjxq †f±i C. Ae¯
vb †f±i D. bvj †f±i Ans A 28. P Ges Q `ywU †f±i †Kv‡bv we›`y‡Z wµqv Ki‡j Zv‡`i jwäi gvb n‡e- [Zcb, AvwRR] i. P 2 + Q2 ii. P 2 + Q2 + 2PQ iii. P 2 + Q2 + 2PQ cos wb‡Pi †KvbwU mwVK? A. i B. i I ii C. ii I iii D. i, ii I iii Ans D 29. †Kv‡bv †f±i R †K hw` `ywU ci¯úi j¤^ Dcvs‡k wefvwRZ Kiv nq Zvn‡j R Gi mv‡_- [Zcb, AvwRR] i. †Kv‡Y Dcvs‡ki gvb X = R cos ii. (90° – ) †Kv‡Y Dcvs‡ki gvb Y = R sin iii. †Kv‡Y Dcvs‡ki gvb Y = R sin wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans A 30. 5N Gi GKwU ej‡K ci¯úi j¤^fv‡e `ywU Dcvsk OY I OX Øviv wb‡`©k Kiv †h‡Z cv‡i- [Mwb, mykvšÍ] Y 5N O X 3N OX Gi gvb KZ? A. 2N B. 3N C. 4N D. 5N S C info 5 2 = 3 2 + OX2 + 2 3 OX cos900 OX2=16 OX=4N 31. wP‡Î P we›`y‡Z 10 N Gi `ywU mgvb ej wµqv Ki‡Q| G‡`i ga¨eZ©x †Kvb 120°| [Mwb, mykvšÍ] 120° p 10N 10N jwä e‡ji gvb KZ ? A. 5N B. 10N C. 17N D. 20N S B info = 120 P = Q = R = 10N, = 2 , = 2 32. GKwU e¯‘i Avw` †eM u| Gi Dci GKwU aªæeK ej F, t †m‡KÛ a‡i cÖ‡qvM Kivq aªæe Z¡iY a m„wó n‡jv| ejwU Avw` †e‡Mi w`‡K nq| †kl †eM v wbY©‡qi Rb¨ wb‡P GKwU †f±i wPÎ AsKb Kiv n‡jv| [Mwb, mykvšÍ] u v x ‡f±i wP‡Î x Gi •`N©¨ KZ? A. F B. Ft C. at D. u + at ` S C info v = u + at, v = u + x x = at 33. GKwU b`x‡Z †¯ªv‡Zi †eM 5kmh–1 Ges GKwU †b․Kvi †eM 10kmh–1 | †¯ªv‡Zi mv‡_ KZ wWMÖx †KvY K‡i †b․Kv Pvjv‡j †b․KvwU Aci cv‡o wVK †mvRvmywR †cu․Qv‡e? [Mwb, mykvšÍ] A. 150° B. 100° C. 120° D. 130° S C info = cos –1 ( †mªv‡Zi †eM †b․Kvi †eM ) = cos –1 – 5 10 = 120 34. XOZ Z‡ji mgvšÍivj Ges 3 ˆ i + ˆ j + 4ˆk †f±‡ii mv‡_ j¤^ GKK †f±iwU n‡e- [Mwb, mykvšÍ] A. 4 ˆ i – 3 ˆk B. 1 5 (4 ˆ i – 3 ˆk) C. 1 5 (3 ˆ i – 4 ˆk) D. 3 ˆ i – 4 ˆk S B info XOZ Z‡ji mgvšÍivj Ges 3 ˆ i + ˆ j + 4ˆk †f±‡ii mv‡_ j¤^ †f±i = 4 ˆ i – 3 ˆk j¤^ GKK †f±i = 4 ˆ i – 3 ˆk 4 2 + (–3) 2 = 1 5 (3 ˆ i – 4 ˆk) wb‡Pi Z_¨ †_‡K wb‡Pi 35 I 36 bs cÖ‡kœi DËi `vI: GKwU mij †`vjK‡K Abyf~wgK ej H Øviv w¯
i Kiv n‡jv| wP‡Î †`vjK wc‡Ûi Dci ej¸‡jvi wµqv †`Lv‡bv n‡q‡Q| †`vj‡Ki myZvi Uvb T| †`vjK wc‡Ûi IRb W| ● 30° T H W 35. †Kvb Z_¨wU mZ¨- [Mwb, mykvšÍ] A. H = T cos 30° B. T = H sin 30° C. W = T cos 30° D. W = T sin 30° S C info Dj¤^ Dcvsk, H = T sin300 Abyf~wgK Dcvsk, W = T cos300 H 30 W T 36. GKwU †f±i A , B ev C †K m‡e©v”P KqwU Dcvs‡k fvM Kiv hvq? [Mwb, mykvšÍ] A. `yBwU B. AmsL¨ C. PviwU D. QqwU Ans B 37. A hw` GKwU †f±i nq Z‡e Gi mwVK †f±‡ii iæc bq †KvbwU? [†gvK‡Q` m¨vi] A. A B. A C. A D. |A| Ans D 38. j¤^ GKK †f±i ˆn = B A | | B A -Gi Rb¨ wb‡Pi †Kvb wPÎwU mwVK? [†gvK‡Q` m¨vi] n ˆ B A B. n ˆ A B n ˆ A B D. n ˆ B A Ans C

ASPECT PHYSICS cÖ_g cÎ †f±i 97 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 39. †¯‥jvi ev †f±i ¸Y‡b `yÕwU †f±‡ii AšÍfz©³ †Kv‡Yi †¶‡Î wb‡Pi †Kvb m¤úK©wU mwVK? [†gvK‡Q` m¨vi] A.0° ≤ ≤ 90° B.0° ≤ ≤ 180° C. 180° ≤ ≤ 0° D. 90° ≤ ≤ 0° Ans B 40. `yÕwU †f±‡ii mgwó I cv_©‡K¨i gvb GKB nq ZLb hLb †f±i `yÕwUi ga¨eZ©x †KvY- [†gvK‡Q` m¨vi] A. 0° B. 90° C. 180° D. 360° Ans B 41. hLb × A= 0 ZLb A-‡K ejv nq - [†gvK‡Q` m¨vi] A. AN~Y©bkxj †f±i B. AmsiÿYkxj †f±i C. mwjbqWvj D. Ai¶Ykxj †f±i Ans A c„w_exi N~Y©b A‡¶i mgvšÍivj `w¶Y ‡_‡K DËi w`‡K GKwU †f±i A Ges †Zvgvi Ae¯’v‡b Abyf~wg‡Ki mv‡_ j¤^ GKwU †f±i B| wb‡Pi cÖkœ¸‡jvi DËi `vI| 42. A B Gi w`K n‡e- [cÖvgvwbK m¨vi] A. c~e© w`‡K B. cwðg w`‡K C. DËi w`‡K D. Dc‡ii w`‡K S A info C = A B C ‡f±iwU c~e©w`‡K Aew¯
Z B A C c~e© (E) DËi (N) Dci 43. A. B m‡e©v”P n‡e hw` †Zvgvi Ae¯’vb- [cÖvgvwbK m¨vi] A. DËi †giæ‡Z nq B. `w¶Y †giæ‡Z nq C. welye †iLvi Dci nq D. 45° A¶vs‡k nq S A info B I A Gi ga¨eZ©x †Kvb 0° n‡j A. B m‡e©v”P n‡e| B Gi Ae¯
vb DËi †giæ‡Z n‡j A. B m‡e©v”P n‡e| 44. 3ms–1 †e‡M ‡`u․‡o hvevi mgq GKRb ‡jvK 6ms–1 †e‡M j¤^fv‡e cwZZ e„wói m¤§yLxb n‡jv| e„wó n‡Z i¶v †c‡Z n‡j Zv‡K KZ †Kv‡Y QvZv ai‡Z n‡e? [BKivg m¨vi] A. 20°34 B. 45° C. 30° D. 34°26 S Blank info = tan–1 3 6 = 26.56 = 2633 45. `ywU †f±i ci¯úi KZ wWwMÖ †Kv‡Y wµqv Ki‡j G‡`i †¯‥jvi ¸Ydj I †f±i ¸Yd‡ji gvb mgvb n‡e? [kvgmyi m¨vi] A. 0 wWwMÖ B. 45 wWwMÖ C. 90 wWwMÖ D. 180 wWwMÖ S B info | A. B|= | A B| ABcos = AB sin ev, sin/cos = 1 ev, tan = 1 = tan45° ev, = 45° STEP 4 MCQ CONCEPT TEST WRITTEN cÖkœ 01. wZbwU †f±i A , B I C Giƒc †hb A .B = 0 Ges A .C = 0, A †f±iwU †KvbwUi mgvšÍivj? A. B B. C C. B × C D. B . C 02. |a × b | + |a . b | = 144; |a | = 3 n‡j |b | = ? A. 16 B. 8 C. 3 D. 4 03. wb‡Pi †KvbwU †¯‥jvi ivwk? A. wefe B. cÖvej¨ C. †eM D. fi‡eM 04. `ywU mggv‡bi †f±i GKwU we›`y‡Z wµqvkxj| G‡`i jwäi gvb †h‡Kv‡bv GKwU †f±‡ii gv‡bi mgvb| †f±i؇qi ga¨eZx© †KvY KZ? A. 180˚ B. 0˚ C. 120˚ D. 90˚ 05. mggv‡bi `ywU †f±‡ii ga¨eZx© †KvY 70˚ n‡j, jwäi w`K n‡eA. 70˚ B. 30˚ C. 35˚ D. 45˚ 06. wb‡Pi †Kvb gv‡bi Rb¨ AGes B ci¯úi wecÖZxc †f±i? A. A = 2ˆ i, B = – 05ˆ i B. A = ˆ i, B =1 – ˆ i C. A = 2ˆ i, B = – 0.5ˆ i D. A = ˆ i, B = 0.5ˆ i 07. ˆ j.( 2 ˆ i – 3 ˆ j + ˆk) Gi gvb KZ? A. 2 B. 3 C. 1 D. 3 08. `ywU †f±‡ii †¯‥jvi ¸Ydj 18 GKK| G‡`i †f±i ¸Yd‡ji gvb 63 GKK| †f±i؇qi ga¨eZ©x †KvY KZ? A. 600 B. 900 C. 300 D. 1200 09. A , B I C wZbwU †f±i ivwk n‡j Ges C = A B n‡j C Gi w`K n‡e- A. A eivei B. B eivei C. A I BGi mgZ‡ji j¤^ eivei D. A I B Gi mgZj eivei 10. `ywU e‡ji jwäi m‡e©v”P gvb 14 N Ges me©wb¤œ gvb 2N| ej `ywU ci¯ú‡ii mv‡_ 90 ‡Kv‡Y †Kv‡bv GKwU KYvi Dci wµqv Ki‡j jwäA. 16 N B. 12 N C. 100 N D. 10 N 11. a= ˆ i + 2ˆ j + ˆk Ges b= 3ˆ i + ˆ j – 4 ˆk `ywU †f±i ivwk n‡j, | a– 2 b| = KZ? A. 104 B. 105 C. 106 D. 107 12. `ywU w`K ivwki cÖ‡Z¨KwUi gvb 10 GKK| G‡`i jwäi gvb 10 2GKK n‡j Zv‡`i ga¨eZx© †KvY KZ? A. 0 B. 60 C. 90 D.120 13. A , B , C wZbwU †f±i| A . B = 0 Ges A . C = 0 AZGe B Gi mgvšÍivj n‡eA. B C B. B C. C D. B. C 14. a Gi †Kvb gv‡bi Rb¨ A I B mgvšÍivj n‡e †hLv‡b A= 5ˆ i + 2ˆ j + 3ˆk Ges B = 15ˆ i + aˆ j + 9ˆk A. 3 B. 2 C. 6 D. 10 15. `yBwU ej hvi GKwU 10 wbDUb wewkó Ges ejØq 120˚ †Kv‡Y wµqv Ki‡j jwäi gvb D‡jøwLZ ejwUi mgvb nq, Aci ejwUi gvb KZ? A. 20 wbDUb B. 0 A_ev 10 wbDUb C. 15 wbDUb D. 5 wbDUb 16. A= ˆ i + 2ˆ j + ˆk †f±iwUi B = ˆ i + ˆ j †f±i Awfgy‡L AskK KZ? A. 3 B. 6 C. 3 2 D. 7 2 17. GKRb mvB‡Kj Av‡ivnx mgZj iv¯Ívi Dci w`‡q KZ †e‡M Pj‡j 6 ms–1 †e‡Mi e„wói †duvUv Zvi Mv‡q 450 †Kv‡Y co‡Q? A. 6 ms–1 B. 10 ms–1 C. 50 ms–1 D. 3 ms–1 18. GKwU †e‡Mi Avbyf~wgK I Dj¤^ Dcvs‡ki gvb h_vµ‡g 60ms-1 I 80ms-1 †eMwU KZ? A. 100ms-1 B. 120ms-1 C. 140ms-1 D. 100 2ms-1

98 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 19. `yÕwU †f±‡ii µm ¸Ydj k~b¨ n‡j †f±i `yÕwUi ga¨eZ©x †KvY n‡e – A. 180 B. 120 C. 90 D. 0 20. GKwU Kbvi Dci F = (5 ˆ i + 3ˆ j –2 ˆk) N ej cÖ‡qv‡M KbvwUi r= (3 ˆ i –2 ˆ j + ˆk) m mib nq| ej Øviv m¤úvw`Z KvR KZ? A. 7 joule B. 8 joule C. 4 joule D. 0 joule OMR SHEET 07. 14. 01. 08. 15. 02. 09. 16. 03. 10. 17. 04. 11. 18. 05. 12. 19. 06. 13. 20. WRITTEN PART 21. |A + B | = |A – B | n‡j A I B Gi ga¨eZ©x †KvY wbY©q K‡i †`LvI †h, †f±iØq ci¯úi j¤^| DËi:....................................................................................... 22. 3kg f‡ii GKwU MwZkxj KYvi MwZ‡eM v = 2 i + 2 j – k | KYvi Ae¯’vb †f±i r = i + j n‡j g~jwe›`y mv‡c‡ÿ Gi †K․wYK fi‡eM wbY©q Ki| DËi:....................................................................................... 23. Uªwj e¨v‡Mi nvZj j¤^v ivLv nq †Kb? e¨vL¨v K‡iv| DËi:....................................................................................... 24. `ywU mgvb †f±i‡K †hvM Ki‡j †Kvb Ae¯’vq I‡`i jwä GKwU †f±‡ii gv‡bi 2 ¸Y n‡e? 25. `ywU Amgvb mgRvZxq †f±‡ii jwä k~b¨ n‡Z cv‡i wKbv e¨vL¨v K‡iv| DËi:....................................................................................... 26. hw` A = 2x2 i + 3yzj – xz 2 k Ges = 2z – x 3 y nq, Zvn‡j (1, 1, –1) we›`y‡Z A , wbY©q Ki| 27. †f±i P = 2i – 3j + k -Gi Dci Q = 4i + 5k -Gi j¤^ Awf‡ÿ‡ci gvb wbY©q Ki| DËi:....................................................................................... 28. †b․Kvi ¸Y hZ j¤^ †b․Kv ZZ mvg‡bi w`‡K GwM‡q hvq- e¨vL¨v K‡iv| DËi:....................................................................................... 29. NÈvq 40km †e‡M c~e©w`‡K Pjgvb GKwU Mvwoi PvjK NÈvq 40 3km †e‡M GKwU UªvK‡K DËi w`‡K Pj‡Z †`L‡Z| UªvKwUi cÖK…Z †eM KZ Ges UªvKwU †Kvb w`‡K Pj‡Q? DËi:....................................................................................... 30. a, b I c wZbwU †f±i ivwk nIqv m‡Ë¡ a. b. c †Kvb †f±i ivwk bq †Kbe¨vL¨v Ki? DËi:....................................................................................... MCQ ANSWER ANALYSIS WRITTEN cÖkœ bs DËi e¨vL¨v 01 C A .B = 0 Ges A .C = 0 myZivs A †f±iwU B I C †f±‡ii Dci j¤^| B × C †f±iwUI B I C †f±‡ii Dci j¤^| A I B × C †f±iØq ci¯úi mgvšÍivj| 02 D a 2 b 2 cos2 + a2 b 2 sin2 = 144 a 2 b 2 = 144 b = 144 a 2 = 144 9 = 16 = 4 03 A 04 C R = R 2 + R2 + 2 R.R.cos R 2 = 2R2 + 2R2 + 2R2 cos 2R2 cos = – R 2 cos = – 1 2 = 120˚ 05 C tan = P sin70˚ P + P cos70˚ = sin 70˚ 1 + cos 70˚ = tan–1 (0.7) = 35˚ 06 C AGes B ci¯úi wecÖZxc †f±i n‡j, A = 2ˆ i, B = 0.5ˆ i 07 D A. B = AxBx + AyBy + AzBz = 0.2 + 1.(–3)+ 0.1 = 3 08 C ABcos = 18; ABsin = 63 ABsin ABcos = 6 3 18 = tan–1 6 3 18 = 30˚ Ans. 09 C `ywU ‡f±‡ii µm ¸b‡bi d‡j m„ó jwä ‡f±‡ii w`K †f±i `ywUi mgZ‡ji j¤^ w`‡K Ae¯
vb K‡i| 10 D ejØq j¤^fv‡e wµqv Ki‡j jwäi gvb, R = R 2 max + R2 max 2 = 142 + 22 2 = 10N 11 C 2b = 2 × (3 ˆ i + ˆ j –4 ˆk) = 6 ˆ i + 2ˆ j – 8 ˆk( a– 2 b) = –5 ˆ i + 9ˆk; gvb = 5 2 + 9 = 106 12 C R 2 = P2 + Q2 + 2PQcos (10 2 2 ) = 102 + 102 + 2 102 cos = cos–1 (0) = 90 13 C A. B = 0 A B Ges A. C = 0 A B AZGe, B Gi mgvšÍivj n‡e C A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D A B C D

ASPECT PHYSICS cÖ_g cÎ †f±i 99 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES cÖkœ bs DËi e¨vL¨v 14 C 2 a = 3 9 3a = 18 a = 6 15 A P 2 = P2 + Q2 + 2PQcos120 P 2 = P2 + Q2 – PQ Q 2 – PQ = 0 Q (Q – P) = 0 nq, Q = 0 A_ev, Q = P = 10 16 C AskK Gi gvb, Acosθ = A .Bˆ | B| = 1 + 2 + 0 1 2 + 12 = 3 2 17 A = tan–1 u v 450 = tan-1 u 6 u = 6ms1 18 A GLv‡b, v = VA 2 + VB 2 = 602 + 802 = 100ms–1 19 D AB 0 1 0 ABsin 0 sin (0) 0 20 A w = F. r = (5 i + 3 j – 2 k) (3 i – 2 j + k) = 15 – 6-2 = 7 Joule 21 g‡b Kwi, A I B Gi ga¨eZ©x †KvY ; |A + B | = |A – B | ; ev, |A + B | 2 = |A – B | 2 ; ev, (A + B ) . (A – B ) = (A – B ) . (A – B ); ev, A . A + B . A + A . B + B . B = A . A – B . A – A . B – B . B ; ev, 2(A . B ) = 2(A . B ) ( A . B = B . A ) ev, 4(A . B ) = 0; ev, A . B = 0, (A 0, B 0); ev, AB cos = 0 cos = 0 ; ev, = 90 (A 0, B 0) †f±i؇qi ga¨eZ©x †KvY 90; Kv‡RB †f±i `ywU ci¯úi j¤^| 22 Avgiv Rvwb, •iwLK fi‡eM, P = m v ; P = 3(2i + 2j – k ) = 6i + 6j – 3k ; Avevi, †K․wYK fi‡eM, L = r P ; L = (i + j ) (6i + 6j – 3k ) = i 1 6 j 1 6 k 0 –3 = i (– 3 – 6 0) + j {6 0 – 1 (–3)} + k (6 – 6) = –3i + 3j 23 Uªwj e¨v‡Mi nvZj Øviv Uªwj e¨vM‡K mvg‡bi w`‡K †U‡b wb‡q hvIqvi mgq nvZ‡j cÖhy³ ej `yBwU Dcvs‡k wef³ nq| GKwU Fsin Ges AciwU Fcos| Fsin DcvskwU Dc‡ii w`‡K Kvh©iZ nq, Ges Fcos DcvskwU e¨vM‡K mvg‡bi w`‡K GwM‡q wb‡q hvq| nvZj j¤^v n‡j Gi gvb Kg nq| G Ae¯
vq cos Gi gvb †ewk nq Ges Uªwji †eM aªæe †i‡L Uvb‡Z Kg ej jv‡M| G Kvi‡Y Uªwj e¨v‡Mi nvZj j¤^v ivLv nq| 24 aiv hvK, mgvb †f±i؇qi cÖ‡Z¨‡Ki gvb A Ges G‡`i AšÍfz©³ †KvY |; cÖkœvbymv‡i, G‡`i jwä †f±‡ii cig gvb = 2 A ( 2A)2 = A2 + A2 + 2A.A.cos; ev, 2A2 = 2A2 + 2A2 .cos; ev, cos = 0 = cos90; = 90 myZivs, †f±i؇qi AšÍfz©³ †KvY 90 n‡j Zv‡`i jwä GKwU †f±‡ii gv‡bi 2 ¸Y n‡e| 25 `yBwU Amgvb mgRvZxq †f±‡ii jwä k~b¨ n‡Z cv‡i bv| `yBwU †f±‡ii hw` gvb mgvb nq Ges Zv‡`i w`K hw` wecixZ nq Z‡e Zv‡`i jwä k~b¨ nq| †hgb : A + (– A) = 0| wKš‧ `yBwU Amgvb wKš‧ mgRvZxq †f±i †hvM Ki‡j Zv‡`i mgwó A_©vr jwä k~b¨ n‡Z cv‡i bv| 26 = x i + y j + z k (2z – x3y) = – 3x2 yi – x 3 j + 2k ; GLb, A . = (2x2 i + 3yz j – xz 2 k ) . (– 3x2 yi – x 3 j + 2k ) = – 6x4 y – 3x3 yz – 2xz2 ; (1, 1, –1) we›`y‡Z A . = –6(1)4 (1) – 3(1)3 (1) (–1) – 2(1) (–1)2 = – 6 + 3 – 2 = – 5 27 P I Q -Gi ga¨eZ©x †KvY n‡j P -Gi Dci Q -Gi j¤^ Awf‡ÿc = Qcos; P .Q = PQ cos; Q cos = P . Q P P .Q = (2i – 3j + k ). (4i + 5k ) = 2 0 + (–3) (4) + (1) (5) = –7 Ges | P | = (2) 2 + (–3) 2 + (1) 2 = 14; Q cos = –7 14 28 †b․Kvi MjyB‡q ¸Y †e‡a F e‡j Uvb‡j F ejwU †b․Kvi MwZi w`‡Ki mv‡_ ‡KvY Drcbœ K‡i| GB F e‡ji †b․Kvi MwZ eivei Dcvsk n‡jv F cos Ges Djø¤^ Dcvsk n‡jv F sin | F sin DcvskwU †b․Kvi nvj Øviv cÖkwgZ n‡q hvq| †Kej gvÎ F cos Dcvsk †b․Kvi MwZi mÂvi K‡i| iwk Z_v ¸Y hZ j¤^v nq Gi gvb ZZ Kg nq Ges F cos Gi gvb ZZ e„w× cvq| d‡j †b․Kvi MwZ †e‡o hvq Ges †b․Kv ZZ `ªæZ mvg‡bi w`‡K GwM‡q hvq| 29 g‡b Kwi UªvKwU DËi w`‡Ki mv‡_ †Kv‡Y c~e©w`‡K Pj‡Q| wÎfzR m~Îvbymv‡i Avgiv cvB, V T = V TC + V C; V 2 T = V 2 TC + V 2 C ; ev, VT = V 2 TC + V 2 C= (10 3) 2 + (40) 2 = 402 (4) = 2 40 = 80 kmh–1 Avevi, tan = VC VTC = 40 40 3 = 1 3 = tan30 = 30 30 a, b, c Øviv wZbwU c„_K c„_K ‡f±i ivwk‡K wb‡`©k K‡i hv‡`i gvb I w`K DfqB Av‡Q| Ab¨w`‡K a. b. c Øviv †f±‡ii WU ¸Y wb‡`©k K‡i| Avevi Avgiv Rvwb, †f±‡ii WU ¸‡Yi Øviv cÖvß ivwkwU GKwU †¯‥jvi ivwk| GRb¨B a. b. c GKwU †¯‥jvi ivwk|

144 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES SURVEY TABLE  Kx coe? †Kb coe? †Kv_v n‡Z coe? KZUzKz coe?  TOPICS MAGNETIC DECISION [hv co‡e] MAKING DECISION [†h Kvi‡Y co‡e] VVI For This Year DU JU RU CU GST Engr. HSC Written MCQ THEORY CONCEPT-01 ej I wewfbœ cÖKvi e‡ji cv_©K¨ 10% 20% 10% 10% 30% - 15% - - CONCEPT-02 wbDU‡bi MwZm~Îmg~n Ges Gi mxgve×Zv 5% 10% 5% 5% 20% - 25% - CONCEPT-03 RoZv, NvZej Ges i‡K‡Ui MwZ msµvšÍ Z_¨vejx 20% 30% 10% 10% 40% 5% 60% CONCEPT-04 ¸iæZ¡c~Y© wKQz ivwki GKK Ges gvÎvmg~n 30% 40% 20% 30% 50% 10% 60% MvwYwZK cÖ‡qvM CONCEPT-01 ej, fi I Z¡iY msµvšÍ MvwYwZK cÖ‡qvM 75% 80% 70% 75% 65% 80% 60% CONCEPT-02 wjdU msµvšÍ MvwYwZK cÖ‡qvM 60% 85% 65% 60% 70% 70% 30% - CONCEPT-03 i‡KU msµvšÍ MvwYwZK cÖ‡qvM 75% 70% 65% 70% 80% 80% 75% CONCEPT-04 fi‡e‡Mi msiÿY m~Î msµvšÍ MvwYwZK cÖ‡qvM 80% 80% 70% 75% 90% 65% 60% CONCEPT-05 Nl©Y msµvšÍ MvwYwZK cÖ‡qvM 45% 60% 65% 70% 70% 60% 55% CONCEPT-06 NvZ ej I e‡ji NvZ msµvšÍ MvwYwZK cÖ‡qvM 60% 65% 55% 70% 65% 40% 60% CONCEPT-07 RoZvi åvgK msµvšÍ MvwYwZK cÖ‡qvM 40% 70% 65% 60% 80% 80% 75% CONCEPT-08 †K›`ªgyLx ej I Z¡iY msµvšÍ MvwYwZK cÖ‡qvM 70% 05% 70% 75% 70% 55% 70% CONCEPT-09 †K․wYK †eM I •iwLK †eM msµvšÍ 60% 70% 65% 70% 60% 35% 60% CONCEPT-10 †K․wYK MwZkw³ I †gvU kw³ msµvšÍ MvwYwZK cÖ‡qvM 50% 50% 55% 50% 40% 80% 75% CONCEPT-11 e¨vswKs msµvšÍ MvwYwZK cÖ‡qvM 70% 60% 50% 55% 70% 45% 70% DU. = Dhaka University, JU. = Jahangirnagar University, RU. = Rajshahi University, CU = Chittagong University, GST = General, Science & Technology, Engr. = Engineering. STEP 1 mvRv‡bv me Z_¨ THEORY fwZ© †ivMxi c_¨ Concept 1 ej Ges wewfbœ cÖKvi e‡ji cv_©K¨ ✅ ej : hv w¯’i e¯‘i Dci wµqv K‡i Zv‡K MwZkxj K‡i ev Ki‡Z Pvq ev hv MwZkxj e¯‘i Dci wµqv K‡i Zvi MwZi cwieZ©b K‡i ev Ki‡Z Pvq Zv‡K ej e‡j| ✅ e‡ji •ewkó¨: (1) w`K Av‡Q, (2) †Rvovq †Rvovq wµqv K‡i, (3) †Kvb e¯‘‡Z Z¡iY m„wó Ki‡Z cv‡i, (4) †Kvb e¯‘‡K weK…Z Ki‡Z cv‡i| ✅ e‡ji GKK: wbi‡c¶ AwfKl© c×wZi bvg GKK c×wZi bvg GKK C.G.S dyne C.G.S g-wt = 980 dyne S.I / M.K.S Newton S.I / M.K.S kg-wt = 9.8N F.P.S poundal F.P.S lb-wt = 32 poundal ✅ †g․wjK ej 4 cÖKvi: (1) gnvKl© ej (2) Zwor Pz¤^Kxq ej (3) mej wbDwK¬q ej (4) `ye©j wbDwK¬q ej ✅ Pvi cÖKvi †g․wjK e‡ji Zzjbv: welq gnvKl© ej `ye©j wbDwK¬qvi ej Zwor Pz¤^Kxq ej mej wbDwK¬qvi ej Kvh©KiY MÖvwfUb bvgK KYvi cvi¯úwiK wewbg‡qi d‡j Intermediate vector bosons ‡dvUb (n) ‡gmb ag© AvKl©Y ag©x weKl©Y ag©x AvKl©Y ag©x I weKl©Y ag©x AvKl©Y ag©x cvjøv Amxg G e‡ji gvb KLbI k~b¨ nq bv 10–16m Gi Kg| Z‡e 10–15 m Gi †ekx `~i‡Z¡ G ej Abyf~Z nq bv Amxg 10–15m Z‡e 10–14m `~i‡Z¡ G ej D‡c¶bxq Av‡cw¶K mejZv 1 1030 1039 (kv. Zcb) 1040 (Zdv¾j) 1041 (kv. Zcb) 1042 (Zdv¾j) D`vniY B‡jKUªb I †cÖvU‡bi ga¨Kvi ej (K) wbDwK¬q weUv fv½‡bi Rb¨ `vqx| (L) AwaKvsk †ZRw¯ŒqZv fv½vi Rb¨ (K) w¯’wZ¯’vcK ej; (L) AvYweK MVb; (M) ivmvqwbK wewµqv; (N) P›`ª I m~‡h©i ga¨Kvi ej, B‡jKUªb I †cÖvU‡bi ga¨Kvi ej; (O) w¯cÖs Gi ga¨Kvi ej (K) †cÖvUb I wbDUªb‡K Ave× K‡i wbDwK¬qvm •Zix| Note: (i) †g․wjK ejmg~‡ni mwµqZvi µg: mej wbDwK¬q ej Zwor Pz¤^Kxq ej `ye©j wbDK¬xq ej gnvKl© ej| (ii) cÖ‡dmi Avãym mvjvg, IqvBb evM© I Møv‡mv wZbRb M‡elYv K‡i `ye©j wbDwK¬q ej I Zwor Pz¤^Kxq e‡ji g‡a¨ m¤úK© ¯’vcb K‡i‡Qb, hv mvjvg I IqvBbevM© ZË¡ bv‡g cwiwPZ| wbDUwbqvb ejwe`¨v NEWTONIAN MECHANICS cÖ_g cÎ 04 Aa¨vq c„ôv 144

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 145 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ Zwor Pz¤^Kxq e‡ji D`vniY: N‡i Av‡mv N †i Av †mv Nl©Y ivmvqwbK wewµqv AvYweK MVb ¯úk© ej, w¯’wZ¯’vcK ej ✅ ¯úk© ej: †h ej m„wói Rb¨ `ywU e¯‘i cÖZ¨ÿ ms¯úk© cÖ‡qvRb Zv‡K ¯úk© ej e‡j| ¯úk© e‡ji D`vniY: N‡lwUi ¯úk© N †l wU i ¯úk© Nl©Y ej mv›`ª ej, msN‡l©i d‡j m„ó ej Uvbv ej ¯úk© ej REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. Higgs cÖwµqv GK ai‡bi- [DU. 12-13] A. fi •Zwii cÖwµqv B. kw³ •Zwii cÖwµqv C. B‡jKUªb •Zwii cÖwµqv D. ej •Zwii cÖwµqv Ans A 02. †KvbwU me©v‡c¶v `~e©j ej-[DU. 03-04, KU. 11-12, AFMC. 20-21 ; wm. †ev. 2016] A. gnvKl© ej B. †P․¤^K ej C. Zwor †P․¤^K ej D. wbDwK¬q ej PviwU †g․wjK e‡ji Av‡cw¶K mejZv Zzjbv Ki‡j †`Lv hvq, gnvKl© ej me‡P‡q `ye©j Ges me‡P‡q kw³kvjx ej n‡”Q mej wbDwK¬q ej| 03. evqy‡Z GK ev· Zzjvi IRb Ges 4wU †jvnvi e‡ji IRb cÖwZ‡ÿ‡Î wVK 1kg †`Lv †M‡j- [DU. 03-04] A. Zzjvi fi I †jvnvi ej¸‡jvi fi mgvb B. Zzjvi fi †ewk C. f‡ii Zzjbv e‡ji msL¨vi Dci wbf©i Ki‡e D. Zzjvi fi Kg Ans A STEP 02 ANALYSIS OF JU QUESTION 01. `ye©j wbDwK¬q ej m„wó nq wK‡mi Rb¨? [JU. 18-19] A. weUv ÿq B. †cÖvUb ÿq C. Mvgv ÿq D. wbDUªb ÿq Ans A 02. wm.wR.Gm c×wZ‡Z fi‡e‡Mi GKK n‡jv- [JU. 09-10] A. MÖvg/†m‡KÛ B. MÖvg-†m‡KÛ C. MÖvg-†mw›UwgUvi/†m‡KÛ D. MÖvg/†mw›UwgUvi-†m‡KÛ Ans C 03. GK wbDUb ej ej‡Z wK eyS? [JU. 09-10] A. 1kg1m/sec2 B. 1kg10m/sec2 C. 10kg10m/sec2 D. Dc‡ii †KvbwUB bq Ans A STEP 03 ANALYSIS OF RU QUESTION 01. wb‡Pi ej¸wji g‡a¨ †KvbwU me‡P‡q kw³kvjx ej? [RU :18-19] A. gnvKl© ej B. Zwor Pz¤^Kxq ej C. mej wbDwK¬q ej D. `ye©j wbDwK¬q ej Ans C 02. wbDwK¬qv‡mi g‡a¨ †Kvb KYvi cvi¯úwiK wewbg‡qi Øviv mej wbDwK¬q e‡ji DrcwË nq? [RU. 16-17] A. MÖvwfUb B. wbDwUª‡bv C. †gmb D. B‡jKUªb Ans C 03. 1 wWMÖx mgvb KZ †iwWqvb? [RU. 16-17] A. 0.00165 B. 0.0075 C. 0.0165 D. 0.0175 S D info 360 wWMÖx = 2π †iwWqvb ; 1˚= 180 = 0.0175 †iwWqvb 04. †KvbwU †g․wjK ej bq? [RU. 16-17] A. kw³kvjx wbDwK¬q ej B. gnvKl© ej C. Zwor †P․¤^K ej D. ZworPvjK ej S D info †g․wjK ej 4 wU | h_v : gnvKl© ej, Zwor †P․¤^K ej, `ye©j wbDwK¬q ej, mej wbDwK¬q ej| 05. †Kvb e„‡Ëi e¨vmv‡a©i mgvb e„ËPvc e„‡Ëi †K‡›`ª †h †KvY Drcbœ K‡i Zv‡K e‡j-[RU. 16-17] A. †iwWqvb B. 1 wWMÖx C. 1 †iwWqvb D. †KvbwUB bq Ans C 06. nvZNwoi N›Uvi KuvUvi ch©vqKvj KZ? [RU. 16-17] A. 24 N›Uv B. 12 N›Uv C. 6 N›Uv D. 1 N›Uv Ans B 07. ‡g․wjK e‡ji msL¨v KqwU? [RU. 15-16] A. 3 B. 4 C. 5 D. 2 Ans B 08. wb‡Pi †KvbwU msiÿYkxj e‡ji †ÿ‡Î AmZ¨ bq? [RU. 14-15] A. kw³i AcPq N‡U B. K…ZKvR cybiæ×vi Kiv Am¤¢e bq C. hvwš¿K kw³i msiÿY m~Î Lv‡U bv D. †KvbwUB bq Ans B 09. DcMÖn gnvk~‡b¨ †Kvb bxwZ †g‡b P‡j? [RU. 10-11] A. wbDU‡bi Z…Zxq m~Î B. AwfKl© m~Î C. †Kcjv‡ii m~Î D. †KvbwUB bq Ans C 10. GKwU Nwoi wgwb‡Ui KuvUvi cÖvšÍ Ges ga¨ ¯Í‡ii †K․wYK †eM- [RU. 09-10] A. cÖv‡šÍ †ekx B. ga¨¯’‡j †ekx C. ga¨¯’‡j Kg D. Dfq ¯’‡jB mgvb Ans D STEP 04 ANALYSIS OF CU QUESTION 01. GKwU PjšÍ evBmvB‡K‡ji N~Y©biZ PvKvi †ÿ‡Î wb‡¤œi †KvbwU mwVK n‡e? [CU-A, Set-3. 20-21] A. Abyev`g~jK MwZkw³ B. N~Y©gvb MwZkw³ C. w¯’wZ kw³ D. Abyev`g~jK Ges N~Y©gvb MwZkw³ Df‡qB S D info Abyev`g~jK MwZkw³ gv‡b •iwLK MwZkw³| GKwU e¯‘i •iwLK †eM _vK‡j Zvi Abyev`gyjK MwZkw³ _vK‡e| 02. †KvbwU msi¶Ykxj ej bq? [CU-A, Set-2. 19-20] A. Zwor ej B. AwfKl© ej C. mv›`ª ej D. †P․¤^K ej Nl©Y ej I mv›`ª ej AmsiÿYkxj ej| 03. †K․wYK Z¡i‡Yi GKK Kx? [CU. 15-16] A. rad/min B. rad/s C. rad/s2 D. n/s Ans C 04. gnvKl© ej wbDK¬xq e‡ji Zzjbvq KZ¸Y Zxeª? [CU. 14-15; 12-13; CoU: 14-15] A.1042 B. 1030 C. 1042 D. 1040 E. 1040 Ans C 05. Z¡i‡Yi d‡j e¯‧i MwZkw³ I fi‡eM- [CU. 05-06] A. n«vm cvq B. e„w× cvq C. AcwiewZ©Z _v‡K D. wظY nq E. wZb¸Y nq P = mv = mat Avevi, Ek = 1 2 mv2 = 1 2 m(at)2 06. †cÖvUb I B‡jKUª‡bi g‡a¨ AvK©l‡Yi Rb¨ †Kvb †g․wjK ejwU `vqx? [CU. 04-05] A. kw³kvjx wbDwK¬q ej B. ga¨vKl©Y ej C. `~e©j wbDwK¬q ej D. Zwor Pz¤^Kxq ej E. Zwor PvjK ej e¯‘Z Av‡cw¶K MwZ‡Z cwiågYiZ `ywU AvwnZ KYvi g‡a¨ wµqvkxj ejB n‡”Q Zwor Pz¤^Kxq ej| 07. e¯‧i Dci wµqviZ †gvU ej k~b¨ n‡j e¯‧wUi- [CU. 03-04] A. Z¡iY k~b¨ n‡e B. †eM k~b¨ n‡e Ans A C. MwZkw³ k~b¨ n‡e D. cÖhy³ ej KZ…©K K…Z KvR k~b¨ n‡e

146 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. w¯
wZ¯
vcK e‡ji aib wK? [JnU. 12-13] A. Amsi¶Ykxj ej B. msi¶Ykxj ej C. hvwš¿K ej D. kvwãK ej Ans B 02. `ye©j wbDwK¬q ej Ges ZworPz¤^Kxq e‡ji GKxf~Z ZË¡ Avwe®‥vi K‡ib- [JnU. 10-11] A. g¨v·I‡qj B. mvjvg, I‡qBbevM© Ges Møv‡mv C. AvBb÷vBb D. wbDUb Ans B KU 01. `ywU ej mvg¨ Ae¯
v cÖwZôv Ki‡Z PvB‡j wK n‡e? [KU. 15-16] A. ej `ywUi gvb mgvb Ges Zv‡`i wecixZgyLx n‡Z n‡e B. ej `ywU‡K `ywU mij‡iLv eivei wµqvkxj n‡Z n‡e C. ej `ywUi gvb mgvb Ges Zv‡`i mggyLx n‡Z n‡e D. ej `ywUi jwä 1 Gi AwaK n‡Z n‡e Ans A 02. cÖ‡dmi Avãym mvjvg I w÷‡db IqvBb evM© †Kvb ej `ywU‡K GKxf~Z K‡ib?[KU. 12-13] A. wek¦Rbxb gnvKl© I Zwor †P․¤^K ej B. `ye©j wbDwK¬q ej I mej wbDwK¬q ej C. Zwor Pz¤^K ej I mej wbDwK¬q ej D. Zwor Pz¤^K ej I `~e©j wbDwK¬q ej mvjvg, IqvBb evM© I Møv‡mvi mw¤§wjZ cÖ‡Póvq `ye©j wbDwK¬q ej I Zwor Pz¤^K e‡ji g‡a¨ m¤úK© ¯’vwcZ K‡ib| CoU 01. wbDUbxq ev wPivwqZ ejwe`¨vi †g․wjK ivwk bq †KvbwU? [CoU.18-19] A. ¯’vb B. mgq ev Kvj C. †eM D. fi Ans C 02. gnvKl© e‡ji evnK †KvbwU? [CoU. 17-18] A. †evmb B. MÖvwfUb C. †gmb D. ‡Kvqv›Ub Ans B 03. wbDwK¬q‡bi g‡a¨ †Kvb KYvi cvi¯úwiK wewbg‡qi Øviv mej wbDwK¬q ej wµqvkxj nq? [CoU. 17-18] A. weUv KYv B. †gmb KYv C. †evmb KYv D. Avjdv KYv Ans B IU 01. cvwLi Dov ch©‡eÿY K‡i D‡ovRvnv‡Ri GKwU g‡Wj •Zwi K‡ib †K? [IU. 15-16] A. †Rgm IqvU B. wKwðqvb C. wjIbv‡`©v `v wfw D. Avj nv‡Rb Ans C 02. e¯‧‡K †Kvb A¶‡K †K›`ª K‡i Nyiv‡j †h UK© Drcbœ nq Zvi Rb¨ g~jZ `vqx- [IU. 13-14] A. †K․wYK Z¡iY B. •iwLK Z¡iY C. fi D. †eM Ans A 03. cošÍ e¯‧i †¶‡Î hvwš¿K kw³i msi¶YkxjZvi bxwZ- [IU. 12-13] A. cÖ‡hvR¨ B. cÖ‡hvR¨ bq C. †h †Kvb †¶‡Î cÖ‡hvR¨ D. †KvbwUB bq Ans A BRUR 01. †Kvb KYvi Dci wbU UK© k~b¨| KYvwUi †K․wYK †eM †Kgb n‡e? [BRU. 13-14] A. evo‡e B. Kg‡e C. GKB _vK‡e D. †KvbwUB bq Ans C PART B Analysis of Science & Technology Question SUST 01. 2.010-10 m `~i‡Z¡ Aew¯
Z `ywU B‡j±ª‡bi g‡a¨ gnvKl© ej Ges Zwor ej DfqB wµqv K‡i| AwfKl© e‡ji gvb Zwor e‡ji †P‡q KZ¸Y Kg ev †ewk kw³kvjx ? [SUST. 14-15] A. 1042 ¸Y Kg B. 10-42 ¸Y Kg C. 1042 ¸Y ‡ekx D. 10-42 ¸Y †ekx E. 1011 ¸Y †ekx Ans D 02. w¯
wZ¯
vcK msN‡l© msiwÿZ _v‡K- [SUST. 08-09] A. MwZkw³ I w¯’wZkw³ B. fi‡eM I MwZkw³ C. fi‡eM I w¯’wZkw³ D. †KvbwUB bq Ans B 03. †KvbwU mwVK? [SUST. 06-07] A. msiÿYkxj e‡ji mv‡_ w¯’wZkw³ RwoZ Kiv hvq bv B. AmsiÿYkxj e‡ji mv‡_ w¯’wZkw³ RwoZ Kiv hvq C. msiÿYkxj e‡ji mv‡_ w¯’wZkw³ RwoZ Kiv hvq D. msiÿYkxj I AmsiÿYkxj Dfq cÖKvi e‡ji mv‡_ w¯’wZkw³ RwoZ Kiv hvq Ans C JUST 01. †KvbwU †g․wjK ej bq? [JUST. 14-15, MBSTU. 15-16] A. gnvKl© ej B. AwfKl© ej C. Zwor †P․¤^K ej D. mej wbDwK¬q ej E. `ye©j wbDwK¬q ej Ans B MBSTU 01. †K›`ªgyLx ej Øviv m¤úvw`Z KvRÑ [MBSTU-A, Set-2 19-20] A. m‡e©v”P B. b~¨bZg C. FbvZ¥K D. k~b¨ mi‡Yi w`K cÖwZgyn~‡Z© †K›`ªgyLx e‡ji j¤^ eivei| BSMRSTU 01. abPvR© enbKvix †cÖvUb¸‡jv †Kvb e‡ji Kvi‡Y GK‡Î wbDwK¬qv‡m Ae¯
vb K‡i? [BSMRSTU-H, 19-20] A. gnvKl©xq ej B. Zwor Pz¤^Kxq ej C. mej wbDK¬xq ej D. wbDwK¬q `ye©j ej mej wbDwK¬q ej †cÖvUb I wbDUªb‡K Ave× K‡i wbDwK¬qvm •Zwi K‡i Ges `~e©j wbDwK¬q ej weUv ÿ‡qi Rb¨ `vqx| NSTU 01. AmsiÿYkxj e‡ji D`vniY †KvbwU?[NSTU-A, 19-20 ; BSMRSTU-H, 19-20] A. Nl©Y ej B. •e`y¨wZK ej C. Pz¤^K ej D. AwfKl©R ej Nl©b ej I mv›`ª ej AmsiÿYkxj ej| AwfKl©xq ej, •e`y¨wZK ej, Zwor ej, †P․¤^K ej, gnvKl© ej, Av`k© w¯úªs Gi weK…wZ cÖwZ‡ivax ej n‡jv msiÿYkxj ej| BSFMSTU 01. wbDwK¬qvm MV‡bi Rb¨ `vqx ej †KvbwU? [BSFMSTU-A, 19-20] A. ZworPz¤^Kxq ej B. `ye©j wbDwK¬q ej C. mej wbDwK¬q ej D. gnvKl© ej mej wbDK¬xq ej †cÖvUb I wbDUªb‡K Ave× K‡i wbDwK¬qvm •Zix K‡i| Ges `~e©j wbDK¬xq ej weUv fv½‡bi Rb¨ `vqx| STEP 06 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. †Kvb ai‡bi KYvi wewbg‡qi d‡j gnvKl© ej wµqvkxj nq? [MAT. 16-17] A. wbDUªb B. †gmb C. MÖvwfUb D. †dvUb Ans C DAT 01. †g․wjK ej¸wji g‡a¨ me©v‡cÿv kw³kvjx ej †KvbwU? [DAT; 19-20] A. gnvKl© ej B. `ye©j wbDwK¬q ej C. mej wbDwK¬q ej D. ZvwoZ‡P․¤^K ej gnvKl© e‡ji mv‡c‡ÿ †g․wjK ej¸wji Av‡cwÿK mejZv n‡jv: mej wbDwK¬q ej (1041) ZvwoZ‡P․¤^K ej (1039) `ye©j wbDwK¬q ej (1030) gnvKl© ej (1) 02. ej I e‡ji wµqvKv‡ji ¸Ydj‡K wK e‡j? [DAT. 18-19] A. e‡ji NvZ B. åvgK C. NvZ ej D. kw³ Ans A 03. AvYweK MV‡bi Rb¨ `vqx ej †KvbwU? [DAT. 17-18; w`. †ev. 2015] A. gnvKl© ej B. `ye©j wbDwK¬q ej C. mej wbDwK¬q ej D. Zwor Pz¤^Kxq ej Ans D 04. †Kvb KYvi wewbg‡qi Kvi‡Y Zwor †P․¤^Kxq ej wµqvkxj nq? [DAT. 16-17] A. †evmb B. †dvUb C. †gmb D. MÖvwfUb Ans B AFMC 01. †KvbwU me‡P‡q `~e©j ej †KvbwU? [AFMC. 2020-21] A. mej wbDK¬xq ej B. `~e©j wbDK¬xq ej C. gnvKl© ej D. Zwor †P․¤^Kxq ej S C info me‡P‡q kw³kvjxÑmej wbDwK¬q ej (1041) : me‡P‡q `ye©jÑgnvKl© ej (1)|

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 147 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. †KvbwU msiÿYkxj ej? [Xv. †ev. 2016] A. evqyi evav B. Zwor ej C. Nl©Y ej D. mv›`ª ej Ans B 02. cwieZ©bkxj e‡ji †ÿ‡ÎÑ [iv. †ev. 2017] A. ïay e‡ji gvb cwiewZ©Z nq B. ïay e‡ji w`K cwiewZ©Z nq C. e‡ji gvb I w`K DfqB cwiewZ©Z nq D. e‡ji gvb I w`K DfqB AcwiewZ©Z _v‡K Ans C 03. wb‡Pi †Kvb ejwU wecixZ eMx©q m~Î †g‡b P‡j bv? [iv. †ev. 2016] A. gnvKl© ej B. Zwor ej C. †P․¤^K ej D. mskw³ ej Ans D 04. wb‡Pi †Kvb ejwU me‡P‡q `ye©j ej? [P. †ev. 2016] A. kw³kvjx wbDwK¬q ej B. Zwor †P․¤^K ej C. gnvKl© ej D. `ye©j wbDwK¬q ej Ans C 05. U‡K©i Aci bvg Kx? [w`. †ev. 2015] B. Nl©Y ej B. RoZvi åvgK C. N~Y©b ej D. †K›`ªgyLx ej Ans C 06. msiÿYkxj ej n‡jvÑ [w`. †ev. 2015] i. gnvKl© ej ii. Av`k© w¯úªs ej iii. mv›`ª ej wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i,ii I iii Ans A 07. mej wbDwK¬q e‡ji Rb¨ †Kvb KYv `vqx? [Kz.‡ev 2021; e. †ev. 20217] A. MÖvwfUb B. wbDwUª‡bv C. †gmb D. B‡jKUªb Ans C 08. †Kv‡bv GKwU e¯‧‡K Agm„Y Z‡j w¯
i Ae¯
vq ivLv n‡j Gi Dci †gvU KqwU ej wµqv K‡i? [wm.‡ev 2021] A. 1 B. 2 C. 3 D. 4 Ans B 09. Zwo”Pz¤^Kxq ej †Kvb KYvi cvi¯úwiK wewbg‡qi d‡j Kvh©Ki nq? [wm. †ev. 2015] A. †dvUb B. †gmb C. †cÖvUb D. MÖvwfUb Ans A 10. wb‡Pi †KvbwU Nl©Y e‡ji D`vniY? [wm. †ev. 2015] A. msmw³ ej B. msiÿYkxj ej C. AvmÄb ej D. AmsiÿYkxj ej Ans D 11. Nwoi KuvUvi MwZ †Kvb MwZi D`vniY? [g.‡ev 2021] A. ch©ve„Ë MwZ B. ¯ú›`b MwZ C. mij‣iwLK MwZ D. Av‡cwÿKK MwZ Ans A 12. BD‡iwbqvg, †_vwiqvg Gme †g․wjK c`v‡_©i †ZRw¯…q fv½b N‡U †Kvb e‡ji Kvi‡Y? [g.‡ev 2021] A. gnvKl© ej B. Zwor †P․¤^K ej C. `ye©j wbDwK¬q ej D. mej wbDwK¬q ej Ans C 13. GKK ej- [e.†ev. 2015] A. e¯‘i Ici GKK Z¡iY m„wó K‡i B. GKK f‡ii e¯‘i Ici †h †Kv‡bv Z¡iY m„wó K‡i C. e¯‘i Ici †h †Kv‡bv Z¡iY m„wó K‡i D. GKK f‡ii e¯‘i Ici GKK Z¡iY m„wó K‡i Ans D Concept 2 wbDU‡bi MwZm~Îmg~n Ges mxgve×Zv ✅ wbDU‡bi MwZm~Î : 1687 mv‡j wbDUb Zvi MÖš’ Òwd‡jv‡mvwdqv b¨vPvivwjm wcÖwÝwcqv g¨v_g¨vwUKvÓ †Z e¯‘i fi, MwZ I e‡ji g‡a¨ m¤úK© ¯’vcb K‡i wZbwU m~Î cÖ`vb K‡ib| GB m~θwjB wbDU‡bi MwZ m~Î bv‡g cwiwPZ| 1g m~Î: evwn¨K ej cÖ‡qv‡M e¯‘i Ae¯’vb cwieZ©b bv NUv‡j w¯’i e¯‘ wPiKvj w¯’i _vK‡e Ges MwZkxj e¯‘ mg‡e‡M mijc‡_ Pj‡Z _vK‡e| 2q m~Î: e¯‘i fi‡e‡Mi cwieZ©‡bi nvi e¯‘i Dci cÖhy³ e‡ji mgvbycvwZK Ges ej †h w`‡K wµqv K‡i e¯‘i fi‡e‡Mi cwieZ©bI †m w`‡K N‡U| 3q m~Î: cÖ‡Z¨K wµqviB mgvb I wecixZ cÖwZwµqv Av‡Q| ◈ aviYv cvIqv hvq: 1g m~Î 2q m~Î 3q m~Î 1. RoZv 2. e‡ji msÁv 1. e‡ji AwfgyL 2. cwigvc 3. Z¡i‡Yi m‡½ e‡ji m¤úK© 4. e‡ji GKK 5. e‡ji wbi‡c¶ bxwZ 1. fi‡e‡Mi wbZ¨Zv m~Î ev fi‡e‡Mi msi¶Y wewa 2. wewfbœ cÖKvi wµqv-cÖwZwµqv: Uvb (pull), c„ôUvb (Tension); av°v (push), AvKl©Y-weKl©Y, Nl©Y ej, i‡K‡Ui R¡vjvbxi av°v (Thrust) ✅ m~‡Îi cÖ‡qvM: m~Î cÖ‡qvM fi‡e‡Mi msiÿY m~Î Kvgvb †_‡K †Mvjv †Quvov; i‡K‡Ui MwZ; Av‡ivnxi †b․Kv †_‡K jvd †`Iqv; A¨v_‡j‡Ui js R¨v¤ú †`Iqv| wbDU‡bi MwZi wØZxq m~Î wbDU‡bi MwZi cÖ_g m~Î cÖwZcv`b Kiv hvq; F = ma cÖwZcv`b Kiv hvq| wbDU‡bi MwZi Z…Zxq m~Î i‡K‡Ui DÇqb; †b․Kv Pvjv‡bv; †Uwe‡ji Dci eB; cvwLi AvKv‡k Dov| †Nvovi Mvwo Uvbv; e›`y‡Ki ¸wj; cv‡q nuvUv; ✅ wbDU‡bi MwZi Z…Zxq m~‡Îi D`vniY: †Uwej †Nviv‡Z cvie bv| †Uwej †Nviv‡Z cv i e bv †Uwe‡ji Dci eB _vKv †Nvovi Mvwo Uvbv cv‡q nuvUv i‡K‡Ui DÇqb e›`y‡Ki ¸wj †Quvov †b․Kv Pvjv‡bv ✅ cÖwZcv`b: 2q m~Î n‡Z 1g m~Î cÖwZcv`b Kiv hvq 3q m~Î n‡Z fi‡e‡Mi wbZ¨Zv m~Î 2q m~Î n‡Z F= ma mgxKiY 2q m~Î n‡Z †K›`ªgyLx e‡ji ivwkgvjv

148 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ wbDU‡bi MwZm~‡Îi mxgve×Zv: Zvc mÂvj‡bi †¶‡Î wbDU‡bi m~Îvejx cÖ‡hvR¨ n‡e bv| G‡¶‡Î Zvc MwZ we`¨v (Thermo dynamics) cÖ‡hvR¨| Zij I M¨vm‡K GK K_vq cÖevnx (Fluid) ejv nq| cÖevnxi †¶‡Î wbDU‡bi MwZ m~Îvejx cÖ‡hvR¨ n‡e bv| Gme †¶‡Î cÖevnx ej we`¨v (Fluid mechanics) cÖ‡hvR¨| cÖm½ KvVv‡gv‡K Aek¨B Ro cÖm½ KvVv‡gv n‡Z n‡e| †hgb- w¯’i KvVv‡gv, mg‡e‡M MwZkxj KvVv‡gv BZ¨vw`| Ab¨_vq, wbDU‡bi MwZ m~Îvejx cÖ‡hvR¨ n‡e bv| hLb e¯‘i †eM Av‡jvi †e‡Mi KvQvKvwQ nq ZLb wbDU‡bi m~Î Lv‡U bv, ZLb Av‡cw¶KZvi m~Î cÖ‡hvR¨| †Kb e¯‘i †eM hw` Av‡jvi †e‡Mi Kg nq ZLb wbDU‡bi m~Î cª‡hvR¨| G‡K AbyiƒcZv bxwZ e‡j| wbDU‡bi MwZmg~n eo AvKv‡ii e¯‘i Rb¨ cÖ‡hvR¨ wKš‘ Lye ¶z`ª ¶z`ª e¯‘ †hgb- B‡jKUªb, †cÖvUb, wbDUªb BZ¨vw`i Rb¨ Kvh©Ki bq| Gme †¶‡Î †Kvqv›Uvg ejwe`¨v (Quantum mechanics) cÖ‡hvR¨| g¨v·I‡q‡ji Zwor Pz¤^Kxq ZË¡: Zwor I Pz¤^Kxq NUbvejx Z_v Av‡jvi e¨wZPvi, AceZ©b, †cvjvivqb cÖf„wZ e¨vL¨v `v‡b m¶g n‡jI K…ò e¯‘i wewKiY, Av‡jvK Zwor wµqv, †g․j KZ©…K †iLv eY©vjx DrcwË, K¤úUb wµqv, cigvYyi ¯’vwqZ¡, KwVb e¯‘i Av‡cw¶K Zvc cÖf„wZ †¶‡Î wPivqZ c`v_© weÁvb Z_v wbDU‡bi MwZm~‡Îi mxgve×Zv cixjw¶Z nq| †Kvqv›Uvg ejwe`¨vi mvnv‡h¨ e¨vL¨v Kiv hvq| Zwor †¶‡Î Pv‡R©i MwZ, †P․¤^K †¶‡Î AYyPz¤^‡Ki MwZ BZ¨vw` †¶‡Î wbDU‡bi MwZi m~Îvejx cÖ‡hvR¨ n‡e bv| Gme †¶‡Î Kzj‡¤^i m~Î (Coulomb's law), j‡iÄ m~Î (Lorentz law) BZ¨vw` cÖ‡hvR¨| REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. c„w_exi N~Y©b nVvr †_‡g †M‡j †giæwe›`y‡Z fi n‡e- [DU. 05-06] A. Less B. greater C. the same as before D. vary with latitude Ans C STEP 02 ANALYSIS OF JU QUESTION 01. wµqv I cÖwZwµqvi ga¨eZ©x †Kv‡Yi gvb KZ? [JU. 18-19, KU. 19-20; mKj †ev. 2018] A. 0 B. 60 C. 90 D. 180 Ans D 02. mg †e‡M MwZkxj †Kvb e¯‧i Dci cÖhy³ ej k~b¨ n‡j- [JU. 12-13] A. e¯‘wUi †eM k~b¨ B. e¯‘wUi g›`b n‡e Ans C C. e¯ÍwU mg‡e‡M Pj‡Z _vK‡e D. e¯‘wUi MwZi †Kvb cwieZ©b n‡e bv STEP 03 ANALYSIS OF RU QUESTION 01. mvB‡Kj Av‡ivnx‡K euv‡Ki †K‡›`ªi w`‡K †nj‡Z nq †Kb? [RU. 15-16] A. †K›`ªwegyLxZvi Rb¨ B. †K․wYK †e‡Mi Rb¨ C. •iwLK †e‡Mi Rb¨ D. †K›`ªgyLx e‡ji Rb¨ Ans D 02. 850 kg f‡ii GKwU Mvox mgZ¡i‡Y MwZkxj| Mvoxi Dci wµqviZ jwäej wbY©q Ki‡Z †Kvb m~Î cÖ‡qvM Ki‡Z n‡e? [RU. 14-15] A. wbDU‡bi 1g m~Î B. wbDU‡bi 2q m~Î C. wbDU‡bi 3q m~Î D. fi‡e‡Mi wbZ¨Zv m~Î Ans B 03. GKwU wbw`©ó †e‡M Pjgvb e¯‧i Dci wKQy ej KvR Kiv ïiæ Kij| hw` H mg¯Í ej¸‡jvi †hvMdj k~b¨ nq Z‡e e¯‧wUi- [RU. 13-14] A. MwZ K‡g †_‡g hv‡e B. MwZi w`K cwieZ©b n‡e C. Z¡iY †e‡o hv‡e D. wbw`©ó †e‡M Pj‡Z _vK‡e Ans D 04. †Kvb e¯‧i Dci cÖhy³ ej Øviv K…Z KvR e¯‧wUi MwZkw³ cwieZ©‡bi- [RU. 13-14] A. †P‡q †ekx B. †P‡q Kg C. mgvb D. †KvbwUB bq KvR-kw³ Dccv`¨: †Kvb e¯‘i Dci cÖhy³ ej Øviv K…Z KvR e¯‘wUi MwZkw³ cwieZ©‡bi mgvb| W = k – k0= k 05. Zzwg hw` f~-c„ô n‡Z Dc‡i †h‡Z _vK Z‡e †Zvgvi- [RU. 13-14] A. fi Kg‡e I IRb evo‡e B. fi evo‡e I IRb Kg‡e C. fi AcwiewZ©Z _vK‡e I IRb Kg‡e D. fi Kg‡e I IRb AcwiewZ©Z _vK‡e Ans C 06. wµqv Ges cÖwZwµqv ej me mgqB- [RU. 10-11] A. GKB e¯‘i Dci wµqv K‡i B. `ywU wfbœ e¯‘i Dci wµqv K‡i C. e¯‘i MwZkxj Ae¯’vq _vKvi Dci wbf©ikxj D. e¯‘؇qi mvg¨e¯’vq _vKvi Dci wbf©ikxj wµqv cªwZwµqv ej memgq `ywU wfbœ e¯‘i Dci wµqv K‡i| KLbB GKB e¯‘i Dci wKqv K‡i bv| cÖwZwµqv ejwU ZZ¶YB _vK‡e hZ¶Y wµqv ejwU _vK‡e| 07. wbDU‡bi Z„Zxq m~‡Îi cÖ‡qvM bq †KvbwU? [RU. 10-11] A. i‡KU B. e›`y‡Ki cðvr MwZ C. hvÎxevnx evm D. †KvbwUB bq hvÎxi evm RoZv A_©vr cÖ_g m~·K e¨vL¨v K‡i| STEP 04 ANALYSIS OF CU QUESTION 01. †Kv‡bv KYvi Dci cªhy³ UK© k~b¨ n‡j wb‡Pi †Kvb ivwkwU aªæeK n‡e? [CU-A, Set-4. 20-21] A. ej B. †K․wYK fi‡eM C. •iwLK fi‡eM D. e‡ji NvZ S B info †Kv‡bv wbw`©ó e¨e¯’vi Dci cÖhy³ †bU UK© k~b¨ n‡j e¨e¯’vwUi †gvU †K․wYK fi‡eM msiwÿZ _v‡K| A_©vr †K․wYK fi‡eM aªæeK _v‡K| STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question IU 01. dp dt F n‡jv- [IU. 13-14] A. AvBb÷vB‡bi m~Î B. wbDU‡bi MwZi 2q myÎ C. M¨vwjwjI m~Î D. m¤úK©wU mwVK bq Ans B 02. cošÍ e¯‧i †¶‡Î hvwš¿K kw³i msi¶YkxjZvi bxwZ- [IU. 12-13] A. cÖ‡hvR¨ B. cÖ‡hvR¨ bq C. †h †Kvb †¶‡Î cÖ‡hvR¨ D. †KvbwU bq AwfK‡l©i cÖfv‡e gy³ fv‡e cošÍ e¯‘i †¶‡Î memgq wefe kw³ I MwZkw³ mgvb _v‡K| A_©vr G‡¶‡Î hvwš¿K kw³i msi¶YkxjZvi bxwZ cÖ‡hvR¨| 03. jb †ivjv‡ii †¶‡Î †KvbwU mwVK? [IU. 11-12] A. Uvbv mnR B. †Vjv mnR C. Dfq †¶‡Î GKB D. †KvbwUB bq jb †ivjv‡ii †¶‡Î †Vjv A‡c¶v Uvbv mnRZi| PART B Analysis of Science & Technology Question SUST 01. ei‡di Dc‡i mvB‡Kj Pvjv‡Z Amyweav nIqvi KviY- [SUST. 06-07] A. ei‡di c„ôPvc Kg B. ei‡di Dci evqyi Pvc Kg C. Zvc K‡g hvq D. ei‡di Nl©YRwbZ ej AwZwi³ Kg Ans D BSMRSTU 01. evwn¨K ej k~b¨ n‡j †KvbwU N‡U? [BSMRSTU-C, 19-20] A. MwZ‡eM k~b¨ n‡e B. fi‡eM aªæe n‡e C. cÖwZwµqv ej Amxg n‡e D. RoZvi åvgK k~b¨ n‡e evwn¨K ej k~b¨ n‡j mg‡e‡M Pj‡Z _v‡K A_©vr fi‡eM aªæe _vK‡e| PSTU 01. cošÍ e¯‧i m~Î cix¶vi mvnv‡h¨ cÖgvY K‡ib- [PSTU. 14-15] A. wbDUb B. M¨vwjwjI C. AvBb÷vBb D. †Kcjvi Ans A

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 149 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 06 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. †Kvb wbw`©ó f‡ii e¯‧i MwZkw³, Gi fi‡e‡Mi mv‡_ m¤úK© Kx? [MAT. 16-17] A. eM©g~‡ji mgvbycvwZK B. e‡M©i mgvbycvwZK C. e‡M©i e¨¯ÍvbycvwZK D. mgvbycvwZK Ans B STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. wµqv I cÖwZwµqv ej h_vµ‡g I n‡jÑ [h. †ev. 2017] i. = ii. iii. wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii B. i,iiIiii Ans A Concept 3 RoZv, NvZej Ges i‡K‡Ui MwZ msµvšÍ Z_¨vejx ✅ RoZv: GKwU e¯‘ †hgb Av‡Q wVK †Zgb _vK‡Z PvIqvi †h cÖeYZv ev ag© Zv‡K RoZv e‡j| RoZv 2 cÖKvi ◈ w¯
wZ RoZv (Interia of rest): gy`ªv I Kv‡ci cix¶v nVvr Mvwo Pj‡Z _vK‡j Av‡ivnx wcQ‡bi w`‡K †n‡j c‡o †Nvov nVvr Pj‡Z ïiæ Ki‡j Av‡ivnx †cQ‡b †n‡j cov Kuv‡Pi Rvbvjvq ey‡jU jvMv‡j ey‡j‡Ui mgvb e¨v‡mi GKwU †MvjvKvi wQ`ª Drcbœ nq| wKš‘ Av‡k cv‡ki KuvP †d‡U hvq bv| ◈ MwZ RoZv (Interia of motion): MwZ RoZvi D`vniY: PjšÍ Mvwo nVvr †_‡g †M‡j Av‡ivnx mvg‡bi w`‡K Suz‡K c‡o| MwZ‡Z avegvb †Nvovi wcV n‡Z jvd w`‡q Av‡ivnx cybivq †Nvovi wc‡V wd‡i Av‡m| jsRv¤ú cÖwZ‡hvMxZvq cÖwZ‡hvMxiv wKQz `~i †`․‡o Zvici jvd †`b| ✅ NvZ ej : Lye Aí mg‡qi Rb¨ Lye †ekx gv‡bi ej wµqvkxj n‡j Zv‡K NvZ ej e‡j| D`vniY: e¨vU w`‡q wµ‡KU e‡j AvNvZ, †Uª‡b †Uª‡b msNl©, Kvgvb n‡Z †Mvjv †Qvov, †evgv we‡ùviY cÖf„wZ| ✅ NvZ e‡j †e‡Mi cwieZ©b nq nVvr wKš‘ miY †Zgb nq bv| ◈ gvÎv mgxKiY: ivwk GKK gvÎv mgxKiY ivwk GKK gvÎv mgxKiY ej N [MLT–2 ] NvZ ej N [MLT–2 ] fi‡eM Kgms–1 [MLT–1 ] e‡ji NvZ Ns [MLT–1 ] ✅ i‡K‡Ui MwZ: i‡K‡U R¡vjvbx wn‡m‡e Zij nvB‡Wªv‡Rb _v‡K| i‡K‡U `n‡bi Rb¨ _v‡K Zij Aw·‡Rb| i‡K‡Ui Z¡iY: a = F M = 1 M m t .V ✅ GB mgxKiY †_‡K †`Lv hvq t- i‡K‡Ui fi Kg‡j Z¡iY e„w× cvq| i‡K‡Ui Z¡iY e„w× Ki‡Z n‡j M¨vm wbM©g‡bi nvi evov‡Z n‡e| M¨v‡mi Av‡cw¶K †eM e„w× Ki‡j Z¡iYI e„w× cv‡e| i‡KU hZ Dc‡i DV‡Z _vK‡e ZZB Gi Z¡iY e„w× cv‡e| ✅ †K›`ªgyLx I †K›`ªwegyLx e‡ji e¨envi: iv¯Ívi e¨vswKs (Banking of road) mvB‡Kj Av‡ivnxi euvK †bqv MÖn I DcMÖn¸‡jvi MwZ N~Y©vqgvb fvix hš¿cvwZ Zi‡ji wewfbœ Nb‡Z¡i Dcv`vb c„_K Kiv wewfbœ ai‡bi cv¤ú REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF JU QUESTION 01. †Kvb m¤úK©wU mwVK? [JU. 18-19] A. = I1 B. = I C. = 1 D. = I Ans D 02. GKwU e¯‧i `ywU wfbœ we›`y‡Z `ywU mgvb,mgvšÍivj I wecixZgyLx ej wµqv Ki‡j D³ ejØq‡K wK e‡j? [JU. 09-10] A. UK B. Ø›Ø C. fi‡e‡Mi åvgK D. RoZvi åvgK Ans B STEP 02 ANALYSIS OF CU QUESTION 01. 'M' fi Ges 'a' cÖvšÍwewkó GKwU mylg eM©vK…wZi PvKwZi GKwU K‡Y©i mv‡c‡ÿ Gi RoZvi åvgK: [CU. 18-19] A. Ma2 3 B. Ma2 6 C. Ma2 9 D. Ma2 12 Ans D 02. e¯‧i †K․wYK fi‡e‡Mi cwieZ©‡bi nvi- [CU. 16-17] A. cÖhy³ e‡ji mgvbycvwZK B. †K․wYK Z¡i‡Yi e¨¯ÍvbycvwZK C. cÖhy³ U‡K©i mgvb D. RoZvi åvg‡Ki mgvbycvwZK Ans D STEP 03 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. †¯‥jvi ¸Y‡bi D`vniY †KvbwU? [KU-A, Set-Ka. 19-20] A. KvR B. ej C. UK© D. †K․wYK fi‡eM †f±‡ii ¸Y‡bi †ÿ‡Î: KvR, W = F . s ; UK©, = r F ; †K․wYK fi‡eM, L = r P ; •iwLK †eM, v = r 02. †Kvb e¯‧i Dci U‡K©i jwä k~b¨ n‡j e¯‧wUi †K․wYK fi‡eM [KU. 11-12] A. e„w× †c‡Z _v‡K B. Kg‡Z _v‡K C. msiw¶Z _v‡K D. k~b¨ nq †K․wYK fi‡eM, L = I = . = 0 . = 0 03. wbw`©ó A‡¶i Pviw`‡K N~Yv©qgvb e¯‘‡Z Z¡iY m„wó‡Z cÖhy³ ؇›`i åvgK‡K e‡j- [KU. 11-12] A. Kvcj B. UK© C. RoZvi åvgK D. †K․wYK Z¡iY Ans B 04. Pjgvb MwZ‡Z •iwLK Z¡i‡Yi mv‡_ †hgb ej mswkøó, N~Y©b MwZ‡Z †K․wYK Z¡i‡Yi mv‡_ †Zgwb mswkøó ivwk nj- [KU. 03-04, 09-10] A. UK© B. RoZv C. †K․wYK ej D. Awf‡K›`ª ej Ans A

150 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 05. †Kvb A‡¶i Pvwiw`‡K GKwU e¯‧i RoZvi åvgK wbf©i K‡i- [KU. 09-10] A. e¯‘i †K․wYK †e‡Mi Dci B. e¯‘i ‣iwLK †e‡Mi Dci C. e¯‘i †K․wYK Z¡i‡Yi Dci D. e¯‘i fi web¨v‡mi Dci Ans D IU 01. N~Y©b MwZkw³ E RoZvi åvgK I Ges †K․wYK †eM -Gi ga¨eZ©x m¤úK© n‡”Q - [IU. 15-16] A. E = I B. E = I 2 C. E = 1 2 I D. E = 1 2 I 2 Ans D 02. [L] = [ML2 T –1 ] gvÎv mgxKiYwU- [IU. 13-14] A. fi‡e‡Mi B. •iwLK fi‡e‡Mi C. †K․wYK fi‡e‡Mi D. †KvbwUB bq Ans C 03. e¯‧i fi‡e‡Mi cwieZ©b wbf©ikxj bq? [IU. 12-13] A. e‡ji Dci B. mg‡qi Dci C. f‡ii Dci D. †e‡Mi Dci Ans B 04. gvbyl‡K mvg‡b Pj‡Z mvnvh¨ K‡i? [IU. 12-13] A. Djø¤^ Dcvsk B. Avbyf~wgK Dcvsk C. Djø¤^ Avbyf~wgK Dcvsk D. †KvbwUB bq Ans B 05. [J]=[MLT–1 ] gvÎv mgxKiYwU [IU. 12-13] A. fi‡e‡Mi B. e‡ji NvZ C. †K․wYK fi‡e‡Mi D. •iwLK fi‡eMi Ans B 06. U‡K©i gvb †ekx n‡j †Kvb e¯‧i N~Y©b- [IU. 10-11] A. GKB _v‡K B. K‡g hvq C. †e‡o hvq D. wظY †e‡o hvq Ans C BRUR 01. A‡¶i mv‡c‡¶ N~Y©biZ e¯‧i Dci †h we›`y‡Z ej wµqvkxj H we›`yi Ae¯
vb †f±i I cÖhy³ e‡ji ¸Ydj‡K ejv nq- [BRU. 15-16] A. Ø›Ø B. UK© C. RoZvi åvgK D. †K․wYK Z¡iY Ans B BU 01. e¯‧i fi M, PµMwZi e¨vmva© K Ges RoZvi åvgK I Gi g‡a¨ m¤úK© n‡jv- [BU: 11-12] A. K = I M B. K = I M C. K= IM D. K = M I Ans A 02. †Kvb A¶ mv‡c‡¶ N~Y©biZ †Kvb `„p e¯‧i cÖwZwU KYvi fi Ges Gi A¶ †_‡K Zv‡`i cÖ‡Z¨‡Ki j¤^ `~i‡Z¡i e‡M©i ¸Ydj‡K wK ejv nq? [BU: 10-11] A. PµMwZi e¨vmva© B. †K․wYK fi‡eM C. N~Y©b MwZkw³ D. RoZvi åvgK Ans D PART B Analysis of Science & Technology Question SUST 01. hw` Ae¯
vb †f±i r , fi‡eM P Ges cÖhy³ ej F nq, Z‡e †K․wYK fi‡eM L I UK© Gi ivwk ( L, ) Abyhvqx- [SUST. 17-18] A. ( r F, r P) B. ( r p, r F) C. ( p r, F r) D. ( F r, p r) †K․wYK N~Y©biZ †Kvb e¯‘i e¨vmva© †f±i I •iwLK fi‡e‡Mi †f±i ¸Ydj‡K †K․wYK fi‡eM e‡j| A_©vr L = r p UK©: †Kvb wbw`©ó A‡ÿi Pviw`‡K N~Y©vqgvb †KvY e¯‘‡Z Z¡iY m„wói Rb¨ cÖhy³ ؇›Øi åvgK‡K UK© ev e‡ji åvgK e‡j| A_©vr = r F ( r p, r F) MBSTU 01. i‡KU DÇqb †Kvb m~‡Îi dj? [MBSTU. 14-15] A. wbDU‡bi 1g m~Î B. wbDU‡bi 2q m~Î C. wbDU‡bi 3q m~Î D. †KvbwUB bq Ans C 02. GKwU wb‡iU wmwjÛv‡ii RoZvi åvgK 1 2 Mr2 n‡j PµMwZi e¨vmva© KZ? (M wmwjÛv‡ii fi Ges r e¨vmva©) [MBSTU-C, Set-2 19-20] A. 1 2 M B. 1 2 Mr C. r 2 D. r 2 k = I M = 1 2 Mr2 M = r 2 STEP 04 ANALYSIS OF ENGINEERING & BUTex QUESTION 01. AvaywbK †RU wegvb †Kvb m~Î e¨envi K‡i Pvjbv Kiv nq? [RUET. 10-11; CU. 13-14] A. fi‡e‡Mi wbZ¨Zv m~Î B. wbDU‡bi MwZ m~Î C. AwfKl© m~Î D. None Ans A STEP 05 ANALYSIS OF MEDICAL & DENTAL QUESTION DAT 01. RoZvi åvgK †KvbwUi Dci wbf©i K‡i? [DAT; 19-20] A. †K․wYK Z¡iY B. Pµ MwZi e¨vmva© C. e¯‘i f‡ii eÈb D. N~Y©b MwZ kw³ RoZvi åvgK N~Y©b Aÿ †_‡K e¯‘i f‡ii eÈb I `~i‡Z¡i Dci wbf©i K‡i| STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. e‡ji NvZ n‡”QÑ [iv. †ev. 2017] i. ej I e‡ji wµqvKv‡ji ¸Ydj ii. fi‡e‡Mi cwieZ©b iii. fi‡e‡Mi cwieZ©‡bi nvi wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans A 02. mg‡K․wYK †e‡M N~Y©bkxj e¯‧i N~Y©b MwZkw³ [iv. †ev. 2016] A. RoZvi åvg‡Ki mgvbycvwZK B. RoZvi åvg‡Ki e¨¯ÍvbycvwZK C. RoZvi åvg‡Ki e‡M©i mgvbycvwZK D. RoZvi åvg‡Ki e‡M©i e¨¯ÍvbycvwZK Ans A 03. wb‡Pi †Kvb m¤úK©wU mwVK? [iv. †ev. 2016] A. = B. = C. = D. = Ans C 04. mg‡K․wbK †e‡Mi AveZ©biZ †Kv‡bv `„p e¯‧i MwZkw³ I RoZvi åvg‡Ki AbycvZÑ [P.‡ev. 2019] A. ‡K․wbK †e‡Mi mgvbycvwZK B. ‡K․wbK †e‡Mi e‡M©i mgvbycvwZK C. •ewLK †e‡Mi mgvbycvwZK D. •ewLK †e‡Mi e‡M©i mgvbycvwZK Ans B 05. e‡ji Nv‡Zi mv‡_ †Kvb ivwkwUi mvswL¨K gvb mgvb? [P.‡ev. 2019] A. ‡K․wbK fi‡e‡Mi cwieZ©b B. •iwLK fi‡e‡Mi cwieZ©b C. RoZvi åvgK D. UK© Ans B 06. ej I e‡ji wµqvKvj‡K Kx e‡j? [P.‡ev. 2016] A. NvZ ej B. KvR C. e‡ji NvZ D. UK© Ans C 07. e¨vmva© †f±i I cÖhy³ e‡ji †f±i ¸Yb‡K e‡j [P. †ev. 2017] A. RoZvi åvgK B. UK© C. †K․wYK fi‡eM D. PµMwZi e¨vmva© Ans B 08. U‡K©i Aci bvg Kx? [w`. †ev. 2015] A. Nl©Y ej B. RoZvi åvgK C. N~Y©b ej D. †K›`ªgyLx ej Ans C

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 151 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 09. †Kv‡bv e¯‧i RoZvi åvgK wbf©i K‡i- [h. †ev. 2015] A. †K․wYK †e‡Mi Dci B. †K․wYK fi‡e‡Mi Dci C. •iwLK †e‡Mi Dci D. fi I N~Y©b A‡ÿi Ae¯’v‡bi Dci Ans D 10. w¯
wZ RoZvi D`vniY †KvbwU? [wm.‡ev 2021] A. ‡Nvovi Mvwo Uvbv B. a~wjgq †cvkv‡K AvNvZ Kiv C. eo †b․Kvi ¸Y Uvbv D. e›`yK †_‡K ¸wj †Quvov Ans B 11. †Kv‡bv e¯‧i RoZvi åvgK wbf©i K‡i GiÑ [wm. †ev. 2016; h. †ev. 2015] A. fi I N~Y©b A‡ÿi Ici B. AvqZ‡bi Dci C. †K․wYK †e‡Mi Ici D. †K․wYK fi‡e‡Mi Ici Ans A 12. i‡K‡Ui MwZi Rb¨ [wm. †ev. 2016] i. Av‡cwÿK †eM e„wׇZ Z¡iY e„w× cvq ii. M¨vm wbM©g‡bi nvi e„wׇZ Z¡iYI e„w× cvq iii. i‡KU hZ Dc‡i hvq Z¡iY ZZ K‡g wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i,ii I iii Ans A 13. e‡ji NvZ n‡”Q- [wm. †ev. 2015] A. ej I e‡ji wµqvKv‡ji ¸Ydj B. fi‡e‡Mi cwieZ©b C. fi‡e‡Mi cwieZ©‡bi nvi D. A I B DfqBmwVK Ans D 14. Lye Aí mg‡qi Rb¨ Lye eo gv‡bi ej cÖhy³ n‡j Zv‡K e‡jÑ [e.‡ev. 2019] A. mskw³ ej B. N~Y©b ej C. Zwor ej D. NvZ ej Ans D 15. e‡ji Nv‡Zi GKK wb‡¤œi †Kvb ivwki GK‡Ki Abyi~c? [e. †ev. 2017] A. ej B. fi‡eM C. KvR D. UK© Ans B Concept 4 ¸iæZ¡c~Y© wKQz ivwki GKK Ges gvÎvmg~n ivwk GKK gvÎv ivwk GKK gvÎv Ø›Ø ev Kvcj, UK© N – m ML2T –2 RoZvi åvgK Kgm2 ML2 ؇›Øi åvgK Nm or Joule ML2T –2 †K․wYK †eM rad s –1 T –1 UK© ev e‡ji åvgK Nm or Joule ML2T –2 †K․wYK Z¡iY rad s –2 T –2 †K․wYK fi‡eM kgm2 s –1 ML2T –1 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. †K․wYK fi‡e‡Mi GKK †KvbwU? [DU. 18-19] A. kg m2 s –1 B. kg m2 s –2 C. kg ms–1 D. kg m2 s 2 s –2 Ans A STEP 02 ANALYSIS OF JU QUESTION 01. †K․wYK fi‡e‡Mi gvÎv †KvbwU? [JU-A, Set-G. 2020-21; 18-19, IU. 2006-07] A. [ML2T -2 ] B. [MLT-2 ] C. [MLT-1 ] D. [ML2T -1 ] S D info Avgiv Rvwb, L = rp = mvr [L] = [ML2T –1 ] Ges GKK : kgm2 s –1 02. e‡ji åvg‡Ki gvÎv †KvbwU? [JU-A, Set-B. 2020-21; 10-11; 09-10] A. [ML2T –2 ] B. [MLT–1 ] C. [M–1T –2 ] D. [ML2T –1 ] S A info e‡ji åvgK, = rF e‡ji åvg‡Ki ev e‡ji †gv‡g‡›Ui gvÎv [] = [L] [MLT–2 ]= [ML2T –2 ] 03. †Kvb m¤úK©wU mwVK? [JU. 18-19] A. = I2 B. = I C. = 1 D. = I Ans D 04. e‡ji gvÎv mgxKiY- [JU. 11-12, 10-11, RU. 10-11; RUET. 12-13] A. [ML2 T] B. [MLT–2 ] C. [M2 LT] D. [MLT–1 ] Ans B STEP 03 ANALYSIS OF RU QUESTION 01. UK© I †K․wYK Z¡i‡Yi m¤úK© †KvbwU? [RU. 14-15] A. = I B. = I –1 C. = I 2 D. = I2 Ans A STEP 04 ANALYSIS OF CU QUESTION 01. wb‡Pi †Kvb m¤úK©wU mwVK? [CU-A, Set-2. 2020-21; 18-19] A. L = r × F B. L = F × r C. L = r × p D. L = p × r S C info †K․wYK fi‡eM Ges •iwLK fi‡e‡Mi g‡a¨ mgvbycvwZK m¤úK©| 02. RoZvi åvg‡Ki GKK Kx? [CU-A, Set-2. 2020-21; CU. 16-17] A. kgm–2 B. kgm C. kgm–1 D. kgm2 S D info RoZvi åvgK I = mr2 | m Gi GKK kg Ges r Gi GKK m| 03. •iwLK Z¡iY I †K․wYK Z¡i‡Yi m¤úK© †KvbwU? [CU-A, Set-4. 20-21] A. a =r/ B. a = /r C. a =r2 D. a = r S D info •iwLK ivwk = MwZc‡_i e¨vmva© †K․wYK ivwk| v = r, s = r, a = r 04. NvZe‡ji (Impulse) gvÎv Kx? [CU. 18-19] A. MLT–1 B. MLT–2 C. ML–1 T –1 D. ML–1 T –2 Ans A 05. †K․wYK †e‡Mi GKK †Kvb&wU? A. wgUvi/†m. B. dzU/†m. C. †iwWqvb/ †m. D. jy‡gb/†m. E. wgUvi/†m2 Ans C 06. †K․wYK †e‡Mi gvÎv mgxKiY- [CU. 02-03; KU. 13-14] A. [LT–1 ] B. [T–1 ] C. [LT–2 ] D. [L–1 T] Ans B 07. e„Ëxq MwZi †¶‡Î †K․wYK fi‡e‡Mi ivwk †Kvb&wU? [CU. 12-13] A. mr B. mr2 C. mr 2 D. mr E. mr Ans B STEP 05 ANALYSIS OF DU-7 Clg QUESTION 01. w¯úªs aªæeK Gi GKK †KvbwU? [DU-7Clg.A.19-20] A. Nm B. Nm1 C. Nm2 D. Nm2 w¯úªs aªeK, K = F x = Nm1 STEP 06 ANALYSIS OF GST QUESTION PART A Analysis of General University Question IU 01. U‡K©i gvÎv- [IU. 2004-05; MBSTU. 2019-20; JKKNIU. 2019-20] A. [ML2 T –2 ] B. [MLT3 ] C. [ML2 T] D. [MLT2 ] Ans A BRUR 01. †KvbwU mwVK? [BRU. 13-14] a. F = m dx/dt B. F = x dm/dt C. F = m d/dt(dx/dt) D. F = x dat/dt Ans C BU 01. wb‡Pi †Kvb&wU e‡ji GKK? [BU: 13-14] A. Nm B. kgms–2 C. Nm–1 D. kgms–1 Ans B

152 An Exclusive Parallel Text Book Of Mathematics ASPECT MATH ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES PART B Analysis of Science & Technology Question SUST 01. e¯‧i IRb †Kv_vq me‡P‡q †ewk? [SUST. 05-06] A. †giæ‡Z B. welyexq A‡j C. KK©UµvwšÍ‡Z D. giæ A‡j Ans A 02. f~-c„‡ô †Kvb e¯‧i fi 1kg c„w_exi †K‡›`ª Bnvi fi KZ? [SUST. 05-06] A. 0kg B. 9.87kg C. 1kg D. †KvbwUB bq Ans C JUST 01. wb‡Pi †Kvb `ywUi gvÎv mgxKiY GKB? [JUST. 14-15] A. KvR I ¶gZv B. ؇›Øi åvgK I KvR C. †eM I g›`b D. kw³ I ¶gZv E. ؇›`i åvgK I ¶gZv Ans B MBSTU 01. †KvbwU fi †e‡Mi gvÎv mgxKiY? [MBSTU. 14-15] A. [P] = [M] L T B. [P] = [M] L T C. [P] = 1 M L T D. [P] = [M] L T 2 Ans A 02. SI c×wZ‡Z RoZvi åvg‡Ki GKK †KvbwU? [MBSTU. 15-16] A. kg.m2 B. kg.m C. kg2 .m D. Joule Ans A STEP 07 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. †KvbwU U‡K©i mwVK GKK? [KUET. 14-15] A. Dyne cm-1 B. Nm C. Nm-1 D. Nm.s Ans B STEP 08 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. fi‡e‡Mi gvÎv mgxKiY †KvbwU? [ MAT. 2017-18; MBSTU-A, Set-2 19-20] A. [MLT1 ] B. [MLT2 ] C. [ML2T] D. [ML2T 2 ] fi‡eM, P = mv gvÎv, [MLT–1 ] STEP 09 ANALYSIS OF HSC BOARD QUESTION 01. †K․wYK K¤úv¼ Gi gvÎv †KvbwU? [Xv.†ev. 2019] A. [M0LT] B. [M0L 0T -1 ] C. [M0L -1T] D. [M0LT-1 ] Ans B 02. e‡ji Nv‡Zi GKK njÑ [iv. †ev. 2017] A. kgms-2 B. kgm-1 s -1 C. kgm-2 s D. kgm-1 03. †K․wYK fi‡e‡Mi GKK †KvbwU? [iv. †ev. 2015] A. kgm2 s –2 B. kgms–2 C. kgms–1 D. kgm2 s –1 Ans D 04. †KvbwU †K›`ªgywL e‡ji ivwkgvjv? [iv. †ev. 2015] A. mv 2 r B. mv 2 r C. mv 2 r 2 D. m 2 r Ans B 05. fi‡e‡Mi åvg‡Ki gvÎv †KvbwU? [P.†ev. 2017] A. ML2T -2 B. MLT-2 C. ML2T -1 D. M 0L 2T -2 Ans C 06. ej aªye‡Ki gvÎv †KvbwU? [P. †ev. 2016] A. ML2T 2 B. MLT2 C. MT2 D. MT2 Ans D 07. †KvbwU e‡ji Nv‡Zi gvÎv mgxKiY? [h. †ev. 2017] A. ML1T 2 B. MLT1 C. MLT2 D. M 1LT2 Ans B 08. e‡ji Nv‡Zi GKKÑ [Kz.‡ev. 2019] A. N B. Nm C. Nm-1 D. kgms-1 Ans D STEP 2 mg„× †ewmK MvwYwZK cÖ Ö‡qvM (MATH) kU©KvU †UKwbK Concept 1 ej fi I Z¡iY msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. F = ma = m u 2 2s [e‡ji mKj AsK GB m~‡Î] 02. mg euvav = ma Or, mg Fk = ma [euvav †`Iqv _vK‡j GB m~Î] 03. S = F t1 m t1 2 + t2 a = dv dt; a = v u t ; a = v 2 u 2 2s ; a = v 2 2s [hLb u = 0]; a = 2s t 2 [u = 0]; a = s2 s1 t2 t1 Model Example01 6N Gi GKwU ej 3kg f‡ii GKwU w¯
i e¯‧i Dci wµqv Ki‡j 5sec G e¯‧wU KZ `~iZ¡ AwZµg Ki‡e? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] F = ma a = F m = 6 3 = 2ms –2 Avevi, S = ut + 1 2 at2 S = 1 2 at2 [u = 0] S = 1 2 (5)2 S = 25m F = 2ms t 2 S = Ft2 2m = 6 5 2 2 3 = 25m Model Example02 mgZ¡i‡Y avegvb GKwU e¯‧ Gi MwZi 5 th †m‡K‡Û I 10th †m‡K‡Û h_vµ‡g 50m Ges 100m `~iZ¡ AwZµg K‡i| e¯‧i fi 4kg n‡j, wµqvkxj e‡ji gvb KZ? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] Avgiv Rvwb, sth = u + 1 2 a (2t 1) S5th = u + 1 2 a (2 5 –1) 50 = u + 1 2 a (9).............(i) Avevi, 5 th = u + 1 2 a (2t – 1) S10th = u + 1 2 a (2 10 – 1) 100 = u + 1 2 a 19 .......(ii) (ii) – (i) 100 50 = u + 1 2 19a u 1 2 9a 50 = 1 2 19a 1 2 9a 50 = 19a – 9a 2 = 10a 2 10a 2 = 50 a = 10ms –2 cÖhy³ ej, F = ma 4 10 = 40N a = s2 s1 t2 t1 = 100 – 50 10 – 5 = 50 5 = 10ms –2 ej, F = ma = 4 10 = 40N

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 153 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW START PRACTICE 01. †g‡Si Dci w`‡q Mwo‡q hvIqv 10 gm f‡ii GKwU gv‡e©j 10 sec Pjvi ci †_‡g †Mj| gv‡e©jwUi cÖv_wgK †eM wQj 10 ms–1 | Nl©Y e‡ji gvb KZ? 02. 20 kg fi wewkó GKwU e›`yK †_‡K ¸wj Qyo‡j 10 ms–1 †e‡M wbM©Z nq| ¸wji fi 100 gm Ges ¸wjwU GKwU Kv‡Vi Z³vq 20 cm `–iZ¡ cÖ‡ek K‡i †_‡g †Mj, Gi Dci cÖhy³ evavRwbZ ej KZ? 03. mgZ¡i‡Y avegvb GKwU e¯‧ Gi MwZi 5 th †m‡K‡Û I 8 th †m‡K‡Û h_vµ‡g 0.18m Ges 0.30m `–iZ¡ AwZµg K‡i| e¯‧i fi 3kg n‡j wµqvkxj e‡ji gvb KZ? 04. GKwU 13N IR‡bi I GKwU 12N IR‡bi `yBwU e¯‧ GKwU fiwenxb `woi Øviv Nl©Y wenxb KwcK‡ji Dci SzjšÍ| 13N IR‡bi e¯‧i wb¤œgyLx Z¡iY gy³fv‡e cošÍ e¯‧i Z¡i‡Yi hZ¸Y Zv n‡jvNOW PRACTICE SOLVE : 01. †_‡g hvIqvq, v = 0 F = ma = m u – v t = 10 10–3 10 – 0 10 = 0.01 N 02. †_‡g hvIqvq, v = 0 ¸wji Dci cÖhy³ evavRwbZ ej, F = ma = m u 2 2s = 0.1 (10) 2 2 0.2 = 25 N 03. Z¡iY, a = sm – sn tm – tn = 0.30 – 0.18 8 – 5 = 0.04ms–2 ; e¯‘i Dci cÖhy³ e‡ji †¶‡Î, F = ma = 3 0.04 = 0.12N 04. c«_g e¯‘i fi,m1= 13 9.8; w`¡Zxq e¯‘i fi, m2 = 12 9.8; Avgiv Rvwb, m1 I m2 f‡ii `yBwU e¯‘ GKwU fiwenxb `woi Øviv Nl©Y wenxb KwcK‡ji Dci Swzj‡q w`‡j m1 f‡ii e¯‘i wb¤œgyLx Z¡iY, a = m1 – m2 m1 + m2 g a g = 13 9.8 – 12 9.8 13 9.8 + 12 9.8 = 13 – 12 13 + 12 = 1 25 m1 m2 T T REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. 10kg f‡ii GKwU e¯‧i Dci 2F gv‡bi ej cÖ‡qvM Kivi d‡j e¯‧wUi Z¡iY nq 60 m/s2 | M f‡ii GKwU e¯‧i Dci 5F gv‡bi ej cÖ‡qvM Kivi d‡j hw` e¯‧wUi Z¡iY 50m/s2 nq, Z‡e fi M KZ? [DU. 18-19] A. 3.3kg B. 4.8kg C. 21kg D. 30kg Avgiv Rvwb, ej = fi †eM 2F = 10 60 F = 300N Avevi, 5F = 5 50 M 50 M = 30 kg 02. 30kg f‡ii GKwU w¯
i e¯‧i †eM 2 wgwb‡U e„w× K‡i 36 km/hr G DbœxZ Kivi Rb¨ e¯ÍwUi Dci KZ ej cÖ‡qvM Ki‡Z n‡e? [DU. 16-17] A. 2N B. 2.5N C. 300N D. 5 N †eM e„w×, Δv = 36 3.6 ms–1 = 10ms–1 F = ma = m v t = 30 (10) 2 60 = 2.5N 03. mgvb fi wewkó wZbwU LÛ A,B,C `woi Øviv wP‡Î cÖ`wk©Z iƒ‡c mshy³| C LÛwU F ej Øviv Uvbv n‡j m¤ú~Y© e¨e¯
vwU Z¡wiZ nq| Nl©Y D‡c¶v Ki‡j LÛ B Gi Dci †gvU ej n‡jv- [DU. 14-15] F A B C A. 0 B. F /3 C. F /2 D. 2F /3 ‡h‡nZz fi mgvb Ges Nl©Y ej D‡c¶bxq| Kvh©Ki Z¡iY, a == F m+m+m = F 3m, F1 = m F 3m = F 3 04. GKwU 13 N IR‡bi I GKwU 12 N IR‡bi `yBwU e¯‧ GKwU fiwenxb `woi Øviv Nl©Y wenxb KwcK‡ji Dci Szj¯Í| 13 N IR‡bi e¯‧i wb¤œg–Lx Z¡iY gy³fv‡e co¯Í e¯‧i Z¡i‡Yi hZ¸Y Zv n‡jv- [DU. 14-15] A. 1 12 B. 1 13 C. 1 25 D. 13 25 c«_g e¯‘i fi,m1 = 13 9.8; w`¡Zxq e¯‘i fi, m2 = 12 9.8; we know,m1 I m2 f‡ii `yBwU e¯‘ GKwU fiwenxb `woi Øviv Nl©Y wenxb KwcK‡ji Dci Swzj‡q w`‡j m1 f‡ii e¯‘i wb¤œgyLx Z¡iY, a = m1 m2 m1 + m2 g a g = 13 98 12 98 13 98 + 12 98 = 13 – 12 13 + 12 = 1 25 05. GKwU Kv‡Vi LÛ‡K Avbyf~wg‡Ki mv‡_ 60° †Kv‡b 200N ej Øviv Uvbv n‡”Q| e¯ÍwUi Dci Avbyf~wg‡Ki w`‡K Kvh©Kix ej KZ? [DU. 13-14] A. 200N B. 100N C. 174N D. Zero Fx = FR cos = 200 cos60° = 100N 06. GKwU bj †_‡K 2m/s †e‡M cvwb †ei n‡q GKwU †`qvj‡K j¤^fv‡e AvNvZ Ki‡Q b‡ji cÖ¯
‡”Q` 0.03m2 aiv hvK cvwb †`qvj †_‡K wievDÛ Ki‡Qbv| †`qv‡ji Dci wK cwigvb ej cÖ‡qvM Ki‡Q| (cvwbi NbZ¡ 1000kg/m3 ) [DU. 09-10] A. 1000 N B. 300 N C. 120 N D. 240 N F = ma = A.lv–u t = A.v v t [⸪ l = v] = 0.03 1000.2. 2 1 = 120N 07. GKwU AwZgvbe Zvi cÖwZc‡¶i weiæ‡× 2800 N e„nr cÖ¯ÍiLÛ Qz‡o gv‡i| cÖ¯ÍiLÛwU‡K Avbyf~wgK 15.0 ms–2 Z¡iY w`‡Z n‡j Zv‡K cÖ¯ÍiLÛwU‡Z KZ Avbyf~wgK ej cÖ‡qvM Ki‡Z n‡e? [DU. 05-06] A. 4.29 103N B. 42.0 103N C. 2.7 103N D. 187 N F = ma = w g a = 2800 9.8 15 = 4.29 103

154 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 08. 36 kg f‡ii GKwU e¯‧i Dci KZ ej cÖhy³ n‡j 1 wgwb‡U e¯‧wUi †eM 15 km/hr e„w× cv‡e? [DU. 04-05] A. 2.4 N B. 2.5 N C. 14.4 N D. 28.8 N F = ma = m v u t = 36 15 60 3.6 = 2.5 N 09. GKwU ej 4 kg fi wewkó w¯
i e¯‧i Dci wµqv K‡i| Gi d‡j e¯‧wU 6 †m‡K‡Û 30ms-1 †eM cÖvß nq| e‡ji gvb KZ? [DU. 03-04 ; RU. 12-13, BRU. 15-16] A. 30 N B. 20 N C. 18 N D. None F = ma = t v u m = 6 30 0 4 = 20 N 10. 8 kg f‡ii GKwU e¯‧‡Z 6m/s2 Z¡iY cÖ`vb Ki‡Z n‡j e¯‧wU‡Z wK cwigvY ej cÖ‡qvM Ki‡Z n‡e? [DU. 98-99] A. 18 N B. 28 N C. 38 N D. 48N Z¡iY m„wóKvix ej, F = ma = 8 6 = 48 N 11. hw` 10 gm f‡ii ey‡jU GKwU e›`y‡Ki 20 cm e¨v‡i‡ji g‡a¨ 200 ms1 †eM cvq, Zvn‡j Z¡iY m„wóKvix ej KZ? [DU. 00-01] A. 2103N B. 200N C. 4104N D. 103N Z¡iY m„wóKvix ej, F = ma = m u 2 2s = 0.01 (200) 2 2 0.2 =103N STEP 02 ANALYSIS OF JU QUESTION 01. 50 Kg IR‡bi GKRb c¨vivmyU e¨enviKvix mg‡e‡M wb‡Pi w`‡K bvg‡Qb| Zuvi Dc‡i evqyi evav- [JU-A, Set-D. 19-20] A. 610 N B. 490 N C. 520 N D. 590 N evqyi evav, W = mg = 50 9.8 = 490 N 02. w¯
i cvwb‡Z 500 kg f‡ii GKwU w¯
i †b․Kv‡K b`xi `yB Zx‡i †_‡K `wo w`‡q 60 †Kv‡Y Uvbv n‡”Q| †b․KvwU KZ mg‡q Zx‡ii mgvšÍiv‡j 240 m AwZµg Ki‡e? MšÍe¨¯
‡j †cu․Qv‡bvi Rb¨ `yB w`K n‡Z mgvb Uv‡b †b․KvwU Uvbv nq| (`woi Uv‡bi gvb 12N) [JU-H, Set-1. 19-20] A. 500 s B. 100 2 s C. 5s D. 10 2 m a = 2 12 cos60 500 ms2 = 12 500 ms2 s = 1 2 at2 t = 2s a = 2 240 12 500 = 20000 = 2 10000 = 100 2s 03. 60 Kg f‡ii GKwU e¯‧i Dci KZ ej cÖ‡qvM Ki‡j 1 wgwb‡U Gi †eM 10ms–1 e„w× cv‡e? [JU-A, Set-A. 19-20] A. 20 N B. 15 N C. 5 N D. 10 N Avgiv Rvwb ej, F = ma = m v t = 60 10 60 = 10 N 04. 100 kg f‡ii GKwU Mvox 20m/s †e‡M PjwQj| †eªK †P‡c G‡K 5 †m‡K‡Û _vwg‡q †`qv n‡jv| g›`bKvix ej KZ? [JU. 17-18] A. 400 N B. 500 N C. 600 N D. 300 N g›`bKvix ej, F = m v – u t = 100 (20 – 0) 5 = 400 N 05. 5 U‡bi GKwU UªvK N›Uvq 36km †e‡M Pj‡Q| GwU 4m `~i‡Z¡ _vg‡Z n‡j KZ e‡ji cÖ‡qvRb n‡e? [JU. 16-17] A. 62,500 N B. 63,500 N C. 64,500 N D. 65,500 N u = 36 3.6 = 10ms–1 ; †_‡g hvIqvq, v = 0 ; cÖhy³ ej, F = m u 2 2s = 5000 102 2 4 = 62,500 N 06. GKwU ej 4kg fi wewkó GKwU w¯
i e¯‧i Dci wµqv K‡i| Gi d‡j e¯‧wU 6s G 30m/s †eM cÖvß nq| e‡ji gvb KZ? [JU. 16-17] A. 10N B. 20N C. 30N D. 40N w¯
i e¯‧i, u = 0 ; †L‡jvqvo KZ…©K cÖhy³ Mo ej, F = ma = m t v = 4 6 30 = 20 N 07. 4 gm f‡ii GKwU ey‡jU 400 m/s G QyU‡j †h mge‡j Zv‡K 1s G _vwg‡q †`q Zv KZ? [JU. 11-12] A. 2.6 N B. 1.6 N C. 4.6 N D. 3.6 N †_‡g hvIqvq, v = 0 ; cÖhy³ ej, F = m u t = 4 1000 400 1 = 1.6N STEP 03 ANALYSIS OF RU QUESTION 01. F ej cÖ‡qv‡M GKwU †Uªb s `~i‡Z¡ _v‡g| hw` Avw`‡eM wظY Kiv nq Z‡e †Kvb `~i‡Z¡ †UªbwU _vg‡e? [RU. 17-18] A. S B. 2S C. S4 D. 4S u wظY n‡j s Gi gvb u 2 = 22 = 4 ¸Y n‡e| 02. 30 †KwR f‡ii GKwU e¯‧i Dci KZ ej cÖ‡qvM Ki‡j 1 wgwb‡U Gi †eM 36 kmh-1 e„w× cv‡e? [RU. 16-17] A. 4N B. 5N C. 6N D. 10N †eM e„w×, Δv = 36 3.6 ms–1 = 10ms–1 F = ma = m v t = 30 (10) 60 = 5 N 03. MvQ †_‡K 2 kg f‡ii GKwU dj bx‡Pi w`‡K coj| evZv‡mi evav 7.6 N n‡j djwUi Z¡iY KZ n‡e? [RU. 13-14] A. 9.8ms–2 B. 4.9ms–2 C. 6 ms–2 D. 2.45ms –2 f = ma a = F m = mg – 7.6 2 = 2 9.8 2 = 12 2 = 6ms–2 04. 4.8 U‡bi GKwU UªvK NÈvq 36km †e‡M P‡j‡Q| 4m `~i‡Z¡i g‡a¨ _vgv‡Z n‡j KZ e‡ji cÖ‡qvRb? [RU. 12-13] A. 62500N B. 48000N C. 60N D. 60000N u = 36 3.6 = 10ms–1 F = m u 2 2s = 4.8 103 102 2 4 = 60000N STEP 04 ANALYSIS OF CU QUESTION 01. 4 N Gi GKwU ej 2kg f‡ii GKwU w¯
i e¯‧i Dci wµqv Ki‡j 5 sec G e¯‧wU KZ `~iZ¡ AwZµg Ki‡e? [CU. 13-14] A. 15m B. 10m C. 20m D. 25m E. 5m F = 2ms t 2 ev,S = Ft2 2m = 4 5 2 2 2 = 25m 02. 30kg f‡ii GKwU e¯‧i Dci KZ ej cÖ‡qvM Ki‡j 1 wgwb‡U Gi †eM 36km/h e„w× cv‡e? [CU. 13-14 ; IU-10-11, 13-14] A. 5N B. 300N C. 108N D. 360N E. 10N F = mv t = 30 36 60 3.6 = 5 N 03. 200 kg f‡ii GKwU †gvUi Mvwo 30ms–1 †e‡M Pj‡Q †eª‡Ki mvnv‡h¨ MvwowU‡K 20m `~i‡Z¡ _vwg‡q †`qv nj| evav`vbKvix e‡ji gvb KZ? [CU. 10-11] A. 4500N B. 3000N C. 2500N D. 3500N E. 4000N F = m. v 2 u 2 2S = 200 (30) 2 2 20 = 4500N 04. 40N ej 5kg f‡ii GKwU w¯
i e¯‧i Dci 4†m. wµqv Ki‡j e¯ÍwU KZ †eM cÖvß n‡e? [CU. 08-09] A. 24ms–1 B. 32ms–1 C. 36ms–1 D. 40ms–1 E. 48ms–1 F = ma v = Ft m = 40 4 5 = 32ms–1

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 155 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 05 ANALYSIS OF GST QUESTION 01. f~-mgZ‡ji mv‡_ 30 †Kv‡Y AvbZ c‡_ GKwU 2 kg f‡ii e¯‧‡K 3 ms–2 Z¡i‡Y DVv‡Z n‡j e¯‧wUi Dci KZ wbDUb (N) ej cÖ‡qvM Ki‡Z n‡e? [GST-A. 20-21] A. 6.8 B. 11.8 C. 12.8 D. 15.8 S D info F = ma + mg sin = 2×3+2×9.8×1 2 = 15.8N PART A Analysis of General University Question JnU 01. wbDU‡bi MwZi 2q m~Î F = m a Abymv‡i F = 0 n‡j- [JnU. 16-17] A.v aªæeK B. v = 0 C. v cwieZ©bkxj D. †KvbwUB bq 2q m~Î F = m a Abymv‡i, m dv dt = m a = F, F = 0 n‡j, dv dt = 0, v = constant 02. 1J MwZkw³i †Kvb e¯‧i MwZi wecix‡Z 1N ej cÖ‡qv‡Mi Kiv n‡j e¯‧wU KZ `~iZ¡ AMÖmi n‡q †_‡g hv‡e? [JnU. 10-11] A.1m B. 10m C. 1 10m D. †KvbwUB bq W = FS ev, 1= 1 S ev, S = 1m KU 01. MvQ †_‡K 2 kg -i GKwU dj †mvRv wb‡Pi w`‡K co‡Q, evZv‡mi evav 8.6N n‡j, djwU cZ‡bi Z¡iY KZ? [KU. 13-14] A. 5.5ms–2 B. 5.5ms–1 C. 5.0ms–2 D. 5.0 ms–1 mg – FR = ma ev, a = mg – FR m = 2 9.8 – 8.6 2 = 5.5ms–2 02. 2 †m‡K‡Û 5 dzU †e‡M PjšÍ 2 cvDÛ f‡ii †Kvb e¯‧i Dci 60 cvDÛvj ej cÖ‡qvM K‡i G †eM †m‡K‡Û 60 dzU G DbœxZ Kiv nj| GB mg‡q e¯‧wU KZ c_ AwZµg Ki‡e? [KU. 04-05] A. 30 ft B. 60ft C. 45ft D. 75ft F = ma ev a = 60 2 ev a = 30fts–1 Ges s = v 2 u 2 2a = (60) 2 (5) 2 2 30 ev s = 60fts–1 IU 01. 60N ej 30kg f‡ii GKwU e¯‧i Dci 1 wgwbU wµqv K‡i| e¯‧wUi †e‡Mi cwieZ©bÑ [IU. 14-15] A. 240 ms–1 B. 120 ms–1 C. 60 ms–1 D . 90 ms–1 F = mv t ev, v = Ft m = 60 60 30 = 120ms–1 BU 01. GKwU 1000kg f‡ii †cø‡bi Rb¨ evZv‡mi evav 1800N| †cøbwU mg‡e‡M Pj‡j Gi Dci jwä ej KZ? [BU: 13-14] A. zero B. 11800N C. 1800N D. 9800N Ans C PART B Analysis of Science & Technology Question SUST 01. 1.5 kg f‡ii GKwU ej †mvRv wb‡Pi w`‡K co‡Q| evZv‡mi evav 7.5N n‡j ejwUi Z¡iY KZ m.s–2 ? [SUST-A, 19-20] A. 11.6 B. 14.8 C. 8.4 D. 4.8 E. 2.6 W Fk = ma a = mg Fk m = 1.5 9.8 7.5 1.5 = 4.8 ms2 m = 1.5 kg Fk = 7.5 N a = ? 02. 2kg f‡ii GKwU KYv v = (2 i + 4 j + 2 k) m/s †e‡M PjwQj| evwn¨K ej F-Gi wµqvq KYvwU †_‡g †Mj| F KZ©„K K…Z KvRKZ? [SUST. 18-19] A. 12J B. 24J C. 48J D. –12J E. –24J | v| 2 = 22 + 42 + 22 = 24(ms–1 ) 2 W = MwZ kw³i cwieZ©b = 1 2 mv2 = 24J †h‡nZz, F I s Gi w`K ci¯úi wecixZgyLx| ZvB KvRwU FYvZ¥K| 03. 800 g f‡ii GKwU e¯‧i Dci KZ N ej wµqv Ki‡j e¯‧wUi †eM 4s G (6 i + 3 j + 4 k) m/s n‡Z e„w× †c‡q (12 i – 3 j + 7 k) m/s n‡e? [SUST. 14-15] A. 5.4 B. 4.8 C. 3.6 D. 2.8 E. 1.8 v = (12 i – 3 j + 7 k – 6 i – 3 j – 4 k) = 6 i – 6 j + 3 k |v| = (6) 2 + (– 6) 2 + (3) 2 = 9 F = mv t = 0.8 9 4 = 1.8 04. 72 km/hr †e‡M Pjgvb GKwU Mvwoi PvjK 40.5 m `~‡i GKwU wkï‡K iv¯Ívi Dci †`L‡Z †c‡jb| mv‡_ mv‡_ †eªK Kivq MvwowU wkïwUi 50 cm mvg‡b G‡m †_‡g †Mj| MvwowU _vg‡Z KZ mgq jvM‡jv Ges KZ ej cÖ‡qvM Ki‡Z n‡jv? (Av‡ivnxmn Mvwoi fi 1000 kg) [SUST. 12-13] A.2 s, 5 104 N B. 3 s, 4 103 N C. 4 s, 5 103 N D. 5 s, 3 103 N E. 1.5 s, 6103 N miY S = 40.5 – 0.5 = 40m, Avw`‡eM, u = 72 3.6 = 20m/s †kl‡eM, = 0, F = m(v 2 – u 2 ) 2s = 1000 (202 – 0 2 ) 2 40 = 5 103N Ges F = mv t ev, 5 103 = 1000 20 t ev ,t = 4 †m. 05. 15m/s †e‡M Pjgvb 5kg f‡ii GKwU e¯‧i Dci 10N ej KZ mgq a‡i cÖ‡qvM Ki‡j †mwU 100m `~iæZ¡ AwZµg Ki‡e| [SUST. 11-12] A. 55s B. 7.5s C. 8s D.10s E. 20s F = 2mS t 2 ev, t 2 = 2mS F = 2 5 100 10 ev, t 2 = 100 ev, t = 10s 06. B‡jKUª‡bi fi 9.11 10–31 kg GKwU A¨vKwm‡jU‡i B‡jKUª‡bi Dci 1s mgq 1N ej cÖ‡qvM Kiv n‡j B‡jKUª‡bi MwZ n‡e- [SUST. 10-11] A. 1.10 1024m/s Gi KvQvKvwQ B. 10 1018m/s Gi KvQvKvwQ C. 1.10 1012m/s Gi KvQvKvwQ D. 3 108m/s Gi KvQvKvwQ F = m v t ev, v = Ft m = 1110–6 9.1110–31 = 0.101 1025ms–1 1.10 1024 Gi KvQvKvwQ 07. GKwU †Uªb N›Uvq 36km †e‡M hv‡”Q| 500m Gi gv‡S †UªbwU‡K _vgv‡Z n‡j KZ g›`b cÖ‡qvRb? [SUST. 07-08] A. 0.1 m /s2 B. 1 m/s2 C. 3.16m/s2 D. 10m/s2 u = 36km/h = 36 3.6ms–1 = 10ms–1 Ges 0ms–1 a = v 2 u 2 2s = 0 2 102 2 500 = –0.1ms–2 A_©vr 0.1 m/s–2 g›`b cÖ‡qvRb| 08. 27 kg f‡ii GKwU e¯‧i Dci KZ ej cÖ‡qvM Ki‡j GK wgwb‡U Gi †eM N›Uvq 12 km e„w× cv‡e? [SUST. 07-08] A. 15.5 N B. 20 N C. 2 N D. 1.5 N [v ~ u = 12kmh–1 = 12 3.6ms–1 = 10 3 ms–1 ] F = ma ev F = m v u t = 27 10 3 60 = 1.5N

156 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 09. 22000kg f‡ii e¯‧ w¯
i Ae¯
vq Av‡Q| GKwU ej 10.5 †m e¯‧wUi Dci KvR Kivq e¯‧wUi †eM 13.6 wg/‡m. n‡j ejwUi gvb KZ? [SUST. 06-07] A. 2200N B. 28500 C. 500N D. †KvbwUB bq F = ma = m. v u t = 22000 13.6 – 0 10.5 = 28495.23N 10. 5kg Ges 7kg IR‡bi `ywU e¯‧ GKwU cywji iwki `yB cÖv‡šÍ Szj‡Q| †h Z¡i‡Y fvix e¯‧wU wb‡Pi w`‡K avweZ n‡e Zv nj- [SUST. 04-05] A. 1.36 m/s B. 1.63 m/s C. 9.8 m/s D. 19.6 m/s a = m1 m2 m1 + m2 g = 2 7 + 5 9.8 = 2 12 9.8 = 1.63 m/s–2 11. 6N Gi GKwU ej Avbyf~wgKfv‡e 2kg f‡ii GKwU e¯‧i Ici 4s a‡i KvR Ki‡Q, e¯‧wUi MwZkw³i cwigvY- [SUST. 04-05] A. 12J B. 24 J C. 48 J D. 144 J F = ma a = F m = 6 2 = 3ms–2 v = at = 3 4 = 12ms–1 Ek = 1 2 mv2 = 1 2 2 (12)2 = 144J MBSTU 01. GKwU Mvox Pjv ïiæ Kivi 4 s c‡ii †eM 8 ms1 Ges 7 s c‡ii †eM 23 ms1 n‡j, Mo Z¡iY KZ? [MBSTU-C, Set-2 19-20] A. 8 ms2 B. 10 ms2 C. 12 ms2 D. 5 ms2 a = dv dt = 23 – 8 7 – 4 = 15 3 = 5 ms–2 02. 10kg f‡ii †Kvb e¯‧‡Z 3ms–2 Z¡iY m„wó‡Z KZ e‡ji cÖ‡qvRb n‡e? [MBSTU. 14-15] A. 10N B. 3.33N C. 20N D. 30N F = ma = 10 3 = 30N BSMRSTU 01. 25 MÖvg f‡ii GKwU e¯‧ 600ms–1 †e‡M QzU‡Q| †h mgej Zv‡K 4 †m‡K‡Û _vwg‡q w`‡Z cv‡i Zvi cwigvY KZ? [BSMRSTU. 14-15] A. 3.75N B. 3.75dynes C. 2.75N D. 2.75dynes F = mv t = 0.025 600 4 = 3.75N NSTU 01. kw³kvjx I `ye©j e‡ji AbycvZÑ [NSTU-A, 19-20] A. 102 B. 104 C. 106 D. 102 S Blank info kw³kvjx ej `~e©j ej = 1042 1030 = 1012 STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. 16kg f‡ii GKwU w¯
i e¯‧i Dci 4s e¨vcx 8N ej cÖhy³ nj| D³ e¯‧wUi †e‡Mi cwieZ©b n‡e- [BUET. 12-13] A. 0.5ms–1 B. 2.0ms–1 C. 4.0ms–1 D. 8.0ms–1 F = m(v–u) t ev, v – u = Ft m = 8 4 16 = 2ms–1 KUET 01. 5 U‡bi GKwU UªvK NÈvq 36km †e‡M P‡j‡Q| GwU 4m `~i‡Z¡ _vgv‡Z n‡j KZ e‡ji cÖ‡qvRb n‡e- [KUET. 12-13] A. 62.50KN B. 62.70KN C. 52.50KN D. 62KN F = mv2 2s u = 36 3.6 = 10 = 5 1000 102 2 4 = 62.50KN 02. 900kg f‡ii GKwU †gvUi UªvK NÈvq 60km †e‡M P‡j| †eK †P‡c UªvKwUi 50m `~‡i _vgvb nj| hw` gvwUi Nl©Y RwbZ ej 200N nq, Z‡e †eªK RwbZ e‡ji gvb wbY©q Ki| [KUET. 10-11] A. 2100N B. 2200N C. 2500N D. 2300N E. 3000N v = 60 3.6 = 16.666m/s; F = mv2 2s = 900 (16.666) 2 2 50 = 2500N †eªK RwbZ ej = 2500 – 200 = 2300N STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION DAT 01. 60 kg f‡ii GKwU e¯‘i Ici KZ ej cÖ‡qvM Ki‡j 1 wgwbU ci Gi †eM 10 ms–1 n‡e? [DAT. 18-19; MAT. 16-17] A. 20 N B. 05 N C. 40 N D. 10 N F = ma = m. v t = 60 10 60 = 10N STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. GKwU bj †_‡K 2ms-1 †e‡M cvwb †ei n‡q GKwU †`qvj‡K AvNvZ Ki‡Q| b‡ji cÖ¯
‡”Q` n‡”Q 0.03m2 | aiv hvK cvwb †`qvj †_‡K wd‡i Avm‡Q bv| †`qv‡ji Dci cvwb wK cwigvY ej cÖ‡qvM Ki‡Q? [Xv. †ev. 2016] A. 1000N B. 300N C. 240N D. 120 N S D info F = dp dt = d dt(mv) = d dt(Vv) = v d dt(Ax) = v.A.v = Av2 F = 1000 × 0.03 × 22 = 120 N 02. 1kg f‡ii `ywU e¯‧‡K ci¯úi n‡Z 1m `~‡i ¯
vcb Ki‡j Zviv ci¯úi‡K †h ej Øviv AvKl©Y K‡i Zvi gvb njÑ [iv. †ev. 2017] A. 1N B. 6.6710-7Nm2 kg-2 C. 6.6710-11Nm2 kg-2 D. 6.6710-11N S C info F = 6.67 × 10 –11 × 1 × 1 1 2 = 6.673 × 10–11 Nm2 kg–2 03. 0.25kg f‡ii GKwU wµ‡KU ej 40ms-1 †e‡M AvmwQj| GKRb †L‡jvqvo ejwU‡K 0.2 †m‡K‡Û _vwg‡q w`j| †L‡jvqvo KZ…©K cÖhy³ Mo ej KZ? [iv. †ev. 2017] A. 20N B. 10N C. –20N D. –50N S D info F = ma = m v – u t F = 0.25 × 0 – 40 0.2 N = – 50 N 04. m„ó Z¡i‡Yi gvb A. 2ms2 B. 4.8ms2 C. 6ms2 D. 8ms2 S A info a = F M1 + M2 + M3 = 120 N (25 + 20 + 15) kg = 2 ms–2 wb‡Pi DÏxcKwU co Ges 05 I 06 bs cÖ‡kœi DËi `vI: [h. †ev. 2016] `ywU e¯‘i fi 2kg Ges 5kg| G‡`i †eM h_vµ‡g 6ms1 Ges 4ms1 | 05. cÖ_gwU 2m `~i‡Z¡ _vgv‡bv nj| Gi Z¡iY KZ? A. 9 ms2 B. 1.5 ms2 C. 1.5 ms2 D. 9 ms2 S A info v 2 = v0 2 +2as a = v 2 – v0 2 2s = (0) 2 – (6) 2 2 × 2 = –9ms–2 06. wØZxq e¯‧‡K GKB `~i‡Z¡ _vgv‡Z KZ ej jvM‡e? A. 16 N B. 18 N C. 20 N C. 22 N S C info F = ma = m . v 2 – v0 2 2s = 5 × (0) 2 – (4) 2 2 × 2 = – 20N 07. wP‡Î gm„b AvbZ Zj †e‡q Mwo‡q cov e¯‧i Z¡iYÑ [e.‡ev. 2019] A. g B. g sin C. g cos D. k~b¨ Ans B

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 157 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 2 wjdU msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. a Z¡i‡Y wjdU Dc‡i DVv‡j, F = m(g+ a); Or, a g›`‡b wb‡P bvg‡j, F = (g + a) 02. a Z¡i‡Y wjdU †Kvb wb‡P bvgv‡j, F = m(g – a); Or, a g›`‡b Dc‡i DV‡j, F = m(g a) 03. Lvov Dc‡ii w`‡K mg‡e‡M DVv‡j, F = mg 04. Dc‡i DVv‡Z cÖ‡qvRbxq mgq, g a 2h T 05. Dci n‡Z bvg‡Z cÖ‡qvRbxq mgq g a 2h T Model Example01 GKwU wjdU 1ms 2 Z¡i‡Y wb‡P bvg‡Q wjd‡Ui †g‡Si 1.5m Dci †_‡K GKwU ej‡K w¯
i Ae¯
v †_‡K Qvov nj| g = 10ms 2 n‡j, wjd‡Ui †g‡S ¯úk© Ki‡Z ejwUi KZ mgq jvM‡e? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] wjdU wb‡P bvg‡Q, A_©vr a = 10 1 = 9ms –2 Avevi, h = ut + 1 2 at2 h = 1 2 at2 [u = 0] t 2h a = 2 1.5 9 = 3 9 = 1 3 = 0.57s t = 2h g a = 2 1.5 10 – 1 = 1 3 = 0.57sec Model Example02 GKwU wjdU Dc‡ii w`‡K 2ms 2 Z¡i‡Y DV‡Q| hw` ejwU wjd‡Ui Zjv n‡Z 3m Dci n‡Z †Q‡o †`Iqv nq| Zvn‡j ejwU co‡Z KZ mgq jvM‡e? [g = 10ms 2 ] General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] wjdU Dc‡i DV‡Q, A_©vr a = (10 + 2)ms –2 = 12ms –2 h = ut + 1 2 gt2 h = 1 2 gt2 t = 2h g = 2 3 12 = 6 12 = 1 2 t = 0.707sec. t = 2h g + a = 2 3 10 + 2 = 6 12 = 0.707sec CONCEPTUAL MATH MEx 01 iwki mvnv‡h¨ 2kg f‡ii GKwU e¯‧‡K Lvov Dc‡ii w`‡K 1ms–2 Z¡i‡Y Uvbv n‡”Q| iwki Uvb KZ? F = m(g + a) = 2 (9.8 + 1) 21.6N MEx 02 GKwU wjdU 4ms–2 Z¡i‡Y Dc‡i DV‡Q| wjd‡Ui †g‡Si 3m Dci n‡Z GKwU ej‡K †Q‡o †`qv n‡j GwU KZ mgq ci †g‡S ¯úk© Ki‡e? GLv‡b, cÖK…Z Z¡iY, = g + a = 9.8 + 4 = 13.8 ms–2 h = ut + 1 2 (g + a)t2 50 = 0 + 1 2 13.8 t 2 t = 2.7 †m‡KÛ| MEx 03 GKwU wjdU 2m/s2 Z¡i‡Y wb‡P bvg‡Q| wjd‡Ui †g‡Si 1m Dci †_‡K GKwU ej‡K w¯
i Ae¯
v †_‡K Qvov nj| g = 10m/s2 aiv n‡j wjd‡Ui †g‡S ¯úk© Ki‡Z ejwUi mgq jvM‡eg a h t 2 10 2 2 1 8 2 2 1 = 0.5 sec. NOW START PRACTICE 01. 100 kg f‡ii GKwU wjdU 1.8ms–2 Z¡i‡Y wb‡P bvg‡Q| wbd‡Ui g‡a¨ `uvov‡bv 60kg f‡ii GKRb e¨w³ KZ ej Abyfe Ki‡eb? 02. iwki mvnv‡h¨ 4kg fi wewkó GKwU e¯‧‡K Lvov Dc‡ii w`‡K mg‡e‡M Uvbv n‡”Q| iwki Uvb KZ? 03. 4ms–2 Z¡i‡Y wb¤œMvgx GKwU wjd‡Ui wfZi GKRb †jvK GKwU ej‡K 15ms–1 †e‡M Dc‡ii w`‡K wb‡ÿc Ki‡j GwU 2 †m. ci KZ D”PZvq DV‡e? NOW PRACTICE SOLVE : 01. F = m(g a) = 60 (9.8–1.8) =60 8 = 480 N 2. F = mg = 4 9.8 = 39.2N 03. cÖK…Z Z¡iY = g a = 9.8 4 = 5.8ms–2 h = ut – 1 2 (g – a)t2 15 2 – 1 2 5.8 2 2 = 18.4m REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU 7.0 kg f‡ii e¯‧ GKwU wjd‡Ui †g‡Si Dci w¯
i Ae¯
vq Av‡Q| wjd‡Ui DaŸ©Mvgx Z¡iY 2 m/s2 n‡j e¯‧i Dci †g‡S KZ©„K ej KZ? [DU. 15-16] A. 68.6 N B. 54.6 N C. 82.6 N D. 0.0 N T = F = m (g + a) = 7(9.8 + 2) = 82.6 N 02. GKwU wjd&‡Ui †g‡S‡Z ivLv GKwU IRb gvcvi h‡š¿i Dci GKRb 50kg fi wewkó gvbyl `vwo‡q Av‡Q| wjd&wU w¯
Z Ae¯
v †_‡K 2m/s2 Z¡i‡Y 1 †m‡KÛ a‡i Dc‡ii w`‡K I‡V, Zvi ci mg`ªæwZ‡Z DV‡Z _v‡K| wjdwU Pjvi ci †_‡K IRb gvcvi h‡š¿ KZ fi †`L‡e? (a‡i bvI ga¨Kl©b RwbZ Z¡iY 10m/s2 ) [DU. 09-10] A. First 60 kg and then 0 kg B. Always 50 kg C. First 60 kg and then 50 kg D. Always 60 kg hLb 2ms–2 Z¡iY Dc‡i D‡V ZLb cÖwZwµqv w = m(g + a) = 50(10 + 2) = 600N ZLb w = mg ev, m = w g = 600 10 = 60kg Avi †k‡l m(g + a) n‡e = mg = 50 1 = 50

158 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 03. GKwU wjdU 2m/s2 Z¡i‡Y wb‡P bvg‡Q| wjd‡Ui †g‡Si 1m Dci †_‡K GKwU ej‡K w¯
i Ae¯
v †_‡K Qvov nj| g = 10m/s2 aiv n‡j wjd‡Ui †g‡S ¯úk© Ki‡Z ejwUi mgq jvM‡e- [DU. 02-03] A. 0.5 sec B. 0.6 sec C. 0.7 sec D. 1.1 sec t = 2h g – a = 2 1 10 – 2 = 2 8 = 1 2 = 0.5 sec. STEP 02 ANALYSIS OF JU QUESTION 01. 100 kg f‡ii GK e¨w³ wjdU G `vwo‡q Av‡Q| wjdU& hw` 2ms–2 Z¡i‡Y Dc‡i DV‡Z _v‡K Zvn‡j †jvKwUi Dci DaŸ©gyLx cÖwZwµqv ej n‡e- [JU. 15-16] A. 1000 N B. 1280 N C. 1180 N D. †KvbwUB bq T = m(g + a) = 100(9.8 + 2) = 1180 STEP 03 ANALYSIS OF RU QUESTION 01. GKwU wjdU 2.8 ms–2 Z¡i‡Y bx‡P bvg‡Q| wjd‡Ui g‡a¨ `uvov‡bv GKRb e¨vw³i fi 90 kg n‡j wZwb †h IRb Abyfe Ki‡eb- [RU. 17-18] A. 252 N B. 630 N C. 882 N D. 1134 N wb¤œMvgx wjd‡Ui †¶‡Î, F = m (g – a) = 90 (9.8–2.8) = 630 N STEP 04 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question SUST 01. †Kvb e¯‧i Dci GKwU Constant ej F cÖhy³ n‡”Q Ges Zv a Z¡iY wb‡q GwM‡q Pj‡Q e¯‧i †eM hLb v0, ZLb nVvr e‡ji gvb k~b¨ K‡i †`qv n‡jv, ZLb †_‡K t mg‡q e¯‧ †h `~iZ¡ AwZµg Ki‡e Zv n‡jv- [SUST. 12-13] A. v0 t B. v0t + 1 2 at2 C. v0t – 1 2 at2 D. 1 2 v0t mib S = v0t – 1 2 at2 02. GKwU wjd&U 1 m/s-2 Z¡i‡Y Dc‡i DV‡Q| †Zvgvi IRb AvbygvwbK- [SUST. 05-06] A. 10% e„w× cv‡e B. 10 K‡g hv‡e C. mgvb _vK‡e D. †KvbwU bq g a g h W W ev, g g a W h W ev, 9.8 9.8 1 W h W ev, 9.8 10.8 W h W ev, 1 1.101 W h W kZKiv e„w× = (1.101–1) 100 = 0.10 100% = 10% Concept 3 i‡KU msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. i‡K‡Ui Dci wµqvkxj jwäZ¡iY a = M Vr dt dm – g GLv‡b dt dm n‡jv R¡vjvbx wbM©g‡bi nvi| 02. i‡K‡Ui †ÿ‡Î DaŸ©gyLx av°v ej, F = Vr . dt dm 03. i‡K‡Ui Z¡iY, ) dt dm ( M 1 M F a . Vr 04. t-Zg †m‡K‡Û i‡K‡Ui †eM, Vth = V ln M Mt – gt [Mt = ïay i‡K‡Ui †eM] CONCEPTUAL MATH MEx 01 GKwU i‡KU cÖwZ †m‡K‡Û 60gm R¦vjvwb LiP K‡i Ges H R¦vjvbx M¨vm 90km/s †e‡M i‡KU n‡Z wb®…všÍ nq| i‡K‡Ui Dci KZ ej wµqv K‡i? i‡K‡Ui Dci wµqviZ ej, F = vr dm dt = 90 103 60 10–3 1 = 5.4 103N MEx 02 10,000kg R¡vjvbxmn GKwU i‡K‡Ui fi 15000kg| R¡vjvbx hw` 200 kg/s nv‡i cy‡o i‡K‡Ui mv‡c‡¶ 2000m/s †e‡M wbM©Z nq, Zvn‡j i‡K‡Ui Dci cÖhy³ av°v ev _ªvó Ges R¡vjvbx †kl nevi gyn~‡Z© i‡K‡Ui †eM KZ? F = dm dt . vr = 200 × 2000 = 4×105N, Vth = V ln M Mt – gt = 2000 ln 15000 5000 – 9.8 × 50 = 1707 ms–1 GLv‡b 200 kg R¡vjvbx cy‡o = 1sec 10000kg R¡vjvbx cy‡o = 10000 200 = 50sec NOW START PRACTICE 01. gnvKv‡k Aew¯
Z GKwU kvUj gnvKvk hv‡bi fi 3 103 kg Ges R¡vjvwbi fi 50kg R¡vjvbx 5kgs–1 nv‡i e¨eüZ n‡j Ges 150ms–1 mylg `ªæwZ‡Z wbM©Z n‡j kvUj hv‡bi Dci av°v wbY©q Ki| 02. DaŸ©gyLx hvÎvi cÖ_g †m‡K‡Û GKwU i‡KU wb‡Ri f‡ii 1 1000 Ask nvivq| i‡KU n‡Z wbM©Z M¨v‡mi MwZ‡eM 2000ms–1 n‡j, i‡K‡Ui Z¡iY KZ? 03. 5000kg R¡vjvbxmn GKwU i‡K‡Ui fi 20,000kg. i‡K‡Ui mv‡c‡¶ 3000ms–1 `ªywZ‡Z R¡vjvbx 200kg s–1 nv‡i cy‡o| i‡K‡Ui hw` Lvov Dc‡ii w`‡K wbw¶ß n‡q _v‡K Z‡ei. i‡K‡Ui DaŸ©Mvgx av°v ii. wb‡¶‡ci mgq i‡K‡Ui Dci cÖhy³ ej iii. R¡vjvbx †kl nevi mgq m„ó cÖhy³ jwä ej KZ? NOW PRACTICE SOLVE : 01. F = dm dt .vr = 5 150 = 750N 02. a = vr dm dt . 1 m = 2000 m 1000. 1 m = 2ms–2 03. i. i‡K‡Ui av°v , F = Vr dm dt = 3000 200 = 6 105 N dm dt = 200kgs ii. wb‡¶‡ci mgq i‡K‡Ui Dci cÖhy³ ej = F– m0g = 6105 –(20000 9.8) = 4.04 105 iii. R¡vjvbx †kl nevi mgq m„ó cÖhy³ jwä ej = F–mg = 6105 –(5000 9.8) = 5.51 105N

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 159 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. 10,000kg R¡vjvbxmn GKwU i‡K‡Ui fi 15000kg| R¡vjvbx hw` 200 kg/s nv‡i cy‡o i‡K‡Ui mv‡c‡¶ 2000m/s †e‡M wbM©Z nq, Zvn‡j i‡K‡Ui Dci cÖhy³ av°v ev _ªvó KZ? [DU. 15-16] A. 4 105N B. 4 10–5N C. 4 104N D. 4 106N F = dm dt . vr = 200 × 2000 = 4 × 105N STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. gnvKv‡k Aew¯
Z GKwU kvUj gnvKvk hv‡bi fi 3 × 103 kg Ges R¡vjvwbi fi 50kg R¡vjvwb 5kgs–1 nv‡i e¨eüZ n‡j Ges 150ms–1 mylg `ªæwZ‡Z wbM©Z n‡j kvUj hv‡bi Dci av°v n‡e- [KU. 09-10] A. 945N B. 850N C. 750N D. 650N av°v, F = vr dm dt = 5 × 150 = 750N PART B Analysis of Science & Technology Question JUST 01. GKwU i‡KU cÖwZ †m‡K‡Û 0.07 kg R¡vjvwb LiP K‡i| i‡KU †_‡K wbM©Z M¨v‡mi †eM 100 km/s n‡j i‡K‡Ui Dci KZ ej wµqv K‡i? (GLv‡b AwfKl© e‡ji cÖfve D‡c¶v Kiv †h‡Z cv‡i)| [JUST. 14-15] A. 7 × 103N B. 7N C. 7 × 105N D. 7 × 106N E. 8N F = v. dm dt = 100 103 0.07 = 7 103N MBSTU 01. GKwU i‡KU DaŸ©gyLx hvÎvi cÖ_g 2 †m‡K‡Û Gi f‡ii 1 60 Ask nvivq| i‡KU n‡Z wb®…všÍ M¨v‡mi MwZ‡eM 3600ms–1 n‡j i‡K‡Ui Z¡iY KZ? [MBSTU. 15-16] A. 36ms–2 B. 20.2ms–2 C. 15.2ms–2 D. 30ms–2 a = vr m dm dt – g = 3600 m m 60 2 – g = 3600 60 2 – 9.8 = 30 – 9.81 = 20.2 Concept 4 fi‡e‡Mi msiÿY m~Î msµvšÍ MvwYwZK cÖ‡qvM FORMULA fi‡e‡Mi msi¶Y m~Î: (†Kej msNl© n‡j cÖ‡hvR¨) 01. m1 f‡ii GKwU e¯‘ u1 †e‡M Pj‡Q| Gi fi cwieZ©b n‡q m2 Ges hw` bZzb †eM u2 nq Z‡e, m1u1 = m2u2 02. m f‡ii e¯‘ av°vi ci m1, m2, m3 f‡ii wZbwU UzKivq cwiYZ n‡q, u1, u2, u3 †e‡M Pj‡j G‡`i cÖ‡Z¨‡Ki fi‡eM mgvb n‡e| A_©vr m1u1 = m2u2 = m3u3 03. m1 I m2 f‡ii `ywU e¯‘ h_vµ‡g u1 I u2 †e‡M GKB w`‡K MwZkxj n‡j wgwjZ e¯‘i †eM, v = m1u1 + m2u2 m1 + m2 04. m1 I m2 f‡ii `ywU e¯‘ h_vµ‡g u1 I u2 †e‡M wecixZ w`‡K Pj‡Q| e¯‘Øq wgwjZ nIqvi ci †eM v = m1u1 m2u2 m1 + m2 05. m1 f‡ii GKwU e¯‘ u1 †e‡M m2 f‡ii GKwU w¯’i e¯‘‡K av°v w`‡j, av°vi ci wgwjZ e¯‘؇qi †eM v = m1u1 m1 + m2 06. MV = mv 07. e›`y‡Ki cðvr †e‡Mi Rb¨, V = mv M [g‡b ivL‡e]: hw` cðvr †eM bv e‡j e›`y‡Ki †eM ej‡j, V = mv M CONCEPTUAL MATH MEx 01 5 kg Zv‡ii GKwU ivB‡dj †_‡K 20g f‡ii GKwU ey‡jU 1000 m/s MwZ‡Z Qz‡U hvq| wcQb †_‡K ivB‡d‡ji av°vi †eM KZ? fi‡e‡Mi msiÿY m~Îvbyhvqx, Avw` fi‡eM = †kl fi‡eM ivB‡dj I ¸wji Avw` fi‡e‡Mi mgwó = ivB‡dj I ¸wji †kl fi‡e‡Mi mgwó O = MV + mv 5V + 0.002 1000 = 0 V = –4ms–1 MEx 02 200kg f‡ii GKwU w¯
i †b․Kv †_‡K 20kg Ges 25kg f‡ii 2wU evjK h_vµ‡g 10ms 1 Ges 8ms 1 †e‡M GKB w`‡K jvd w`‡j †b․KvwU KZ †e‡M Pj‡Z ïiæ Ki‡e? Mv = m1u1 + m2u2 v = m1u1 + m2u2 M = 20 10 + 25 8 200 = 2ms–1 MEx 03 5 †gwUªK Ub f‡ii evjy †evSvB UªvK 20ms–1 PjwQj| PjšÍ Ae¯
vq cÖwZ wgwb‡U Uªv‡Ki wQ`ª w`‡q 10kg evjy wb‡P c‡o hvq| 10 wgwbU ci Uªv‡Ki †eM KZ? 20 min ci Uªv‡Ki fi, m2 = 5000 (10 10) = 4900kg m1u1 = m2u2 u2 u2 = 20.4ms –1 MEx 04 Mv‡Qi Wv‡j emv 2kg f‡ii GKwU cvwL‡K 0.02kg f‡ii GKwU ey‡jU 400ms-1 Avbyf~wgK †e‡M AvNvZ K‡i cvwLwUi wfZ‡iB i‡q †Mj| cvwLi †eM KZ? cvwLi fi, m1 = 2kg, ey‡j‡Ui fi, m2 = 0.02kg, cvwLi Avw`†eM u1 = 0ms1 , ey‡j‡Ui Avw`†eM, u2 = 400ms-1 , wgwjZ †eM v = ? m1u1 + m2u2 = (m1+m2)v v = m2u2 m1 + m2 = 400 0.02 2.02 = 3.96ms1 4ms1 MEx 05 2 10g fi Gi GKwU ey‡jU 3 kg f‡ii GKwU e›`yK †_‡K 300 ms–1 †e‡M †ei n‡j e›`y‡Ki cðvr †eMe›`y‡Ki †¶‡Î, MV + mv = 0 e›`y‡Ki cðvr †eM, V = M mv = 3 0.01300 = 1ms–1

160 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW START PRACTICE 01. 4 kg I 6 kg f‡ii `yBwU e¯‧ h_vµ‡g 10 ms–1 Ges 5 ms–1 †e‡M GKB w`‡K MwZkxj| ci¯úi av°v LvIqvi ci e¯‧ `yBwU hy³ Ae¯
vq Pj‡Z _vK‡j, hy³ e¯‧i †eM KZ? 02. w¯
i Ae¯
vq _vKv GKwU e¯‧ we‡ùvi‡Yi d‡j M1 Ges M2 f‡ii `ywU L‡Û wef³ nq Ges LÛ `ywU wecixZ w`‡K h_vµ‡g V1 Ges V2 †eM cÖvß nq| V1Ges V2 Gi AbycvZ KZ n‡e? NOW PRACTICE SOLVE : 01. fi‡e‡Mi msi¶Y m~Î Abyhvqx, m1 u1 + m2 u2 = (m1 + m2 )v hy³ e¯‘i †eM, v = m1 u1 + m2 u2 m1 + m2 = 4 10 + 6 5 4 + 6 = 7ms–1 02. fi‡e‡Mi msi¶Ym~Î Abyhvqx, M1V1 + M2 (–V2) = 0 M1V1 = M2V2 V1 V2 = M2 M1 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. w¯
i Ae¯
vq _vKv GKwU e¯‧ we‡ùvwiZ n‡q m1 I m2 f‡ii `yBwU e¯‧‡Z cwiYZ n‡q h_vµ‡g v1 I v2 †e‡M wecixZ w`‡K Pjgvb| v1 v2 Gi AbycvZ KZ? [DU.A 19-20, 17-18, 12-13, DU. Tech. 20-21, JU. 14-15] A. m1 m2 B. m2 m1 C. m2 m1 D. m2 m1 S B info we‡ùvwiZ nIqvi Av‡M e¯‘wU w¯’i Ae¯’vq GKwU m¤ú~Y© e¯‘ wQj ZvB Avw`‡e‡Mi mgwó 0 (k~b¨)| fi‡e‡Mi wbZ¨Zv m~Î Abymv‡i, (m1 + m2) u = m1v1 m2v2 (m1 + m2) 0 = m1v1 m2v2 m1v1 = m2v2 v1 v2 = m2 m1 02. 4kg I 6kg f‡ii `yBwU e¯Í h_vµ‡g 10m/s Ges 5m/s †e‡M GKB w`‡K MwZkxj| ci¯úi av°v LvIqvi ci e¯Í `yBwU hy³ Ae¯
vq Pj‡Z _vK‡j, hy³ e¯‧i †eM KZ? [DU. 07-08, 13-14] A. 10m/s B. 7m/s C. 6m/s D. 4m/s cÖ_g e¯‘i fi m1 = 4kg, wØZxq e¯‘i fi m2 = 6kg cÖ_g e¯‘i †eM v1 = 10ms–1 , wØZxq e¯‘i †eM v2 = 5ms–1 e¯‘ `yBwU mshy³ Ae¯’vq †eM v =? v = m1v1 + m2v2 m1 + m2 ev, v = 4 10 + 6 5 4 + 6 = 70 10 = 7ms–1 03. 20 m/s †e‡M Pjgvb 1000 kg f‡ii GKwU UªvK 1500 kg f‡ii GKwU w¯
i UªvK‡K av°v w`‡q GKmv‡_ hy³ n‡q †h †e‡M Pj‡Z _vK‡e- [DU. 02-03] A. 12.5 m/s B. 10 m/s C. 8 m/s D. 7.5 m/s 1 2 1 1 . m m m u v 2500 20 1000 1000 1500 20 1000 = 8 m/sec. STEP 02 ANALYSIS OF JU QUESTION 01. 30N ej 5kg f‡ii GKwU e¯‧i Dci 10sec wµqv K‡i| e¯‧wUi fi‡e‡Mi cwieZ©b KZ n‡e? [JU. 17-18] A. 400 kg ms–1 B. 300 kg ms–1 C. 500 kg ms–1 D. 100 kg ms–1 e‡ji NvZ = fi‡e‡Mi cwieZ©b = Ft = 3010 = 300 kg ms–1 02. 5kg f‡ii GKwU e›`yK n‡Z 0.01kg f‡ii GKwU ¸wj 300 ms–1 †e‡M †ei n‡j e›`y‡Ki cðvr †eM KZ n‡e? [JU. 13-14] A. 0.06 ms–1 B. 0.6 ms–1 C. 1800 ms–1 D. 18000 ms–1 fi‡e‡Mi msi¶Ym~Î Abyhvqx, MV = mv V = mv M = 0.01 300 5 = 0.6 ms–1 03. GKwU e¯‧i Dci 5N ej 10s wµqv K‡i| fi‡e‡Mi cwieZ©b KZ? [JU. 10-11] A. 50kgms–1 B. 10kgms–1 C. 60kgms–1 D. 100kgms–1 fi‡e‡Mi cwieZ©b = Ft = 5 10 = 50 STEP 03 ANALYSIS OF RU QUESTION 01. 5kg f‡ii GKwU ivB‡dj n‡Z 20g f‡ii GKwU ey‡jU 1000 ms–1 †e‡M wbM©Z nq| ivB‡d‡ji cðvr †eM KZ?[RU. Moderna, Set-2. 20-21, DU. 12-13] A. 4ms–1 B. 20ms–1 C. 2ms–1 D. 5ms–1 S A info fi‡e‡Mi msiÿY m~Îvbyhvqx,MV= mv V = 0.02 1000 5 = 4ms–1 02. 5 †gwUªK Ub f‡ii evjy †evSvB GKwU UªvK 20ms–1 †e‡M PjšÍ Ae¯
vq Uªv‡K wQ`ª n‡q cÖwZ wgwb‡U 20kg nv‡i evjy UªvK †_‡K wb‡P co‡Z ïiæ Ki‡jv| fi‡eM msiwÿZ n‡j 50 wgwbU ci Uªv‡Ki †eM KZ n‡e? [RU-C, Set-3. 19-20] A. 20ms–1 B. 25ms–1 C. 25.5ms–1 D. †KvbwUB bq GLv‡b, m1 = 5000 Kg ; v1 = 20 ms–1 ; 20 kg cÖwZ wgwb‡U c‡o †M‡j 50 wgwbU ci fi, m2 = 5000 – (20 50) = 5000 – 1000 = 4000 kg; v2 = ? ; Avgiv Rvwb, m1v1 = m2v2 v2 = 5000 20 4000 = 25ms–1 03. 1200 Kg f‡ii GKwU Mvwo 20 m/sec `ªæwZ‡Z PjwQj| AZtci MvwowU 800 Kg f‡ii GKwU w¯
i Mvwo‡K av°v w`j| av°vi ci Mvwo `ywU GKwÎZ n‡q 120 m wcQjv‡q †_‡g †Mj| evav`vbKvix e‡ji gvb KZ? [RU. 16-17] A. 600 N B. 800 N C. 1000 N D. 1200 N fi‡e‡Mi msi¶Ym~Î Abyhvqx, m1v1 + m2v2 = Mu 1200 20 + 800 0 =2000 u u = 12 ms–1 u = 12 ms–1 ; †_‡g hvIqvq, v = 0 ; cÖhy³ ej, F = ma = m u 2 2s = 2000 122 2 120 = 1200 N 04. 108kmhr–1 †e‡M MwZkxj 50kg f‡ii GKwU e¯‧i fi‡eM KZ? [RU. 12-13] A. 5400kgms–1 B. 2700kgms–1 C. 1500kgms–1 D. 2160kgms–1 v = 108 3.6 = 30m/s e¯‘i fi‡eM = mv = 50 30 =1500kgms–1 05. 2000kg f‡ii GKwU UªvK 36km/hr †e‡M GKwU MvQ‡K AvNvZ Ki‡j 0.1sec mg‡q †_‡g †Mj| Uªv‡Ki Dci msN‡l©i Mo ej KZ? [RU. 11-12] A. 2 × 103N B. 2 × 102N C. 2 × 104N D. 2 × 105N F = ma = mv t = 2000 10 – 0 0.1 = 2 × 105N; v = 36 3.6 ms–2 v = 10 06. 1100kg f‡ii GKwU Mvox 50ms–1 †e‡M Avbyf~wgK iv¯Ívi Dci w`‡q Pjvi mgq nVvr 100kg f‡ii GKwU e¯‧ Mvox †_‡K evB‡i c‡o †Mj| Mvwoi eZ©gvb †eM KZ? [RU. 11-12] A. 44ms–1 B. 50 ms–1 C. 55 ms–1 D. 100 ms–1 Mvwoi fi M = 1100kg, Mvwoi †eM V = 50 ms–1 †ei n‡q hvIqvi ci fi m = 1100 – 100 = 1000kg; Mvwoi eZ©gvb v =? MV = mv ev, v = MV m = 1100 50 1000 = 55 ms–1

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 161 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 04 ANALYSIS OF CU QUESTION 01. GKwU KYvi r = 4m, p = 2Nsec Ges r I p Gi g‡a¨ AšÍM©Z †KvY 300 n‡j, KYvwUi †K․wYK fi‡eM KZ? [CU-A, Set-2. 19-20] A. 2 kgm2 s –1 B. 4 kgm2 s –1 C. 4 kg2ms–1 D. 8 kgm2 s –1 L = rpsin = 4 × 2 × sin30 = 4kgm2 s –1 02. 5N ej †Kvb e¯‧i Dci 6s wµqv K‡i| e¯‧i fi‡e‡Mi cwieZ©b KZ? [CU. 15-16] A. 15kgms–1 B. 30 kgms–1 C. 60kgms–1 D. 30 kgms–1 E. 25 kgms–1 fi‡e‡Mi cwieZ©b = 5 × 6 = 30 kg ms–1 03. 5 kg f‡ii GKwU e›`yK †_‡K 10gm f‡ii GKwU ¸wj 400 m/sec †e‡M †ewi‡q †Mj| e›`y‡Ki cðvr †eM KZ? [CU. 07-08] A. 10 m/sec B. 0.8 m/sec C. 1 m/sec D. 12 m/sec E. 2 m/sec MV = mv ev, V = mv M = 0.01 400 5 = 0.8m/sec STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. 10g fi (mass)Gi GKwU ey‡jU 3 kg f‡ii GKwU e›`yK †_‡K 300 ms–1 †e‡M †ei n‡j e›`y‡Ki cðvr †eM- [JnU. 17-18] A. 0.05 ms–1 B. 0.25 ms–1 C. 2.0 ms–1 D. 1.0 ms–1 e›`y‡Ki †¶‡Î, MV + mv = 0; e›`y‡Ki cðvr †eM, V = mv M = 0.01 300 3 = 1ms–1 02. 30Kg fi wewkó GKwU e¯‧i Dci 250N ej 5 †m‡KÛ wµqv Ki‡j e¯‧wUi fi †e‡Mi cwieZ©b KZ n‡e? [JnU. 13-14] A. 1250 Kg.m/sB. 1250m/s2 C. 150Kg.m/s D. 7500m/s2 fi †e‡Mi cwieZ©b = Ft = 250 5 = 1250 IU 01. 6 †KwR f‡ii GKwU e›`yK n‡Z 0.01 †KwR f‡ii GKwU ¸wj 300 wg./†m. †e‡M †ei n‡j e›`y‡Ki cðvr †eM n‡e [IU. 11-12; JU. 13-14] A. 30 wg./†m. B. 0.5 wg./†m. C. 18000 wg./†m. D. 0.6 wg./†m. MV = mv ev, V = mv M = 0.01 300 6 = 0.5m/s PART B Analysis of Science & Technology Question SUST 01. 15 m.s–1 †e‡M Pjgvb 160g f‡ii GKwU ej‡K Zzwg e¨vU w`‡q AvNvZ Ki‡j, ejwU 25 m.s–1 †e‡M wd‡i †Mj| e¨vU-ej msN‡l©i ¯
vwqZ¡Kvj 10 ms n‡j Zzwg M‡o KZ N ej w`‡q AvNvZ K‡iQ? [SUST-B, 19-20] A. 160 B. 640 C. 560 D. 1600 E. 120 F t = m(v u) F = m(v u) t = 0.160 (25 + 15) 10 103 N 640 N 02. 4kg f‡ii GKwU e¯‧ (2i ˆ + 3j ˆ ) m.s–1 †e‡M Ges 6kg f‡ii Aci GKwU e¯‧ (–4i ˆ – 6j ˆ ) m.s–1 †e‡M PjvKvjxb mshy³ n‡j Dnviv GK‡Î KZ m.s–1 †e‡M Pj‡e? [SUST-B, 19-20] A. 5.44 B. 6.88 C. 5.77 D. 6.99 E. 2.88 M1 = 4kg; u1 = 2 2 + 32 = 13 ; m2 = 6 kg; u2 = (4) 2 + (6) 2 = 16 + 36 = 52 m1u1 + m2u2 = (m1 + m2) v 4 13 + 6 52 10 = v v = 5.77 ms1 03. 40kg I 60kg f‡ii `ywU e¯‧ h_vµ‡g 10 m/s I 5m/s †e‡M ci¯úi wecixZ w`K †_‡K Avmvi mgq G‡K Aci‡K av°v w`j| av°vi ci e¯‧Øq GK‡Î hy³ n‡q †KZ m/s †e‡M Pj‡e? [SUST. 16-17, CU. 12-13] A. 1 B. 2 C. 6 D. 7 m1u1 + m2 ( u2) = (m1 + m2)v 40 10 + 60 (–5) = (40 + 60)v 100 = 100v v = 1ms1 04. GKwU gv‡e©j‡K 0.6 m DuPz †Uwe‡ji cÖvšÍ †_‡K †UvKv w`‡j gv‡e©jwU 5.0 m/s †eM AR©b K‡i| gv‡e©jwU †Uwe‡ji cÖvšÍ n‡Z KZ m `~‡i gvwU‡Z co‡e? [SUST. 14-15] A. 0.6 B. 0.8 C. 1.75 D. 2.35 E. 14.7 gx2 = 2u2 y ev, x = 2u2 y g = 2 (5) 2 0.6 9.8 = 1.749 = 1.75 05. m I 2m f‡ii `ywwU †MvjK wecixZ w`K †_‡K h_vµ‡g v Ges v †e‡M G‡m ci¯úi‡K av°v w`‡q‡Q| Gici m Ges 2m Gi MwZ‡eM h_vµ‡g- [SUST. 08-09] A. 0, 0 B. –5v/3, v/3 C. 5v/3, v/3 D. 4v/3, 2v/3 v = m1u1 + m2u2 m1 + m2 = mv + 2m(v) m + 2m = mv 2mv 3m = mv 3m = – v 3 06. GKwU †MvjK Zvi mgvb f‡ii w¯
i Ae¯
vq Aci GKwU †MvjK‡K av°v w`‡q‡Q| av°v †`qvi c~‡e© †MvjK `ywUi †eM h_vµ‡g v I 0 n‡j, av°v †`qvi ci Zv‡`i †eM KZ? [SUST. 07-08] A. 0 Ges v B. v /2 Ges v /2 C. –v Ges 0 D. –v Ges v †kl †eM V = ? ev, mv = 2 m.V; mv + 0 = (m + m) V ev, v = 2V ev, 2V = v ev, V = v 2 07. 400kg f‡ii GKwU w¯
i †b․Kvi MjyB †_‡K 20kg Ges 25kg f‡ii `ywU †Q‡j h_vµ‡g 10m/s Ges 8m/s †e‡M GKB w`‡K jvd w`‡j †b․KvwU KZ †e‡M Pj‡Z ïiæ Ki‡e? [SUST. 06-07] A. 0 m/s B. 1.0 m/s C. 1.5 m/s D. 2.0 m/s jvd †`qvi c‡i †b․Kvi †eM = v1 m1u1 + m2u2 + m3u3 = m1v1 + m2v2 + m3v3 v1 + 20 10 + 25 8 400v1 = 400 v1 = –400 400 = –1ms–1 A_©vr †b․KvwU 1ms–1 †e‡M wcQ‡bi w`‡K Pj‡Z ïiæ Ki‡e| JUST 01. 15 kg f‡ii GKwU e›`yK n‡Z 0.01 kg f‡ii GKwU ¸wj 300 ms–1 †e‡M †ei n‡q †Mj| e›`y‡Ki cðvr‡eM KZ? [JUST-C, 19-20] A. 0.1 ms–1 B. 1.8 ms–1 C. 0.5 ms–1 D. 0.2 ms–1 cðvr †eM, V = mv M = 1 100 ×300 15 = 0.2 ms–1 02. 4kg f‡ii GKwU e›`yK n‡Z 0.005 kg f‡ii GKwU ¸wj 200 ms–1 †e‡M †ei n‡j e›`y‡Ki cðvr †eM KZ? [JUST. 15-16] A. 0.21 ms–1 B. 0.23 ms–1 C. 0.25 ms–1 D. 0.30 ms–1 MV = mv ev, v = mv M ev, v = 0.005×200 4 = 0.25ms–1 MBSTU 01. 40 Kg Ges 60 Kg f‡ii `yBwU e¯‧ ci¯úi wecixZ w`‡K h_vµ‡g 10 ms1 Ges 5 ms1 †e‡M hvIqvi c‡_ G‡K Aci‡K av°v w`j| av°vi ci e¯‧ `ywU GK mv‡_ hy³ †_‡K KZ †e‡M Pj‡Z _vK‡e? [MBSTU-C, Set-2 19-20] A. 2.8 ms1 B. 2.0 ms1 C. 1.0 ms1 D. 0.5 ms1 v = m1v1 – m2v2 m1 + m2 = 40 10 – 60 5 40 + 60 = 100 100 = 1 ms–1 02. 10gm f‡ii GKwU ey‡jU 6 kg f‡ii GKwU e›`yK †_‡K 300ms–1 †e‡M wbw¶ß n‡jv| e›`yKwUi cðvr †eM n‡jv- [MBSTU. 14-15] A. -0.5ms–1 B. 0.5ms–1 C. - 5ms–1 D . 5ms–1 MV = mv ev, V = 0.01300 6 = 0.5ms–1

162 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 03. hw` e›`y‡Ki fi ¸wji f‡ii 100 ¸Y nq, Z‡e e›`y‡Ki cðv‡eM ¸wji †e‡Mi-[MBSTU. 14-15] A. mgvb B. 1 10 Ask C. 1 100 Ask D. 10 ¸Y MV = mv ev, V = 1 100 v BSFMSTU 01. L = r p n‡jv- [BSFMSTU-A, 19-20] A. †K․wYK Z¡iY B. UK© C. †K․wYK fi‡eM D. •iwLK fi‡eM Ans C STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION CKRuet. Combind 01. 4kg f‡ii GKwU cvwL Mv‡Q e‡m Av‡Q| 20gm f‡ii GKwU ¸wj 200m/sec †e‡M cvwLwU‡K AvNvZ Kij| cvwLwUi Avbyf~wgK †eM KZ n‡e hw` ¸wjwU cvwLi kix‡i †_‡K hvq? [CKRUET. 2020-21] A. 2 m/sec B. 1 m/sec C. 10 m/sec D. 9.95 m/sec E. 0.995 m/sec S E info M × 0 + m × v2 = (M + m) V 0 + 20 × 10 –3 × 200 = (4 + 20 × 10–3 ) × V V = 0.995 ms–1 KUET 01. hw` 5kg f‡ii GKwU e›`yK †_‡K 20gm f‡ii GKwU ¸wj 100m/s MwZ‡Z †Qvov nq Z‡e e›`y‡Ki cðvr †eM KZ? [KUET. 06-07] A. 4ms–1 B. 4000ms–1 C. 40cms–1 D. 4cms–1 E. 400ms–1 MV = mv ev, v = mv M = 100 0.02 5 = 0.4ms–1 = 40cms–1 STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. 10 kg e¯‧ (object) hw` 12 ms1 G P‡j, Z‡e Gi MwZ‡eM n‡e- [MAT. 2020-21] A. 120 kgms1 B. 10 kgms1 C. 12 kgms1 D. 1.2 kgms1 S A info †Kv‡bv e¯‘i fi I †e‡Mi ¸Ydj Øviv fi‡eM cwigvc Kiv hvq| fi‡eM (Momentum) = fi †eM = 10 12 kgms1 = 120 kgms1 02. 10kg f‡ii GKwU e¯‧ 12ms–1 †e‡M Pj‡j fi‡e‡Mi cwigvY KZ? [MAT. 15-16] A. 12 kgms–1 B. 10 kgms–1 C. 1.2 kgms –1 D. 120 kgms–1 p = mv = 10 12 = 120 kgms–1 STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. Aw¯
wZ¯
vcK msN‡l© msiwÿZ nqÑ [Xv. †ev. 2018] A. MwZ kw³ B. w¯’wZ kw³ C. †K․wYK fi‡eM D. fi‡eM Ans D 02. mgAvqZ‡bi GKwU †jŠn †MvjK I GKwU †Uwbm e‡ji fi‡eM mgvb n‡j-[Xv. †ev. 2016] A. †j․n †Mvj‡Ki MwZkw³ †ewk B. †Uwbm e‡ji MwZkw³ †ewk Ans B C. Df‡qi MwZkw³ mgvb D. MwZkw³i Dci fi‡e‡Mi cÖfve †bB 03. †K․wYK fi‡eM n‡jv [w`. †ev. 2017] A. e¨vmva© †f±i I •iwLK fi‡e‡Mi †f±i ¸Y‡bi mgvb B. RoZvi åvgK I •iwLK †e‡Mi ¸Y‡bi mgvb C. •iwLK fi‡eM I †K․wYK fi‡e‡Mi †f±i MyY‡bi mgvb D. •iwLK †eM I e¨vmva© †f±‡ii ¸Y‡bi mgvb Ans A 04. †K․wYK fi‡e‡Mi cwieZ©‡bi nvi- [w`. †ev. 2017] A. e‡ji mgvb B. †K․wYK Z¡i‡Yi mgvb C. U‡K©i mgvb D. RoZvi åvg‡Ki mgvb Ans C 05. ‡Kv‡bv e¯‧i fi‡eM 40kgms1 ej‡Z †evSvq? [h. †ev. 2016] i. e¯‘i fi 1 kg n‡j Gi †eM 40ms1 ii. e¯‘i fi 40 kg n‡j Gi †eM 10ms1 iii. e¯‘i fi 6.3kg n‡j Gi †eM 6.36ms1 wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i,ii I iii Ans B wb‡Pi DÏxcKwU jÿ¨ Ki Ges 06 I 07 bs cÖ‡kœi DËi `vI: [wm. †ev. 2017] A I B e¯‘Øq ci¯ú‡ii wecixZ w`‡K GKB †iLv eivei P‡j msNl© NUvq| msN‡l©i ci Zviv wbR wbR MwZc‡_i wecixZ w`‡K Pj‡Q| 06. msN‡l©i ci B e¯‧i †eM KZ? A. 2.50 ms-1 B. 4.17 ms-1 C. 5.83 ms-1 D. 12.50 ms-1 S A info VB = 2m1 m1 + m2 u1 + m2 – m1 m2 + m1 u2 = 2 × 5 5 + 6 × 4 + 6 – 5 6 + 5 × 5 = 4.17 ms–1 07. Dc‡iv³ msN‡l©i †ÿ‡Î i. fi‡eM msiwÿZ n‡e ii. MwZkw³ msiwÿZ n‡e iii. msNl©wU Aw¯’wZ¯’vcK n‡e wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i,iiIiii Ans B 08. 3kg I 5kg f‡ii e¯‧Øq 30kgm1 I 50kgm-1 fi‡eM wb‡q GKB w`‡K Pj‡Q| [e. †ev. 2016] A. msN‡l©i ci Zviv GKB w`‡K Pj‡e B. msN‡l©i ci Zviv wecixZ w`‡K Pj‡e C. msN‡l©i ci Zviv w¯’i n‡q hv‡e D. Zv‡`i g‡a¨ †Kv‡bv msNl© NU‡e bv Ans D 09. mgvb f‡ii `ywU e¯‧i g‡a¨ w¯
wZ¯
vcK msNl© n‡j wb‡Pi †KvbwU mZ¨? GLv‡b 1g e¯‧i Avw` I †kl †eM u1I 1 Ges 2q e¯‧i Avw` I †kl †eM u2I v2| [e. †ev. 2015] A. u1 = v2 B. u1 = v 1 C. u1 = u2 D. u2= v 2 Ans A 10. Aw¯
wZ¯
vcK msN‡l© msiwÿZ nqÑ [mKj †ev. 2018] A. MwZkw³ B. fi‡eM C. †K․wYK fi‡eM D. w¯’wZkw³ Ans B Concept 5 Nl©Y msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. w¯’wZ Nl©Y ¸YvsK, s= Fs R Ges R=mg,s = f(s)(max) FN 02. k= f(k) FN , k= Fk R ,k= Fk mg 03. MZxq Nl©Y ¸bvsK, k= R Fk = mg Fk = a g 04. wbðj †KvY/Nl©Y †KvY, f = tan1 (s) CONCEPTUAL MATH MEx 01 GKwU †Uwe‡ji Dci 1kg f‡ii GKwU eB Av‡Q| †Uwe‡ji Zj eivei 3N ej cÖ‡qvM Ki‡j eBwU Pjvi Dcµg nq| †Uwej I eB‡qi g‡a¨ w¯
wZ Nl©Y ¸Yv¼ wbY©q Ki| k = Fk R k = 3 9.8 k MEx 02 GKwU †ev‡W©i Dci GKwU w¯
i e¯‧ Av‡Q| †evW© I e¯‧i ga¨Kvi Nl©Y ¸YvsK 0.3 n‡j wbðj †KvY/Nl©Y †KvY KZ? tan = s = tan–1 (s) = tan–1 (0.3) = 1642

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 163 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW START PRACTICE 01. hw` w¯
wZ NlY© ¸Yv¼ 0.4 nq Z‡e 2kg f‡ii †Kvb w¯
i e¯‧‡K KZ ej cÖ‡qvM Ki‡j Pjvi Dcµg nq? 02. `ywU Z‡ji ga¨Kvi w¯
i Nl©Y ¸bvsK 1 3 n‡j Nl©Y †KvY KZ? 03. 3kg f‡ii GKwU eøK‡K GKwU Avbyf~wgK Z‡ji Dci w`‡q KZ e‡j Uvb‡j e¯‧wU mg‡e‡M Pj‡e? (MZxq Nl©Y ¸YvsK = 0.1) 04. GKwU †ev‡W©i Dci GKwU e¯‧ w¯
i Ae¯
vq Av‡Q| †evW© I e¯‧i ga¨Kvi w¯
Z Nl©Y ¸YvsK 3 n‡j wbðj †KvY KZ? 05. ‡Kvb †g‡S‡Z ¯
vwcZ 400N Gi GKwU Kv‡Vi eø‡Ki Dci 160N ej cÖ‡qvM Ki‡j GwU Pjvi Dcµg nq ‡g‡S I Kv‡Vi eø‡Ki ga¨eZx© Nl©Y ¸YvsK KZ? NOW PRACTICE SOLVE : 01. Fk = mgk = 2 9.8 0.4 = 7.84N 02. Nl©Y †KvY, s = tan–1 (s) = tan–1 ( 1 3 ) = 30 03. F = s mg = 0.1 × 9.8 × 3 = 2.94 N 04. f = tan–1 s=tan–1 ( 3) = 60 05. s = Fs R = 160 400 = 0.4 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. 50N Gi GKwU Avbyf~wgK ej GKwU 0.50 kg f‡ii AvqZvKvi e¯‧‡K GKwU Djø¤^ †`Iqv‡j av°v w`‡”Q| e¯‧wU Avw`‡Z w¯
i wQj| hw` •¯
wZK I MZxq Nl©Y ¸Yv¼ h_vµ‡g 2 = 0.6 Ges k = 0.8 nq, Z‡e m/s2 GK‡K e¯‧wUi Z¡iY KZ? [DU. 18-19] A. 1.8 B. 2.0 C. 6.0 D. 8.0 F =ma a = F m = W – Fk m = mg – (kR) m = 0.50 9.8 – (0.8 5) 0.5g = 1.8ms –2 02. 100 µl Gi GKwU e„wói †dvUv c„w_ex c„‡ôi w`‡K 9.8ms–2 Z¡i‡Y co‡Z _vK‡j ga¨vKl©‡Yi Kvi‡b Zvi MwZ‡eM evo‡Z _v‡K, Aci w`‡K evZv‡mi Nl©Y ej MwZ Kwg‡q †`qvi †Póv K‡i| GK ch©v‡q †dvUvwU 50ms–1 P~ovšÍ w¯’i MwZ‡eM cÖvß nj| H mg‡q Zvi Dci Nl©Y e‡ji cwigvb KZ? [DU. 07-08] A. 980 N B. 980 10–3 N C. 980 10–6 N D. 980 10–9 N †h‡nZz †dvUvwU 50 ms–1 P~ovšÍ w¯’i MwZ‡eM cÖvß n‡q‡Q| ZvB F = mg = 100 10–6 9.8 = 98010–6 N 03. f~wgi mv‡_ me©wbgœ 300 †Kv‡Y AvbZ Z‡j ¯
vwcZ e¯‧ wcQ‡j †b‡g hvq| Nl©Yv¼ KZ? [DU. 03-04] A. sin 300 B. tan 300 C. cos 300 D. 1 s = tans ev, s = tan300 STEP 02 ANALYSIS OF JU QUESTION 01. MZxq Nl©Y †KvY n‡jv- [JU. 09-10] A. tan–1Fk R B. tan–1R Fk C. tan–1Fs R D. tan–1Fk R 2 Ans A 02. GKwU Kv‡Vi Z³vi Dci Aew¯
Z GKwU B‡Ui wbðj †KvY 40| BU I Z³vi ga¨Kvi w¯
wZ Nl©Y MybvsK KZ? [JU. 17-18] A. 0.87 B. 0.85 C. 0.84 D. 0.97 Nl©Y ¸YvsK, s = tan s = tan 40˚ = 0.84 03. †Kvb †g‡S‡Z ¯
vwcZ 400 N Gi GKwU Kv‡Vi e‡Ûi Dci Abyf‚wgKfv‡e 160 N ej cÖ‡qvM Ki‡j GwU Pjvi Dcµg nq| †g‡S I Kv‡Vi e‡Ûi ga¨eZx© NlYvsK n‡e- [JU. 13-14] A. 4.00 B. 0.40 C. 1.60 D. 0.16 Kv‡Vi e‡Ûi IR‡bi gvb, mg = 400 jwä ej = Nl©Y ej + Z¡iY ej F = fk + ma = k mg + ma 160 = k 400 + 0 k = 0.40 04. GKwU †Uwe‡ji Dci 1kg f‡ii GKwU eB Av‡Q| †Uwe‡ji Zj eivei 3N ej cÖ‡qvM Ki‡j eBwU Pjvi Dcµg nq| †Uwej I eB Gi gv‡S w¯
wZ Nl©Y ¸YvsK KZ? [JU. 11-12] A. 0.3 B. 1.4 C. 2.5 D. 3.1 s = Fk R = Fk mg = 3 1 9.8 = 0.3 STEP 03 ANALYSIS OF RU QUESTION 01. `ywU Z‡ji ga¨Kvi w¯
i Nl©Y ¸bvsK 1 3 n‡j Nl©Y †KvY KZ? [RU. 17-18] A. 25˚ B. 30˚ C. 45˚ D. 60˚ Nl©Y †KvY, s = tan–1 (s ) = tan–1 ( 1 3 ) = 30 STEP 04 ANALYSIS OF CU QUESTION 01. 10kg f‡ii Dci cÖhy³ ej 20N I Nl©Y ej 5N. Z¡iY n‡e- [CU. 03-04] A. 2ms–2 B. 1.5ms–2 C. 1.75ms–2 D. 1.08ms–2 R – F = ma ev, 20 – 5 =10a ev, a = 1.5ms–2 STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. 3kg f‡ii GKwU eøK‡K GKwU Avbyf~wgK Z‡ji Dci w`‡q KZ e‡j Uvb‡j e¯‧wU mg‡e‡M Pj‡e? (MZxq Nl©Y ¸YvsK = 0.1) [JnU. 12-13, 09-10, 06-07, RU. 15-16] A. 2.94N B. 3N C. 1.94N D. 0.94N F = smg = 0.1 9.8 3 = 2.94N IU 01. Nl©Y ej 5N n‡j 5kg f‡ii GKwU e¯‧‡K 5ms–2 Z¡i‡Y Ki‡Z ej cÖ‡qvM- [IU. 15-16] A. 15N B. 20N C. 25N D. 30N F = 5 + ma = 5 + 5 5 = 30 N PART B Analysis of Science & Technology Question SUST 01. 500 kg f‡ii GKwU Mvox 60 km/hr †e‡M P‡j| †eªK †P‡c MvoxwU‡K 50 m `~‡i _vgv‡bv n‡jv| hw` iv¯Ívi Nl©YRwbZ ej 100 N nq, Z‡e †eªKRwbZ e‡ji gvb KZ N? [SUST. 14-15] A. 510 B. 720 C. 828 D. 1190 E. 1290 F = mv2 2s = 500 60 3.6 2 2 50 = 1388.889 †eªK RwbZ ej = 1388.889 – 100 = 1288.889 = 1290

164 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. 80kg f‡ii GKwU ev·‡K 600N Avbyf~wgK e‡j †g‡Si Dci w`‡q Uvbv n‡”Q| PjšÍ Ae¯
vq ev· I †g‡Si ga¨eZ©x Nl©Y mnM 0.50. ev‡·i Z¡iY KZ? [SUST. 12-13] A.1.2ms–2 B. 1.5ms–2 C. 2.60ms–2 D. 4.5ms–2 E. 5.0ms–2 a = F – smg m = 600 – (0.5 9.8 80) 80 = 2.6ms–2 F mg mg 90- F 03. wP‡Î m f‡ii GKwU eøK AvbZ Zj †e‡q Dc‡ii w`‡K DV‡Q| eøKwU mg‡e‡M Pj‡e F = ? [SUST. 12-13] A. mg(1– sin) B. mg (1– cos) C. mgsin D. mgcos E. 0 F mg = cos (90 – ) ev, F = mgsin JUST 01. wcw”Qj ei‡di Dci 1kg IR‡bi GKwU cv_i 2ms–1 †e‡M Pjvi 10 sec ci Nl©‡Yi d‡j †_‡g †Mj| GLv‡b Nl©Y ej KZ wbDUb? [JUST-A, Set-Ka 19-20] A. 2.0 B. 0.02 C. 0.2 D. 2.2 a = v–u t = 0–2 10 = – 0.2 ms–2 evav`vbKvix ej/Nl©b ej, F = ma = 1×0.2 = 0.2N 02. m wK‡jvMÖvg f‡ii GKwU Mvox v MwZ‡Z r e¨vmv‡a©i mgZj e„ËvKvi c‡_ Pj‡Q? hw` Mvwoi PvKvi mv‡_ iv¯Ívi Nl©YmnM s nq Zvn‡j MvwowU wbqš¿Y bv nvwi‡q m‡e©v”P KZ `ªæwZ‡Z Pj‡Z cvi‡e? [JUST. 15-16] A. srs/ 2 B. s 2 rs/ 2 C. (srg) D. srg E. s 2 r/s Ans C 03. `yBwU Z‡ji ga¨Kvi w¯
i Nl©Y †Kvb 60| Zv‡`i Nl©Y ¸bvsK KZ? [JUST. 15-16] A. 3 B. 1.70 C. 1.76 D. 2 E. 2 s = tan = tan60 = 3 HSTU 01. †Kvb †g‡S‡Z ¯
vwcZ 400 kg Gi GKwU Kv‡Vi eø‡Ki Dci Avbyf~wgKfv‡e 100 N ej cÖ‡qvM Ki‡j GwU Pjvi Dcµg nq| †g‡S I Kv‡Vi eø‡Ki ga¨eZ©x Nl©YvsK KZ? [HSTU. 15-16] A. 4 B. 2 C. 1 D. DËi †bB = F mg = 160 400 9.8 = 0.04 PSTU 01. 10gm f‡ii GKwU †MvjK 1.2m DuPz Xvjy Z³v †e‡q †ei nIqvi mgq †eM wQj 4 m/s| Nl©YRwbZ Kv‡Ri cwigvY KZ? [PUST-A, 19-20] A. 0.0376 J B. 0.197 J C. 0.1176 J D. 0.08 J v 2 = v0 2 + 2as = 0 + 2as s = 16 2a = 8 a GLb, wPÎ n‡Z, mgcos – f = ma mgscos – fs = mas [Dfq c‡ÿ s Øviv ¸Y K‡i] fs = mgscos – mas Wf = mgs 1.2 x – ma 8 a = {(0.01 9.8 1.2) – (0.01 8)} J = 0.0376 J myZivs Nl©Y ej Øviv K…ZKvR = 0.0376 J STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. 1000kg f‡ii GKwU D‡ovRvnvR w¯
i †e‡M †mvRv c‡_ DÇqb Ki‡Q| evZv‡mi Nl©Y ej 1800N D‡ovRvnv‡Ri Dci cÖhy³ bxU ej n‡e- [BUET. 12-13] A. 0N B. 11800N C. 1800N D. 9800N F- NlY© ej = ma ev, F = NlY© ej = 1800N 02. Avbyf~wgK †g‡S‡Z w¯
ive¯
vq 800N IR‡bi GKwU Szwo‡K miv‡Z Kgc‡¶ 200N Avbyf~wgK av°vi cÖ‡qvRb| w¯
ive¯
vq Nl©Y mn‡Mi gvb- [BUET. 12-13] A. 0.25 B. 0.125 C. 0.50 D. 4.00 F = smg ev, s = 200 800 = 0.25 KUET 01. 73kg f‡ii GKwU ev· †K 543N Avbyf~wgK e‡j †g‡Si Dci w`‡q Uvbv n‡”Q| ev·wU hLb P‡j ZLb ev· I †g‡Si ga¨eZ©x Nl©Y mnM 0.53 ev‡·i Z¡iY KZ? [KUET. 09-10] A. 2.24ms–2 B. 0.224ms–2 C. 4.84ms–2 D. 0.448ms–2 E. 3.38ms–2 a = FR – smg m = 543 – 0.53 73 9.8 73 = 2.24ms–2 02. 500kg f‡ii GKwU Mvwo 50kg f‡ii 5 Rb hvÎx wb‡q 30m/s †e‡M PjšÍ Ae¯
vq nVvr eÜ n‡q 100m `~‡i wM‡q †_‡g hvq| Nl©Y RwbZ e‡ji gvb- [KUET. 09-10] A. 500N B. 750N C. 1000N D. 1500N E. 3375N F = mv2 2s = (500 + 50 5) (30) 2 2 100 = 3375N STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. Nl©Y ej I e¯‧i †e‡Mi ga¨Kvi †KvY KZ? [Xv. †ev. 2017] A. B. x 2 C. x 4 D. 0 Ans A 02. `ywU mgvb f‡ii e¯‧i g‡a¨ w¯
wZ¯
vcK msNl© NU‡jÑ [iv.‡ev. 2019] i. msN‡l©i c~‡e©i I c‡ii †gvU fi‡eM GKB _vK‡e ii. msN‡l©i c~‡e©i I c‡ii †gvU MwZkw³ GKB _vK‡e iii. msN‡l©i ci e¯‘Øq †eM wewbgq Ki‡e wb‡Pi †KvbwU mwVKÑ A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans D 03. Avw`‡Z X †MvjKwU v †e‡M mivmwi Y †Mvj‡Ki w`‡K MwZkxj| Y †MvjKwU w¯’i Ae¯’vq i‡q‡Q| †MvjK `ywUi w¯’wZ¯’vcK msNl© nq| Kx N‡U? [P. †ev. 2016] X A. 1 2 v †e‡M Wvbw`‡K MwZkxj B. v †e‡M evg w`‡K MwZkxj C. 1 2 v †e‡M evgw`‡K MwZkxj D. †_‡g hvq Y 1 2 v †e‡M Wvbw`‡K MwZkxj w¯’i _v‡K 1 2 v †e‡M Wvbw`‡K MwZkxj v †e‡M Wvbw`‡K MwZkxj Ans D wb‡Pi DÏxcKwU co Ges 04 I 05 bs cÖ‡kœi DËi `vI : 1500kg f‡ii GKwU Mvwo 400N Nl©Y ejhy³ †mvRv iv¯Ívq 5ms-2 mgZ¡i‡Y P‡j| 04. Mvwoi BwÄb KZ…©K cÖhy³ ejÑ [w`.‡ev. 2019] A. 0.4KN B. 7.1KN C. 7.5KN D. 7.9KN S D info F – = maF = ma + = 1500 × 5 + 40˚ = 7900 N = 7.9 KN 05. •iwLK †eM I †K․wYK †e‡Mi g‡a¨ m¤úK© n‡jvÑ [w`.‡ev. 2019] A. = v r B. v = . r C. = v . r D. v = r Ans D 06. mvB‡K‡ji †eM I PvKvi Nl©‡bi ga¨eZx© †KvY KZ? [Kz.‡ev. 2019] A. 0 B. 90 C. 180 D. 360 Ans C mgcos w = mg 1.2m

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 165 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 6 NvZ ej I e‡ji NvZ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. NvZ ej = ma t v u m t mv mu F = fi‡e‡Mi cwieZ©‡bi nvi e‡ji NvZ, J F.t mv ~ mu = P = fi‡e‡Mi cwieZ©b| †eMØq wecixZ gyLx n‡j e‡ji NvZ J = m (v + u) ‡eMØq GKB w`‡K n‡j e‡ji NvZ J = m(v u) CONCEPTUAL MATH MEx 01 30N ej 5kg f‡ii GKwU e¯‧i Dci 10sec wµqv K‡i| e¯‧wUi fi‡e‡Mi cwieZ©b KZ n‡e? e‡ji NvZ = fi‡e‡Mi cwieZ©b = Ft = 30 10 = 300kgms–1 MEx 02 S Avbyf~wgK w`‡K MwZkxj 2kg f‡ii GKwU †j․n †MvjK 5ms-1 ‡e‡M GKwU †`qv‡j j¤^fv‡e av°v †L‡q 3ms-1 ‡e‡M wecixZ w`‡K wd‡i †Mj| e‡ji NvZe‡ji NvZ, J = m(v + u) (†eMØq wecixZ w`‡K) = 2(5 + 3) =2 8 =16kgms–1 NOW START PRACTICE 01. 40N ej 10kg f‡ii GKwU e¯‧i Dci 10sec wµqv K‡i| e¯‧wUi fi‡e‡Mi cwieZ©b KZ n‡e? 02. 10gm f‡ii GKwU wµ‡KU ej 10ms-1 †e‡M G‡m e¨v‡U jv‡M| ejwU GKB MwZ‡Z e¨v‡Ui AvNv‡Z †diZ cvVv‡bv nq| ejwUi mv‡_ e¨v‡Ui msNl©Kvj 10–3 s n‡j e‡ji Dci e¨v‡Ui cÖ‡qvMK…Z ej KZ? NOW PRACTICE SOLVE : 01. e‡ji NvZ = fi‡e‡Mi cwieZ©b Ft = 40 10 = 400kgms–1 02. Ft = m (v+u) F = 10 103 (10 + 10) 103 = 200N REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. 30 ms-1 †e‡M AvMZ 250 g f‡ii GKwU wµ‡KU ej‡K GKRb †L‡jvqvo K¨vP a‡i 0.1 †m‡KÛ mg‡qi g‡a¨ _vwg‡q w`j| †L‡jvqvo KZ…©K ejwUi Dci cÖhy³ Mo ej KZ? [DU. 10-11] A. 7.5N B. 75N C. 2.5N D. 25N Ft = m(v u) ev, F = 0.250(30 0) 0.1 = 75N STEP 02 ANALYSIS OF JU QUESTION 01. 60 Kg f‡ii GKwU e¯‧ 0.2ms–1 Abyf‚wgK †e‡M GKwU Luvov †`qv‡j av°v w`‡q 0.1ms–1 †e‡M wecixZ w`‡K wd‡i †Mj| e‡ji NvZ- kgms–1 ? [JU-A, Set-A. 19-20] A. 0.010 B. –0.015 C. 0.2015 D. –0.115 S Blank info e‡ji NvZ, J = m(v u) = 60 (0.2 + 0.1) = 18 kgms1 02. 0.5 kg f‡ii GKwU wµ‡KU ej 30ms–1 †e‡M wM‡q j¤^fv‡e GKwU e¨vU‡K AvNvZ Kij Ges wecixZ w`‡K 20 ms–1 †e‡M cÖwZwÿß n‡jv| e‡ji Øviv e¨v‡Ui Dci cÖhy³ e‡ji NvZ- [JU-A, Set-D. 19-20] A. 0.5 Ks B. 1.0 Ns C. 25 Ns D. 50 Ns J = m(v u) = 0.5 (30 + 20) = 25 Ns 03. 16N Gi GKwU ej 4 kg f‡ii Dci 2s wµqv K‡i| e¯‧wUi †e‡Mi cwieZ©b n‡e- [JU. 10-11] A. 16 ms–1 B. 8 ms–1 C. 10 ms–1 D. 20 ms–1 fi‡e‡Mi cwieZ©b, mv = Ft [ e‡ji NvZ = fi‡e‡Mi cwieZ©b] †e‡Mi cwieZ©b, v = Ft m = 16 2 4 = 8 ms–1 04. 8N GKwU ej 4kg f‡ii Dci 4 †m. wµqv K‡i| e¯‧wUi †e‡Mi cwieZ©b- [JU. 07-08] A. 8ms–1 B.16ms–1 C. 10ms–1 D. 20ms–1 Ft = m(v) ev, v = Ft m = 8 4 4 = 8ms–1 STEP 03 ANALYSIS OF RU QUESTION 01. 50 kg f‡ii GKwU w¯
i e¯‧i Ici 5 sec e¨vcx 10 N ej cÖhy³ n‡jv| D³ e¯‧wUi †e‡Mi cwieZ©b n‡eÑ [RU. Sinovac, Set-1. 20-21] A. 1 ms1 B. 10 ms1 C. 50 ms1 D. 0.5 ms1 S A info fi‡e‡Mi cwieZ©b, mv = Ft v = 10 5 50 = 1ms–1 02. Abyf‚wgK w`‡K MwZkxj 5 kg f‡ii GKwU †j․n †MvjK 5ms–1 †e‡M GKwU †`qv‡j j¤^fv‡e av°v †L‡q 3 ms–1 †e‡M wecixZ w`‡K wd‡i †M‡j| e‡ji NvZ KZ?[RU. 13-14] A. 30 kg ms–1 B. 40 kg ms–1 C. 10 kg ms–1 D. 16 kg ms–1 e‡ji NvZ, J = m (v–u) = 5{5–(–3)} = 40 kgms–1 STEP 04 ANALYSIS OF GST QUESTION 01. 5 kg f‡ii GKwU e¯‧ 1.2 ms–1 †e‡M GKwU †`qv‡j j¤^fv‡e av°v †L‡q 0.8 ms–1 †e‡M wecixZ w`‡K wd‡i Avm‡j e‡ji NvZ KZ n‡e? [GST-A. 20-21] A. 4 B. 5 C. 6 D. 10 S D info e‡ji NvZ = fi‡e‡Mi cwieZ©b Ft = mv – mu = 5×1.2 – 5× ( – 0.8) = 10 Ns PART A Analysis of General University Question KU 01. wµ‡KU †Ljvq †evjv‡ii nvZ †_‡K wbw¶ß GKwU ej 0.4 †m‡K‡Û 20 dzU †m.-1 †e‡M cÖvß nq| ejwUi IRb 16 cvDÛ n‡j Gi Dci cÖhy³ e‡ji gvb KZ? [KU. 03-04] A. 20 B. 30 C. 45 D. 25 Ft = m(vu) ev,F = m(v u) t = w g v u t ev F = 16 32 (20 0) 0.4 = 25 cvDÛvj|

166 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES CoU 01. 16N Gi GKwU ej 4kg f‡ii Dci 4sec wµqv K‡i| e¯‧wUi †e‡Mi cwieZ©b wbY©q Ki| [CoU. 15-16] A. 16ms–1 B. 64ms–1 C. 32ms–1 D. 24ms–1 Ft = m(vu) ev, (v u) = Ft m = 16×4 4 = 16 ms–1 PART B Analysis of Science & Technology Question SUST 01. 1.82 10–19N ej GKwU B‡jKUª‡bi Dci 1.6 10 –9 s a‡i wµqv Ki‡j †e‡Mi cwieZ©b KZ ms–1 n‡e? [SUST. 15-16] A. 32 B. 160 C. 320 D. 480 E. 3200 m(v u) = Ft ev, (v u) = Ft m = 1.82 10–19 1.6 10–9 9.1 10–31 = 320 02. GKwU e¯Íi fi‡eM p Ges fi m n‡j p 2 /m Gi GKK wb‡Pi Kvi GKK Gi mgvb n‡e? [SUST. 05-06] A. Z¡iY B. kw³ C. ej D. ÿgZv fi‡eM Gi GKK, P = kgms–1 | f‡ii GKK m = kg p 2 m = kg2m 2 s –2 kg = kgm2 s –2 hv kw³i GKK| Ek = p 2 2m MBSTU 01. 60N ej 30kg f‡ii GKwU e¯‧i Dci 1min wµqv K‡i| e¯‧wUi †e‡Mi cwieZ©b KZ? [MBSTU. 15-16] A. 120m/s B. 125m/s C. 128 m/s D. 130m/s Ft = mv ev v = Ft m = 6060 30 = 120m/s STEP 05 ANALYSIS OF ENGINEERING & BUTex QUESTION RUET 01. GKwU e¯‧i Dci 5N ej 10s wµqv K‡i| fi‡e‡Mi cwieZ©b KZ? [RUET. 14-15] A. 40kgms–1 B. 50kgs–1 C. 45kgms–1 D. 49kgms–1 fi‡e‡Mi cwieZ©b = Ft = 5 10 = 50kgms–1 STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. e‡ji NvZ n‡”QÑ [iv. †ev. 2017; wm. †ev. 2015] i. ej I e‡ji wµqvKv‡ji ¸Ydj ii. fi‡e‡Mi cwieZ©b iii. fi‡e‡Mi cwieZ©‡bi nvi wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans A 02. 4N ej GKwU e¯‧i Ici 1 sec e¨vcx wµqv Ki‡j fi‡e‡Mi cwieZ©b KZ? [mKj †evW© 2018] A. 2 kgms–1 B. 4 kgms–1 C. 8 kgms–1 D. 16 kgms–1 S B info J = F × t = 4 × 1 = 4 kms–1 Concept 7 RoZv åvgK msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. GKwU `‡Ûi •`‡N©¨i Awfj¤^ I ga¨we›`y, I = M 12 L 2 06. PµMwZi e¨vmva©, K = I M , c„w_exi PµMwZi e¨vmva© = 2 5 R 02. `‡Ûi •`‡N©¨i Awfj¤^ I cÖvšÍ we›`y I = 1 3 ML 2 07. dvcv †Mvj‡Ki, I = 1 5 mr2 03. wb‡iU wmwjÛvi, N~Y©b A¶ wbR A¶, I = 1 2 Mr2 08. e„ËvKvi PvKwZi e¨v‡mi mv‡c‡ÿ RoZvi åvgK, I = 1 4 mr2 04. wb‡iU ev c„w_exi RoZvi åvgK, 2 MR 5 2 I 09. wis Gi A‡ÿ mv‡c‡ÿ RoZvi åvgK I = Mr2 05. e„ËvKvi PvKwZi e¨v‡mi mv‡c‡ÿ RoZvi ågK, I = 1 4 Mr2 10. I = 12 M (l 2 + b2 ) AvqZvKvi e¯‘ b l CONCEPTUAL MATH MEx 01 †Kvb A¶ mv‡c‡¶ GKwU e¯‧i RoZvi åvgK 200 kg.m2 | D³ A¶ mv‡c‡¶ e¯‧wUi PµMwZi e¨vmva© KZ? (e¯‧wUi IRb 19.6 N). e¯‘i fi, m = 19.6 9.8 = 2kg ; RoZvi åvgK,I = mK2 K 2 = I m K = I m = 200 2 = 10 m NOW START PRACTICE 01. fvi‡K›`ªMvgx Ges Z‡ji mwnZ j¤^ eivei A¶ mv‡c‡¶ GKwU AvqZvKvi cv‡Zi RoZvi åvgK 8kgm2 | cvZwUi cÖ¯
 1m Ges fi 24 kg n‡j •`N©¨ KZ? 02. GKwU PvKvi fi 10kg Ges RoZvi åvgK 2.5kgm2 n‡j PµMwZi e¨vmva© n‡e03. GKwU KwVb ‡ej‡bi fi 5.0 kg Ges e¨vmva© 3.0 cm| †ej‡bi A‡¶i mv‡c‡¶ Gi RoZvi åvgK KZ?

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 167 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW PRACTICE SOLVE : 01. RoZvi åvgK, I = 1 12 m(l 2 + b2 ) 8 = 1 12 24 (l 2 + 12 ) l = 3m 03. I = 1/2 mr2 ev I = 1/2×5 (0.03)2 = 2.25 10–3 kg m2 02. I = mK2 ev, K = m m I r 0.5 10 2.5 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. fi †K›`ªMvgx Ges Z‡ji mwnZ j¤^ eivei A¶ mv‡c‡¶ GKwU AvqZvKvi cv‡Zi RoZvi åvgK 5 kg/m2 | cvZwUi cÖ¯
 1m Ges fi 12 kg n‡j •`N©¨ KZ? [DU. 05-06] A. 5 m B. 2.5 m C. 2 m D. 60 m I = M 12(l2 + b2 ) ev, 5 = 12 12(l 2 + 12 ) 5 = l2 + 1 ev, l 2 = 5 – 1ev, l = 2m 02. GKwU KwVb †ej‡bi fi 5.0 kg Ges e¨vmva© 3.0 cm| †ej‡bi A‡¶i mv‡c‡¶ Gi RoZvi åvgK KZ? [DU. 05-06] A. 4.50 x 10–3 kg m2 B. 2.25 x 10–3 kg m2 C. 1.50 x 10–3 kg m2 D. 0.38 x 10–3 kg m2 I = 1 2 mr2 ev, I = 1 2 × 5 (.03)2 = 2.25 10–3 kg m2 STEP 02 ANALYSIS OF JU QUESTION 01. GKwU PvKvi fi 5kg Ges PµMwZi e¨vmva© 25cm| Gi RoZvi åvgK KZ? [JU. 18-19] A. 0.4125kg–m 2 B. 0.7125kg–m 2 C.0.3125kg–m 2 D. 0.9125kg–m 2 I = MK2 =5 (0.25)2 = 0.3125kgm2 02. GKwU duvcv wmwjÛv‡ii fi M I e¨vmva© R| R¨vwgwZK A¶ mv‡c‡¶ Gi RoZvi åvgK KZ? [JU. 17-18] A. 1 2 MR2 B. MR2 C. 3 2 MR2 D. MR2 S B info duvcv wmwjÛv‡ii R¨vwgwZK A¶ mv‡c¶ Gi RoZvi åvgK = MR2 wb‡iU wmwjÛv‡ii GK cÖv¯Í w`‡q I Zvi •`‡N©¨i j¤^fv‡e AwZµvšÍ A‡¶i mv‡c‡¶ RoZvi åvgK = 1 3 MR2 wb‡iU wmwjÛv‡ii ga¨we›`y w`‡q I Zvi •`‡N©¨i Awfj¤^fv‡e AwZµvšÍ A‡¶i mv‡c‡¶ RoZvi åvgK = 1 2 MR2 03. GKwU N~Y©vqgvb wcZ‡ji †Mvj‡Ki fi 0.02kg. N~Y©b A¶n‡Z Gi `~iZ¡ 1m n‡j, A¶ mv‡c‡¶ RoZvi åvgK n‡e- [JU. 12-13] A. 2.0kgm2 B. 20kgm2 C. 0.2kgm2 D. 0.02kgm2 I = mr2 = 0.02 1 2 = 0.02kgm2 04. GKwU e„ËvKvi PvKwZi c„‡ôi Awfj¤^ eivei PvKwZi †K›`ª w`‡q MgbKvix A‡¶i mv‡c‡¶ RoZvi åvgK n‡jv- ( ) MR2 2 | e„ËvKvi PvKwZi c„‡ôi Awfj¤^ fv‡e MgbKvix ¯úk©‡Ki mv‡c‡¶ RoZvi åvgK KZ? [JU. 12-13] A.1.5MR2 B. 0.75MR2 C. 0.5R2 D. None mgvšÍivj A¶ Dccv`¨ Abymv‡i, I = IG + MR2 = MR2 2 + MR2 =1.5MR2 05. GKwU PvKvi fi 10kg Ges RoZvi åvgK 2.5kgm–2 n‡j PµMwZi e¨vmva© n‡e- [JU. 11-12] A. 1m B. 0.5m C. 1.5m D. 5m I = mK2 ev, r = I m = 2.5 10 = 0.5m STEP 03 ANALYSIS OF RU QUESTION 01. GKwU e¯‧i RoZvi åvgK 9000 g cm2 | e¯‧wUi fi 10 g n‡j, PµMwZi e¨vmva© KZ? [RU. 17-18] A. 900 cm B. 30 cm C. 10 cm D. 90000 cm RoZvi åvgK,I = mK2 K 2 = I m K = I m = 9000 10–4 g 10g = 30 10–2 m = 30cm 02. A‡¶i Dci Aew¯
Z PviwU we›`y-f‡ii fi 1 kg, 2 kg, 3 kg I 4 kg| H A¶ mv‡c‡¶ f‡ii web¨vmwUi RoZvi åvgK- [RU. 16-17] A. 10 kg m2 B. 30 kg m2 C. 0 kg m2 D. †KvbwUB bq A‡¶i Dci, r = 0 RoZvi åvgK, I = 0 03. GKwU PvKvi fi 5kg Ges PµMwZi e¨vmva© 0.5m n‡j Zvi RoZvi åvgK KZ? [RU. 16-17] A. 0.2kg m2 B. 0.4 kg m2 C. 0.6kg m2 D. 0.8 kg m2 RoZvi åvgK, I = 1 2 mK2 = 1 2 × 5 (0.5)2 = 0.6 kgm2 04. 5 kg f‡ii GKwU PvKvi PµMwZi e¨vmva© 20 cm| Gi Dci 0.4N.m UK© cÖhy³ n‡j, †K․wYK Z¡iY KZ n‡e? [RU. 13-14] A. 2 rad s–2 B. 0.5 rad s–2 C. 2degs–2 D. 0.5degs–2 = I ev, = mr 2 = 0.4 5 (0.20) 2 = 2rad s–2 STEP 04 ANALYSIS OF CU QUESTION 01. GKwU PvKvi fi 10 kg Ges PµMwZi e¨vmva© 0.5 m Gi RoZvi åvgK †KvbwU? [CU. 13-14; JUST. 12-13; w`. †ev. 2015] A. 1.5 kg m2 B. 2.5 kg m2 C. 2.5 N D. 0 RoZvi åvgK, I = mK2 = 10 0.52 = 2.5 kg – m 2 STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. M f‡ii I r e¨vmv‡a©i GKwU wb‡iU wmwjÛv‡ii wbR A‡¶i mv‡c‡¶ RoZvi åvgK n‡”Q- [JnU. 17-18] A. 12 Mr B . M 3r C. M 12r D. 1 2 Mr2 Ans D 02. l ‣`‡N©¨i GKwU miæ`‡Ûi cÖvšÍ w`‡q Ges Gi •`‡N©¨i mv‡_ j¤^fv‡e AwZµgKvix A‡¶i mv‡c‡¶ Zvi Pµ MwZi e¨vmva©- [JnU. 11-12] A. 1 3 ml2 B. 1 3 ml2 C. 1 2 3 D. 1 3 l Ans D PART B Analysis of Science & Technology Question SUST 01. 3kg f‡ii GKwU e¯‧i fi‡K‡›`ªi ga¨ w`‡q MgbKvix GKwU A‡ÿi mv‡c‡ÿ e¯‧wUi RoZvi åvgK 2.5kg.m2 ; GB Aÿ †_‡K 1.2m j¤^ `~i‡Z¡ Aew¯
Z mgvšÍivj A‡ÿi mv‡c‡ÿ e¯‧wUi RoZvi åvgK, kg.m2 GK‡K, KZ n‡e? [SUST. 18-19] A. 3.94 B. 2.12 C. 5.51 D. 6.82 E. 1.95 I = 1 + Mh2 = (2.5 + 3 1.22 )Kg m 2 = 6.82kg – m 2

168 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. 20 cm e¨vmv‡a©i GKwU wb‡iU wmwjÛvi wbR A¶‡K †K›`ª K‡i Nyi‡Q| wmwjÛviwUi PµMwZi e¨vmva© KZ cm? [15-16] A. 10.1 B. 14.1 C. 18.1 D. 20.0 E. 20.1 k = r 2 = 20 2 = 14.1 03. GKwU PvKvi fi 20kg Ges PµMwZi e¨vmva© 0.7m| Gi RoZvi åvgK KZ? [JUST. 14-15] A. 9.8 kgm2 B. 2.5 kgm2 C. 1.10 kgm2 D. 1.4 kgm2 E. 0.35 kgm2 I = mk2 = 20 × (0.7)2 = 9.8 04. †Kvb A¶ mv‡c‡¶ 20N e¯‧i RoZvi åvgK 100kgm2 n‡j Gi PµMwZi e¨vmva© n‡e- [SUST. 02-03] A. 5m B. 5m C. 80m D. 120m I = mK2 ev, K = I m = 100 20 = 5m JUST 01. GKwU PvKvi fi 20 kg Ges PµMwZi e¨vmva© 0.5 m| PvKvwU‡Z 2rads–2 †K․wYK Z¡iY m„wó Ki‡Z KZ gv‡bi UK© cÖ‡qvM Ki‡Z n‡e? [JUST-C, 19-20] A. 50 Nm B. 0.5 Nm C. 10 Nm D. 100 Nm RoZvi åvgK, I = Mk2 = 20 × (0.5)2 = 5 kgm2 = I = 5 × 2 = 10 N-m MBSTU 01. †Kvb Aÿ mv‡c‡ÿ GKwU †j․n wbwg©Z e¯‧i PµMwZi e¨vmva© 0.5 m| e¯‧wUi fi 0.5 kg n‡j Gi RoZvi åvgK KZ? [MBSTU-C, Set-2 19-20] A. 0.050 Kgm2 B. 0.125 Kgm2 C. 0.025 Kgm2 D. †KvbwUB bq I = mk2 = 0.5 (0.5)2 = 0.125 kgm2 02. †KvbwU RoZvi åvgK msµvšÍ mgvšÍivj Aÿ Dccv`¨? [MBSTU-A, Set-2 19-20] A. Iz = Ix + Iy B. I = Ig + MK2C. I = Ig + MK D. I = Ig + Mh2 Iz = Ix + Iy n‡j j¤^ Aÿ Dccv`¨| 03. GKwU `„p e¯‧i PµMwZi e¨vmva© †KvbwU? [MBSTU. 15-16] A. K = I M B. K = M I C. K = I M D. K = M I I = MK2 ev, K 2 = I M ev, K = I M STEP 06 ANALYSIS OF GST QUESTION CUET 01. wbR N~Y©b A‡¶i mv‡c‡¶ `ywU e¯‧i RoZvi åvgK I Ges 2I| hw` Zv‡ii N~Y©b MwZkw³ mgvb nq, Zv‡`i †K․wYK fi‡e‡Mi AbycvZ KZ? [CUET. 14-15] A. 1:2 B. 2 :1 C. 1: 2 D. 2:1 L1 L2 = L1 L2 = I 2I = 1 2 ;, L1 : L2 = 1 : 2 STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. wP‡Î e‡ji åvgK n‡j †KvbwU mwVK? [Xv.‡ev 2021] X Y O r d F A. = r × F B. = d × F C. = r , F sin D. = d F sin Ans A wb‡Pi DÏxcKwU co Ges 02 I 03bs cÖ‡kœi DËi `vI: A x B O Q dx P wP‡Î miæ I mylg `‡Ði fi M Ges •`N©¨ L| 02. `ÐwUi dx As‡ki fi †KvbwU? [Xv.‡ev 2021] A. ML–1 dx B. LM–1 dx C. MLx D. Mx–1 dx Ans A 03. AB N~Y©b A‡ÿi mv‡c‡ÿ `ÛwUi RoZvi åvg‡Ki mgvKwjZ iƒc †KvbwU? [Xv.‡ev 2021] A. M L L 0 x 2 dx B. M L L/2 – L/2 x 2 dx C. M L L/2 – L/2 x 3 dx D. M L L/2 – L/2 x 2 2 dx Ans B 04. †KvbwU mgvšÍivj Aÿ Dccv`¨? [Xv. †ev. 2019] A. L = Ix + Iy B. I = IG + MK C. I = IG + Mh2 D. IG = I + MK2 Ans C 05. 2400J MwZkw³wewkó GKwU PvKv cÖwZ wgwb‡U 602 evi Ny‡i| PvKvwUi RoZvi åvgK KZ? [Xv.†ev. 2019] A. 0.605kgm2 B. 0.828kgm2 C. 1.21kgm2 D. 76.14kgm2 S C info Ek = 1 2 I 2 I = 2Ek 2 = 2 × 2400 2 × 602 60 2 = 1.21 kgm2 06. GKwU PvKvi fi 6kg Ges †Kv‡bv Aÿ mv‡c‡ÿ PµMwZi e¨vmva© 30cm| PvKvwU‡Z 3rads2 Z¡iY m„wó Ki‡Z KZ gv‡bi UK© cÖ‡qvM Ki‡Z n‡e? [Xv. †ev. 2016] A. 1.62 Nm B. 1.8 Nm C. 16.2 Nm D. 18 Nm S A info = I = MK2 × = 6 × 30 100 2 × 3 = 1.62 Nm 07. M f‡ii I l •`‡N©¨i miæ mylg `‡Ði ga¨we›`y w`‡q •`‡N©¨i Awfj¤^fv‡e MgbKvix A‡ÿi mv‡c‡ÿ RoZvi åvgK KZ? [iv.‡ev 2021] A. Ml 2 12 B. Ml 2 3 C. Ml 2 2 D. Ml2 Ans A 08. e„ËvKvi PvKwZi c„‡ôi Awfj¤^fv‡e MgbKvix ¯úk©‡Ki mv‡c‡ÿ PvKwZi RoZvi åvgK bx‡Pi †KvbwU? [P.‡ev 2021] A. I = 3 2 Mr2 B. I = Mr2 C. I = Mr2 2 D. I = Mr2 4 Ans D wb‡Pi DÏxcKwU co Ges 09 I 10 bs cÖ‡kœi DËi `vI : [P. †ev. 2017] 09. wb‡iU PvKwZi : XY A‡ÿi mv‡c‡ÿ PµMwZi e¨vmv‡a©i åvgK A. B. C. r D. S A info Ix = mk2 = 1 4 mr2 k = r 2 10. AB A‡ÿi mv‡c‡ÿ PvKwZi RoZvi åvgK KZ n‡e? A. 5 4 mr2 B. C. D. S A info IAB = Ix + Mr2 = 1 4 mr2 = 5 4 mr2

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 169 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 11. †Kv‡bv miy I mylg `‡Ûi GKcÖvšÍ w`‡q j¤^fv‡e MgbKvix A‡ÿi mv‡c‡ÿ RoZvi åvgK H `‡Ûi •`‡N©¨i ga¨we›`y w`‡q j¤^fv‡e MgbKvix A‡ÿi mv‡c‡ÿ RoZvi åvg‡Ki KZ ¸Y? [P. †ev. 2016] A. 4 B. 2 C. 1 2 D. 1 4 S A info cÖvšÍ w`‡q MgbKvix A‡ÿi mv‡c‡ÿ RoZvi åvgK, I1 = Ml 2 3 Ges ga¨we›`y w`‡q MgbKvix A‡ÿi mv‡c‡ÿ RoZvi åvgK, I2 = Ml 2 12 I1 I2 = Ml 2 3 Ml 2 12 = 1 3 × 12 = 4 I1 = 4I2 12. GKwU XvKvi fi Ges e¨vmva© h_vµ‡g 2.5kg Ges 2m PvKvwU‡Z10rads–2 †K․wYK Z¡iY m„wó Ki‡Z cÖ‡qvRbxq UK©- [w`.‡ev 2021] A. 10Nm B. 25Nm C. 50Nm D. 100Nm e¨vL¨v: = I = mr2 × = 2.5 ×22 ×10 = 100 Nm wb‡Pi DÏxcKwU co Ges 13 I 14 bs cÖ‡kœi DËi `vI: AB `ÛwU XY A‡ÿi mv‡c‡ÿ N~Y©bkxj| `‡Ûi †gvU •`N©¨ 2m Ges †gvU fi 2kg| X Y A B wPÎ-K X Y A B wPÎ-L 13. wPÎ K Gi RoZvi åvgK I1 Ges wPÎÑL Gi RoZvi åvgK I2 n‡j, †KvbwU mwVK? [h.‡ev. 2019] A. I1: I2=1:1 B. I1: I2=1:2 C. I1: I2=4:1 D. I1: I2=1:4 S C info I1 = mℓ2 3 ; I2 = mℓ2 12 I1 I2 = 1 3 × 12 I1 : I2 = 4 : 1 14. wPÎ ÑL G PµMwZi e¨vmv‡a©i gvbÑ [h.‡ev. 2019] A. B. C. D. Ans A 15. M f‡ii Ges R e¨vmv‡a©i GKwU PvKwZ Zvi †K›`ª w`‡q j¤^fv‡e MgbKvix †Kv‡bv A‡ÿi mv‡c‡ÿ Nyi‡Q| PvKwZi RoZvi åvgK KZ? [h. †ev. 2016] A. MR2 2 B. MR2 C. 3 2 MR2 D. 2 MR2 Ans A 16. N~Y©vqgvb e¯‧i RoZv cwigvc Kiv nq †KvbwU Øviv? [h. †ev. 2016] A. mr B. mr 2 C. rp D. rF Ans B 17. GKwU PvKvi RoZvi åvgK 10 kgm2 | PvKvwU‡Z 10 rads–2 †K․wYK Z¡iY m„wó Ki‡Z KZ UK© cÖ‡qvM Ki‡Z n‡e? [h.‡ev. 2015] A. 10 Nm B. 100 Nm C. 150 Nm D. 200 Nm S B info = I = 10 × 10 = 100 Nm 18. e‡ji åvg‡Ki mgxKiY- [Kz.‡ev 2021] i. = r F ii. = I a iii. = d L dt wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans D 19. wP‡Î AB e¯‧wU O †K †K›`ª K‡i XY A‡ÿi PZzw`©‡K Nyi‡Z cv‡i| Zvn‡j- [Kz.‡ev 2021] Y A V C V 90 V X B V N d V i. = d F ii. = r F iii. = r Fsin wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans C 20. cvZjv e„ËvKvi PvKwZi PµMwZi e¨vmva© n‡jv- [Kz. †ev. 2015] A. K = l 2 B. K = l 3 C. K = r 2 D. K = r 12 Ans C 21. †MvjvKvi PvKwZ ev cv‡Zi c„‡ôi Awfj¤^fv‡e MgbKvix A‡ÿi mv‡c‡ÿ RoZvi åvg‡Ki †ÿ‡Î †KvbwU mwVK? [wm.‡ev 2021] A. 2Mr2 B. Mr2 C. Mr2 2 D. Mr2 4 Ans C 22. GKwU PvKvi RoZvi åvgK 5kgm2 | PvKvwU‡K 105 J N~Y©b MwZkw³‡Z Nyi‡Z KZ †K․wYK †e‡Mi cÖ‡qvRb n‡e? [g.‡ev 2021] A. 20 rad/sec B. 89.44 rad/sec C. 141.42rad/sec D. 200rad/sec S D info Ek = 1 2 I 2 105 = 1 2 5 2 2 102 nads1 23. e„ËvKvi PvKwZi c„‡ôi Awfj¤^fv‡e MgbKvix ¯úk©‡Ki mv‡c‡ÿ PvKwZi RoZvi åvgK bx‡Pi †KvbwU? [e.‡ev 2021] A. I = 3 2 Mr2 B. I = Mr2 C. I = Mr2 2 D. I = Mr2 4 Ans C wb‡Pi DÏxc‡Ki Av‡jv‡K 24 I 25bs cÖ‡kœi DËi `vI| 5kg f‡ii I 0.2m e¨vmv‡a©i GKwU e„ËvKvi cvZ XY fvi‡K›`ªMvgx I c„‡ôi mv‡_ j¤^ A‡ÿi Pvwiw`‡K Nyi‡Q| XY Aÿ I cvZwUi e¨vm AB Gi mv‡c‡ÿ h_vµ‡g RoZvi åvgK Ixy I IAB [e. †ev. 2016] 24. Ixy = KZ? A. 0.1kgm2 B. 0.2kgm2 C. 1kgm2 D. 2kgm2 S A info Ixy = mr 2 2 = 5 × (0.2) 2 2 = 0.1 kgm2 25. IAB, Ixy GiÑ A. A‡a©K B. mgvb C. 1.5 ¸Y D. 2 ¸Y S A info IAB = mr2 4 Ges Ixy = mr2 2 IAB = Ixy 2 26. m fi I r e¨vmv‡a©i e„ËvKvi PvKwZi †h‡Kv‡bv e¨v‡mi mv‡c‡ÿ RoZvi åvg‡Ki gvb †KvbwU? [mKj. †ev. 2014] A. 3 2 mr2 B. mr2 C. 1 2 mr2 D. 1 4 mr2 Ans D

170 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 8 †K›`ªgyLx ej I Z¡iY msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. †K›`ªgyLx ej, F = mv2 r = m 2 r = m 2 2 T 4 r 02. †K›`ªgyLx Z¡iY, a = v 2 r = 2 r = 2 2 T 4 r 03. kxl© we›`y‡Z Uvb T = mv2 r mg 04. e„‡Ëi †K‡›`ªi mv‡_ GK ‡j‡f‡j Uvb T = mv2 r = m 2 r = m 2 2 T 4 r 05. e„‡Ëi me©wbgœ we›`y‡Z myZvi Uvb T = mv2 r + mg mv2 r -mg +mg mv2 r mv2 r +mg (i) (ii) (iii) mv2 r Gi w`K mg Gi w`K CONCEPTUAL MATH MEx 01 GKwU K…wÎg DcMÖn 7000km e¨vmva©wewkó e„ËvKvi K¶c‡_ c„w_ex‡K cÖ`w¶Y Ki‡Q| DcMÖnwUi ch©vqKvj 2 h n‡j †K›`ªgyLx Z¡iY KZ? †K›`ªgyLx Z¡iY, a = 2 r = 2 T 2 r = 2 2 3600 2 7 106 = 5.325 m/s2 MEx 02 †evi-Gi nvB‡Wªv‡Rb cigvYyi g‡Wj Abyhvqx, GKwU B‡jKUªb GKwU †cÖvU‡bi Pviw`‡K 5.2 10–11m e¨vmv‡a©i GKwU e„Ëc‡_ 2.18 106 ms–1 †e‡M cÖ`w¶Y K‡i| B‡jKUª‡bi fi 9.1 10–31 kg n‡j, †K›`ªgyLx e‡ji gvb KZ? †K›`ªgyLx ej, F = mv 2 r = 9.1 10–31 (2.18 106 ) 2 5.2 10–11 = 8.3167 10–8 N NOW START PRACTICE 01. c„w_exi wb‡Ri A‡¶i Dci N~Y©‡bi d‡j wbi¶‡iLvi †h †Kvb GKwU we›`yi Z¡iY KZ n‡e? [c„w_exi e¨vmva© = 6.4 106m] 02. 50 gm f‡ii GKwU e¯‧‡K 40 cm `xN© GKwU myZvi GK cÖv‡šÍ †eu‡a e„Ëc‡_ cÖwZ †m‡K‡Û 4 evi Nyiv‡bv n‡”Q| †K›`ªgyLx Z¡iY KZ? 03. e„ËvKvi c‡_ 72 km/h mg`ªæwZ‡Z Pjgvb †Kvb Mvoxi †K›`ªgyLx Z¡iY 1 m/s2 n‡j e„ËvKvi c‡_i e¨vmva© KZ? 04. 2kg f‡ii GKwU cv_i‡K 9.8m `xN© GKwU myZvi mvnv‡h¨ †e‡a Avbyf~wgK Z‡j Nyiv‡bv n‡”Q| myZvwU m‡ev”©P 19.6N Uvb mn¨ Ki‡Z cv‡i| myZv bv wQ‡o cv_iwU‡K m‡ev”P© KZ `yªwZ‡Z Nyiv‡bv †h‡Z cv‡i| NOW PRACTICE SOLVE : 01. Z¡iY, a = 2 r = 2 T 2 r = 2 24 60 60 2 6.4 106 = 3.4 10–2 ms–2 02. †K›`ªgyLx Z¡iY, a = 2 r = 2N t 2 r = 2 4 1 2 0.40 = 252.66 ms–2 03. v = 1 20 3.6 72 ms ; †K›`ªgyLx Z¡iY, a = r v 2 e„ËvKvi c‡_i e¨vmva© , r = m a v 400 1 20 2 2 04. F = r mv2 or v = ms 1 9.8 2 19.6 9.8 m Fr REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU e¯‧ 12 m e¨vmv‡a©i GKwU e„ËvKvi c‡_ Pjgvb Av‡Q| GKwU gyn~‡Z© e„ËvKvi c‡_ Gi `ªæwZ 6 m/s Ges GwU 4 m/s2 nv‡i e„w× cv‡”Q| H gyn~‡Z© e¯‧wUi Z¡i‡Yi gvb KZ? [DU. 20-21] A. 2 m/s2 B. 3 m/s2 C. 4 m/s2 D. 5 m/s2 S D info GLv‡b, e„ËvKvi c‡_i ¯úk©K eivei Z¡iY (tangential acceleration) aT = 4 m/s2 Ges †K›`ªgyLx Z¡iY (centrifugal Acceleration) ac = v 2 r = 6 2 12 = 3 m/s2 ; jw×Z¡iY, a = aT 2 + ac 2 = 4 2 +32 = 5m/s2 02. e„ËvKvi c‡_ 72 km/h mg`ªæwZ‡Z Pjgvb †Kvb Mvoxi †K›`ªgyLx Z¡iY 1 m/s2 n‡j e„ËvKvi c‡_i e¨vmva© KZ? [DU. 17-18] A. 150 m B. 300 m C. 400 m D. 200 m v = 72 3.6 = 20ms–1 ; †K›`ªgyLx Z¡iY, a = v 2 r e„ËvKvi c‡_i e¨vmva© , r = v 2 a = 20 2 1 = 400m 03. Nwoi wgwb‡Ui KuvUvi †K․wYK †e‡Mi gvb- [DU. 15-16; KU. 17-18] A. 60/ rad/s B. 1800/ rad/s C. rad/s D. /1800 rad/s †K․wYK †eM, = 2 T = 2 3600 = /1800 rads–1 04. m f‡ii GKwU e¯‧ r e¨vmv‡a©i e„ËvKvi c‡_ mg`ªæwZ‡Z Pj‡Q| e„ËvKvi MwZi ch©vqKvj T. e¯‧wUi Dci †K›`ªgyLx e‡ji gvb KZ? [DU. 12-13] A. 4 2 mr T 2 B. 4 2 mr T C. 4mr 2 T 2 D. mr2 e¯‘i fi = m; e¨vmva© = r,ch©vqKvj = T,‡K›`ªgyLx ej F =? F = m 2 r = m 2 T 2 r = 4 2mr T 2

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 171 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 05. GKwU KYv 40 cm e¨vmv‡a©i e„ËvKvi c‡_ cÖwZ wgwb‡U 45 evi AveZ©b K‡i| KYvwUi †K›`ªgyLx Z¡iY- [DU. 04-05] A. 8.88 m/s2 B. 1.41 m/s2 C. 35.55 m/s2 D.2.82 m/s2 a = 2 r = 2n t 2 r = 2 45 3.14 60 2 0.4 = 8.88 m/sec2 06. 3.0g f‡ii GKwU e¯‧ 2.0 m e¨vmv‡a©i GKwU e„‡Ë mgnv‡i Pj‡Q| hw` e¯‧wU †m‡K‡Û 4.0 evi e„‡Ëi Pvwiw`‡K †Nv‡i Z‡e e¯‧wUi Dci cÖ‡qvRbxq ej KZ? [DU. 01-02] A. 7.6 N B. 3.8 N C. 4.8 N D. 4.2 N F = m 2 r = m 2n t 2 r = 3 10–3 2 4 3.14 1 2 2 = 3.8 N STEP 02 ANALYSIS OF JU QUESTION 01. 100 gm f‡ii GKwU e¯‧‡K 40 cm `xN© GKwU myZvi GKcÖv‡¯Í †eu‡a e„Ëc‡_ 20 m/s mg`ªæwZ‡Z Nyiv‡bv n‡”Q| †K›`ªgyLx Z¡iY KZ? [JU. 17-18] A. 1200 m/s2 B. 1000 m/s2 C. 1800 m/s2 D. 800 m/s2 †K›`ªgyLx Z¡iY, a = v 2 r = 202 0.4 = 1000 m/s2 02. 0.250 kg f‡ii GKwU cv_i LÛ‡K 0.75m j¤^v GKwU myZvi GK cÖv‡¯Í †e‡a e„ËvKvi c‡_ cÖwZ wgwb‡U 90 evi Nyiv‡j myZvi Dci KZ Uvb co‡e? [JU. 16-17] A. 16.66 N B.17.66 N C. 18.66 N D. 19.66 N †K›`ªgyLx ej, F = m 2 r = m 2N t 2 r = 0.25 2 90 60 2 0.75 = 16.66 N STEP 03 ANALYSIS OF RU QUESTION 01. 50 MÖvg f‡ii GKwU e¯‧‡K GKwU 20 cm •`‡N©¨i myZvi GKcÖv‡šÍ †eu‡a e„ËvKvi c‡_ cÖwZ †m‡K‡Û 3 evi Nyiv‡bv n‡”Q| hw` e¯‧wUi fi AcwiewZ©Z †i‡L myZvi •`N©¨ wظb K‡i Nyiv‡bv nvi A‡a©K Kiv n‡j Z¡iY KZ ¸Y n‡e? [RU. 15-16] A. 0.5 ¸Y B. 2 ¸Y C. AcwiewZ©Z _vK‡e D. 4 ¸Y Ans C 02. 50ft ‣`‡N©¨i GKwU myZvi GK cÖv‡šÍ GKwU KYv 50ft/s mg`ªæwZ‡Z N~Y©vqgvb| KYvwUi Dci †K›`ªgyLx Z¡iY KZ? [RU. 10-11] A. 200ft/s2 B. 50ft/s2 C. 10ft/s2 D. 25ft/s2 a = v 2 r = 502 50 = 50ft/s2 STEP 04 ANALYSIS OF CU QUESTION 01. hLb e„ËvKvi c‡_ N~Y©vqgvb GKwU †Uª‡bi MwZ wظY Kiv nq ZLb †UªbwUi †K›`ªgyLx ej [CU-A, Set-2. 19-20] A. mgvb _v‡K B. 2 ¸Y e„w× cvq C. 8 ¸Y e„w× cvq D. 4 ¸Y e„w× cvq †K›`ªgyLx ej, F = mv2 r , F v 2 F2 = 22 F1 = 4F1 02. 0.2kg f‡ii GKwU cv_i‡K 0.6m j¤^v GKwU myZvi mvnv‡h¨ †e‡a Avbyf~wgK e„ËvKvi c‡_ cÖwZ †m‡K‡Û 2.5 evi Nyivb n‡”Q| myZvi Uvb KZ? [CU. 10-11] A. 30N B. 29.6N C. 28.8N D. 27N E. 26N T = F = m 2 r 2 = 0.2 (2 2.5 3.14)2 0.6 = 29.6N 03. e„ËvKvi c‡_ 30ms–1 mg`ªæwZ‡Z Pjgvb †Kvb e¯Íi †K›`ªgyLx Z¡iY 2ms–2 n‡j e„ËvKvi c‡_i e¨vmva© KZ? [CU. 08-09] A.300m B. 350m C. 400m D. 450m E. 500m a = v 2 r ev, r = v 2 a = (30) 2 2 = 450m STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. e„ËvKvi c‡_ N~Y©biZ e¯‧i Dci wµqviZ ej- [CoU. 12-13] A. F = mv r B. F = mv r 2 C. F = mv2 r D. F = m 2 v r Ans C 02. 100 gm f‡ii GKwU e¯‧‡K 40 cm `xN© GKwU myZvi GK cÖv‡šÍ †eu‡a mg`ªæwZ‡Z Nyiv‡bv n‡”Q| †K›`ªgyLx ej KZ n‡e? [JnU. 10-11] A. 100 N B. 80 N C. 88 N D. 110 N F = mv2 r = 0.1 (20) 2 0.4 = 100N 03. c„w_ex N~Y©‡bi Rb¨ welye †iLvq Aew¯
Z †Kvb e¯‧i †K›`ªgyLx Z¡iY KZ? [c„w_exi e¨vmva© R = 6400 km] [JnU. 07-08] A. 9.8 m/sec2 B. 4.9 m/sec2 C. 0.068 m/sec2 D. 0.034 m/sec2 g = GM R 2 = 6.67 6 1024 (6.4 106 ) 2 10–11 = 9.8 m/s2 KU 01. 0.15 kg f‡ii GKwU cv_i LÛ‡K 0.75 m j¤^v GKwU myZvi GK cÖv‡šÍ †e‡a e„ËvKvi c‡_ cÖwZ wgwb‡U 90 evi Nyiv‡j myZvi Dci Uvb n‡e- [KU. 07-08] A. 8.84 N B. 11.12 N C. 9.98 N D. 7.83 N T = m 2 r = 0.15 2n t 2 0.75 = 0.15 2 3.14 90 60 2 0.75= 9.98 N 02. 4 kg f‡ii GKwU e¯‧ 12 m/s wbw`©ó †e‡M 6 m e¨vmv‡a©i e„ËvKv‡i Nyi‡Q, GLv‡b †K›`ªgyLx e‡ji gvb- [KU. 04-05] A. 2 N B. 20 N C. 66 N D. 96 N T ev F = mv 2 r = 4 (12) 2 6 = 96 N BRUR 01. 0.1 kg fi wewkó GKwU cv_i LÛ 0.8 m ‣`‡N©¨i myZvi GK cÖv‡šÍ †eu‡a e„ËvKvi c‡_ Nyiv‡bv n‡jv| cv_i LÛwU cÖwZ †m‡K‡Û 2 evi AveZ©b Ki‡j myZvi Uvb KZ? [BRU. 15-16] A. 10.63 N B. 11.63 N C. 12.63 N D. 13.63 N myZvi Uvb T = m 2 r = m 2n t 2 r = 0.1 2 2 1 2 0.8 = 12.63N PART B Analysis of Science & Technology Question SUST 01. GKwU KYvi Ae¯
vb †f±i (5i ˆ + 3j ˆ + k ˆ )m| Zvi Dci (–3i ˆ + 7j ˆ )N ej cÖhy³ n‡j U‡K©i gvb KZ? N.m? [SUST-A, 19-20] A. 40.65 B. 44.65 C. 24.65 D. 34.65 E. 54.65 r = 5 i + 3 j + k ; F = 3 i + 7 j UK©, = | r F|; r F = i 5 3 j 3 7 k 1 0 = i(0 7) j(0 + 3) + k(35 + 9) = 7 i 3 j + 44 k | r F| = (7) 2 + (3) 2 + 442 = 44.65 Nm

172 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. 20 g fi wewkó †Kvb e¯‧‡K 5m `xN© myZvi mvnv‡h¨ e„ËvKvi c‡_ Nyiv‡bv n‡”Q| e¯‧wU 6s G 30 wU c~Y© AveZ©b K‡i| myZvi Uvb KZ N? [SUST-B, 19-20] A. 15.92 B. 29.22 C. 98.7 D. 49.12 E. 10.09 F = m 2 r = m 2 T 2 r T = t N = 20 103 2 3.14 0.2 2 5= 6 30 = 0.2 sec = 98.7 N 03. 0.02kg f‡ii GKwU Mvwo 60 km/hr †e‡M P‡j| †eªK †P‡c MvwowU‡K 50m `~‡i _vgv‡bv n‡jv| hw` iv¯Ívi Nl©YRwbZ ej 100N nq, Z‡e †eªKRwbZ e‡ji gvb KZ N? [SUST. 14-15] A. †K›`ªgyLx 0.03 B. †K›`ªwegyLx 0.03 C. †K›`ªgyLx 0.09 D. †K›`ªwegyLx 0.09 †K›`ªwegyLx ej, F = m 2 r = 0.02 (3) 2 0.5= 0.09 2N 04. nvB‡Wªv‡Rb cigvYy‡Z wbDwK¬qvm‡K †K›`ª K‡i e„ËvKvi c‡_ GKwU B‡jKUªb 2.0 10–6m/s •iwLK `ªæwZ‡Z Nyi‡Q| cigvYyi e¨vmva© 5.210–11 n‡j B‡j±ª‡bi †K›`ªgyLx Z¡iY KZ? [SUST. 08-09] A. 9.69 10–2 m/s2 B. 6.89 10–2 m/s2 C. 5.96 10–2 m/s2 D. 7.69 10–2 m/s2 a = v 2 r = (2.0 10–6 ) 2 5.2 10–11 = 7.69 × 10–2 ms–2 05. †K․wYK †eM wb‡q r e¨vmv‡a©i e„ËvKvi c‡_ AveZ©biZ GKwU KYvi †K›`ªgyLx Z¡iY wb‡Pi †KvbwU? [SUST. 07-08] A. 2 /r B. 2 r C. r 2 D. r a = 2 r 06. r e¨vmv‡a©i e„ËvKvi c‡_ N~Y©vqgvb GKwU KYvi `ªæwZ v n‡j KYvwUi †K›`ªgyLx Z¡iY KZ? [SUST. 06-07] A. v 2 /r B. v/r2 C. v 2 r D. vr Ans A 07. GKwU 500kg f‡ii †nwjKÞvi Gi †eøW 30m2 †ÿÎdj Ny‡i Ges v MwZcÖvß nq| evZv‡mi NbZ¡ 1.3kgm3 Ges g = 10m/s2 n‡j †eM v Gi gvb- [SUST. 04-05] A. 11 m/s–1 B. 100 m/s–1 C. 50 m/s–1 D. 1 m/s–1 A = 300m2 r 2 = 300 r = 300 = 9.772 F = mv2 r mg v = rg = 9.772 1.3 10 = 11.2ms–1 08. 0.250 kg f‡ii GKwU cv_i LÛ‡K 1.2 m j¤^v GKwU myZvi GK cÖv‡šÍ †eu‡a e„ËvKvi c‡_ cÖwZ wgwb‡U 90 evi Nyiv‡j myZvi Dci KZ Uvb co‡e? [SUST. 05-06] A. 26.65 N B. 33.30 N C. 17.75 N D. 66.60 N T= m 2 r = m 2n t 2 r = 0.250 2 90 3.14 60 2 1.2 = 26.647 09. GKwU ej 9m e¨vmva© wewkó e„ËvKvi c‡_ 8ms–1 †e‡M Nyiv‡j Zvi Z¡iY KZ? [SUST. 06-07] A. k~b¨ B. 8ms–2 C. 9ms–2 D. 11ms–2 S Blank info a = v 2 r = 8 2 9 = 7.1ms–2 10. 6.0 kg f‡ii GKwU e¯‧‡K 3.0 m `xN© GKwU myZvi cÖv‡šÍ †eu‡a 2.0 ms–1 †e‡M Nyiv‡bv n‡”Q| myZvi Dci Uvb KZ wbDUb (N) n‡e? [SUST. 15-16] A. 4 B. 6 C. 6.5 D. 8 E. 9 Uvb, T = mv2 r = 6 2 2 3 = 8 JUST 01. GKwU cvwb fwZ© evjwZ 1 wgUvi j¤^v iwk‡Z †eu‡a me©wb¤œ KZ †e‡M Nyiv‡j evjwZ n‡Z cvwb co‡e bv? [JUST-A, Set-Ka 19-20] A. 9.8 ms–1 B. 4.9 ms–1 C. 3.13 ms–1 D. 1.67 ms–1 v = gr = 9.8 × 1 = 3.13 ms–1 02. 0.25kg f‡ii †Kvb cv_i LÛ‡K 0.75m j¤^v m~Zvi GK cÖv‡šÍ †eu‡a e„ËvKvi c‡_ cÖwZ wgwb‡U 90 evi Nyiv‡j m~Zvi Uvb KZ n‡e? [JUST. 14-15] A. 32.76 N B. 16.67 N C. 38 N D. 36.5 N E. 36 N F = m 2 r = 0.25 2n t 2 r = 0.25 2 90 60 2 0.75 = 16.67N MBSTU 01. GKwU B‡jKUªb cigvYyi wbDwK¬qv‡mi Pvwiw`‡K 0.53 Av‡g©‡÷ªvs e¨vmv‡a©i GKwU e„ËvKvi c‡_ 4 106 m/s †e‡M cÖ`wÿY K‡i| B‡j±ª‡bi †K›`ªgyLx e‡ji gvb KZ? [MBSTU-C, Set-2 19-20] A. 2.74 109 N B. 2.75 107 N C. 1.46 107 N D. 2.91 109 N Fc = mv2 r = 9.1 1031 (4 106 ) 2 0.53 1010 N = 2.75 107 N BSMRSTU 01. 0.1 kg f‡i GKwU cv_i‡K 0.5m j¤^v GKwU myZvi mvnv‡h¨ e„ËvKvi c‡_ Nyiv‡bv n‡”Q| cv_iwU cÖwZ wgwb‡U 30 evi c~Y© N~Y©b m¤úbœ K‡i| myZvi Uvb KZ? [BSMRSTU-B, 19-20] A. 0.5 N B. 1 N C. 1.4 N D. 4 N Avgiv Rvwb Uvb, T = m 2 r = 0.1 2 30 60 2 0.5 = 0.4935 0.5 N HSTU 01. †Kvb e‡ji Kvi‡Y gvLb‡K `ya †_‡K c„_K Kiv hvq? [HSTU. 14-15] A. †K›`ªgyLx ej B. wegyLx ej C. Z¡iY D. †K›`ªwegyLx ej Ans D PSTU 01. †Kvb ejwU m„wó Kivi Rb¨ iv¯Ívi e¨vswKs Kiv nq? [PSTU. 14-15] A. †K›`ªwegyLx ej B. †K․wYK ej C. †K›`ªgyLx ej D. Nl©Y ej eµ c‡_ Pjvi mgq †K›`ªgyLx e‡ji cÖ‡qvRb| GB ej m„wó Kivi Rb¨ iv¯Ívi e¨vswKs Kiv nq| STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. †ev‡ii nvB‡Wªv‡Rb cigvby g‡W‡j GKwU B‡j±ªb GKwU †cÖvU‡bi Pvwiw`K 5.2 10–11m e¨vmv‡a©i GKwU e„ËvKvi c‡_ Nyi‡Q| †eM 2.18 10– 31ms–1 n‡j †K›`ªgyLx ej KZ n‡e? [BUET. 12-13] A. 3.81 10–6 N B. 8.32 10–8 N C. 2.17 10–47 N D. 8.32 1026 N F = mv2 r = 9.1 10–31 (2.18 106 ) 2 5.2 10–11 = 8.32 10–8 N 02. 0.5kg f‡ii GKwU e¯‧‡K 0.5m `xN© GKwU myZvi mvnv‡h¨ e„ËvKvi c‡_ Nyiv‡bv n‡jv| hw` e¯‧wU cÖwZ †m‡K‡Û 40 evi e„ËvKvi c‡_ AveZ©b K‡i Z‡e †K․wYK fi †e‡Mi gvb KZ? [BUET. 08-09] A. 0.314kg–m 2 /sec B. 3.14kg–m 2 /sec C. 31.4kg–m 2 /sec D. 31.4gm–cm2 /sec E. 314gm–cm2 /sec L= mr2 = 0.5 (0.5)2 2 40 = 31.42kg – m 2 /sec

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 173 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES CKRuet. Combind 01. GKwU iwki GK cÖvšÍ GKwU 9 lb Gi fi †Kv‡bv gm„Y cywji gva¨‡g wb¤œgyLx gvbvi mgq iwkwUi Ab¨cÖv‡šÍi 6 lb Gi GKwU fi †U‡b Zz‡j| wm‡÷‡gi Z¡iY I iwki Uvb KZ? [CKRUET. 2020-21] A. 6.4ft/sec2 ;100 lb B. 3.2ft/sec2 ;92 lb C. 6.4ft/sec2 ;230.4lb D. 3.2ft/sec2 ;100 lb E. 3.2ft/sec2 ;330 lb S C info m1 e¯‘ Z¡i‡Y bvg‡Q, m2 e¯‘ Z¡i‡Y DV‡Qw1 – T = m1 9 × 32 – T = 9 288 – T = 9 .......(i) Avevi, T – w2 = m2 T – 6 × 32 = 6 T – 192 = 6 ..........(ii) (i)+ (ii) 96 = 15 = 6.4 fts–2 T = 192 + 6 × 6.4 = 230.4 lb KUET 01. 0.150kg f‡ii GKwU cv_i LÛ‡K 0.75m j¤^v GKwU myZvi GK cÖv‡šÍ †eu‡a e„ËvKvi c‡_ cÖwZ wgwb‡U 90 evi Nyiv‡j myZvi Dci Uvb wbY©q Ki| [KUET. 11-12] A. 9.99N B. 9.90N C. 9.99KN D. 9.955 E. 9.98N F = m 2 r = m 2n t 2 r = 0.150 2 90 60 2 0.75 = 9.99N 02. 0.5kg f‡ii GKwU e¯‧ 0.4m j¤^v myZvi †kl cÖv‡šÍ e„ËvKvi c‡_ N~Y©qgvb| hw` e¯‧wU †m‡K‡Û 40 evi Ny‡i Zvn‡j †K․wYK fi‡eM wbY©q Ki| [KUET. 07-08] A. 20.11gm-cm2 /s B. 20.11kg-m 2 /s C. 2.011kgm2 /s D. 201.1gm-cm2 /s E. 201kg cm2 /s L = mr2 = 0.5 (0.4)2 2 40 = 20.106 kg – m 2 /sec STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. (i) F = mv2 r (ii) F = m 2 r 2 (iii) L = mvr cÖZxK¸‡jv cÖPwjZ A_© enb Ki‡j, †Kvb m¤úK© mwVK? [Xv. †ev. 2016] A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans C 02. †K›`ªgyLx e‡ji mwVK ivwkgvjv †KvbwU? [w`.‡ev 2021] A. mv+r B. m 2 r C. m 2 r 2 D. mv r 2 F = mv2 r = m 2 r 2 r = m 2 r 03. GKwU evjwZ‡Z wKQz cwigvY cvwb wb‡q r e¨vmv‡a©i e„Ëc‡_ Djø¤^ Z‡j Nyiv‡bv n‡”Q, N~Y©‡bi †K․wYK MwZ KZ n‡j evjwZi cvwb evB‡i †ei n‡e bv? [g.‡ev 2021] A. r g B. g r C. rg D. rg †K›`ªgyLx ej = IRb mv2 r = mg v 2 = rg V = rg Concept 9 †KŠwYK †eM I •iwLK †eM FORMULA 01. v = .r 03. = 0 + 2 t 05. = I 07. L = I = mr 2 = mvr = rp 09. 2 = 0 2 + 2 02. s = r 04. = r × F 06. = 0 + t 08. a = r 10. = 0t + 1 2 t 2 CONCEPTUAL MATH MEx 01 4kg f‡ii GKwU evjK bvMi †`vjvq P‡o 20m e¨v‡mi e„ËvKvi c‡_ 6rpm †K․wYK †e‡M Nyi‡Q| evjKwUi †K․wYK fi‡eM wbY©q Ki? L = I = mr2 × 2n T = 4 × (10)2 × 2×6 60 = 80 kgms–1 MEx 02 10kg f‡ii GKwU PvKvi Dci 10N-m UK© cÖ‡qvM Kiv n‡j 4rads-2 ‡K․wYK Z¡ib m„wó nq| PvKvwUi N~Y©b RoZv I PµMwZi e¨vmva© wbY©q Ki? = I I = = 10 4 = 2.5kg–m 2 Ges I = mr2 ev, 2.5 = 10r2 ev r = 0.5m NOW START PRACTICE 01. GKwU •e`y¨wZK cvLv wgwb‡U 1500 evi †Nv‡i| myBP eÜ Kivi 4 wgwbU ci eÜ n‡q hvq| †_‡g hvevi c~‡e© cvLvwU KZevi Nyi‡e? 02. GKwU •e`y¨wZK cvLvi myBP Ab Ki‡j `kevi c~Y© N~Y©‡bi ci †K․wbK †eM 20 rad/sec nq| †K․wbK Z¡iY KZ? 03. Nwoi wgwb‡Ui KuvUvi †K․wYK †e‡Mi gvb KZ? 04. w¯
ive¯
v n‡Z GKwU KYv‡K 3.14 rad/sec2 mg-‡K․wbK Z¡i‡Y e„ËvKvi c‡_ Nyiv‡j 10 †m‡K‡Û KYvwU KZ †K․wbK †eM jvf Ki‡e? NOW PRACTICE SOLVE : 01. = 1 + 2 2 t = 1500 2 602 240 = 6000 N = 2 = 3000 rev/evi 02. 1 = 0, 2 = 20 rad/sec N = 10 evi = 2N = 2.10 =20 = 2 20 (20) 0 2 2 2 2 1 2 2 ev, 2 3.18rad /sec 10 03. †K․wYK †eM, = T 2 = 3600 2 = /1800 rads–1 [†m‡K‡Ûi KvUvi = 30 rads–1 ; NÈvi KvUvi = 21600 rads–1 ] 04. 2=1+t 2=0+3.14×10=31.4 rads1 m1 m2 T T

174 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU e¯‧ m e¨vmv‡a©i e„ËvKvi c‡_ 4.0 m/s mg`ªæwZ‡Z Nyi‡Q| GKevi Ny‡i Avm‡Z e¯‧wUi KZ mgq jvM‡e? [DU.A 19-20] A. 2/ 2 s B. 2 /2s C. /2s D. 2 /4s S B info 2r = vt t = 2r v s ; t = 2 4 s ; t = 2 2 s ; t = 2 2 s 02. GKwU KYv 2.0 e¨vmv‡a©i e„ËvKvi c‡_ cÖwZ wgwb‡U 30 evi AveZ©b K‡i| Gi •iwLK †eM KZ? [DU. 14-15] A. ms–1 B. 2 ms–1 C. 4 ms–1 D. 0.5 ms–1 •iwLK †eM, v = r = 2 30 60 2 = 2 ms–1 03. GKwU •e`y¨wZK cvLvi m~BP AbKi‡j `kevi c~b© N~b©‡bi ci †K․wYK †eM 20 rad/sec nq| †K․wYK Z¡iY KZ? [DU. 09-10; JU. 17-18] A. 1.83 rad/s2 B. 8.13 rad/s2 C. 3.18 rad/s2 D. 5.17 rad/s2 = 2N = 2.10 =20 = 2 2 – 1 2 2 = (20) 2 – 0 2 2 20 ev, = 10 = 3.183 rad/sec 04. GKwU cvLv cÖwZ wgwb‡U 60 evi †Nv‡i| cvLvwUi †K․wYK †eM KZ? [DU. 08-09; CU. 13-14] A. rad/s B. 2 rad/s C. 4 rad/s D. 2 rad/s †K․wYK †eM, = 2N t = 2 60 60 = 2 rad/s STEP 02 ANALYSIS OF JU QUESTION 01. GKwU Nwoi †m‡K‡Ûi KvuUvi †K․wYK †eM KZ? [JU-H, Set-1. 19-20] A. rad s–1 B. 2 rad s–1 C. 3 rad s–1 D. 30 rad s–1 = 2 T = 2 60 = 30 rads1 02. GKwU KYv 4.5m e„ËvKvi c‡_ cÖwZ wgwb‡U 225 evi AveZ©b K‡i| Gi •iwLK †eM KZ? [JU. 16-17] A. 107 ms–1 B. 106 ms–1 C. 108 ms–1 D. 109 ms–1 •iwLK †eM, v = r = 2N t r = 2 225 60 4.5 = 106ms–1 03. nvZNwoi KvuUvi †K․wYK †eM N›Uvi KvUvi Rb¨ [JU. 16-17] A. /180 rad s–1 B. /270 rad s–1 C. /21600 rad s–1 D. /190 rad s–1 †K․wYK †eM, = 2N t = 2 12 3600 = 21600 rads–1 04. GKwU KYv 1.5m e¨vmv‡a©i e„ËvKvi c‡_ cÖwZ wgwb‡U 120 evi AveZ©b K‡i| Gi •iwLK †eM KZ n‡e? [JU. 15-16] A. 16.84 ms–1 B. 17.84 ms–1 C. 18.84 ms–1 D. 19.84 ms–1 •iwLK †eM, v = r = 2 N t r = 2 120 60 1.5 = 18.84ms–1 05. GKwU †`qvj Nwoi wgwb‡Ui KvUvi •`N©¨ 18 cm n‡j Gi †K․wYK †eM n‡e [JU. 13-14] A. 1.7410–3 rad/s B. 3.1310–4 rad/s C. 1810–4 rad/s D. 10–4 rad/s = 2 T = 2 3600 = 1.74 103 06. GKwU N~Y©biZ KYvi e¨vmva© †f±i r = i – j + k Ges cÖhy³ ejF = 2i – j n‡j U‡K©i gvb n‡e- [JU. 12-13] A. 2 GKK B. 6 GKK C. 4 GKK D. 8GKK r = i – j + k , F = 2i – j , UK© F r = i 1 2 j –1 –1 k 1 0 = i + 2j + k (–1 + 2) = i + 2j + k U‡K©i gvb = 1 2 + 22 + 12 = 6 07. c„w_exi Pvwiw`‡K Puv‡`i K¶ c‡_i e¨vmva© 3.85 105 km K¶ c_ GK evi cÖ`w¶Y Ki‡Z mgq jv‡M 27.3 w`b| Puv‡`i †K․wYK †eM KZ? [JU. 11-12] A. 12.66410–6 rads–1 B. 22.64410–6 rads–1 C. 2.66510–6 rads–1 D. †KvbwUBbq = 2 T = 2 27.3 24 60 60 = 2.665 10–6 rad/s STEP 03 ANALYSIS OF RU QUESTION 01. GKwU Nwoi N›Uvi KuvUvi •`N©¨ 2cm| hw` Nwoi †m‡K‡Ûi KuvUvi •`N©¨ 3 cm nq, Z‡e †m‡KÛ I N›Uvi KuvUvi cÖv‡šÍi •iwLK †e‡Mi AbycvZ KZ n‡e? [RU. 17-18] A. 5/2 B. 1/2 C. 2/3 D. 3/2 •iwLK †eM,v = r = 2 T r ; Nwoi KuvUvi cÖv‡¯Íi †K․wYK †eM constant _v‡K| •iwLK †e‡Mi AbycvZ = vs vh = rs rh = 3 2 02. GKwU •e`y¨wZK cvLv wgwb‡U 3000 evi Ny‡i| myBP eÜ Kivi 4 wgwbU ci cvLvwU eÜ n‡q hvq| †_‡g hvevi Av‡M cvLvwU KZevi Nyi‡e? [RU. 16-17] A. 1500 B. 3000 C. 4500 D. 6000 i = 3000 rev/min = 3000 2 60 rads–1 ; _vgv‡bv n‡j, f = 0 †K․wYK miY, = f + i 2 t = 3000 2 2 60 4 60= 12000π rad Again, N~Y©b msL¨v, N = 2 = 12000 2 = 6000 evi| 03. Puv‡`i •iwLK †eM †KvbwU? [RU. 16-17] A. 2.044 kms–1 B. 1.55 kms–1 C.1.022 kms–1 D. 5 103 kms–1 Ans C 04. 50 MÖvg f‡ii GKwU e¯‧ cÖwZ †m‡K‡Û e„Ë c‡_ 3 evi Ny‡i, †K․wYK †eM KZ? [RU. 15-16] A. 2 B. 3 C. 4 D. 6 †K․wYK †eM, = 2 N t = 2 3 1 = 6 05. •iwLK MwZi †¶‡Î e‡ji †h f~wgKv †K․wYK MwZi †¶‡Î †KvbwUi †mB f~wgKv- [RU. 10-11] A. Ø›Ø B. †K․wYK fi‡eM C. †K․wYK Z¡iY D. UK© •iwLK MwZi †¶‡Î e‡ji †h f~wgKv, †K․wYK MwZi †¶‡Î U‡K©iI GKB f~wgKv| 06. GKwU KYv 4.5m e¨vmv‡a©i e„ËvKvi c‡_ cÖwZ 225 evi AveZ©b K‡i Gi •iwLK †eM KZ? [RU. 10-11] A. 106ms–1 B. 14.13.72ms–1 C. 1012.5ms–1 D. 50ms–1 v = r = 2n t ; r = 2 225 3.14 4.5 60 = 106ms–1 07. 500gm f‡ii GKwU e¯‧ 2m e¨vmv‡a©i e„ËvKvi c‡_ AveZ©b Ki‡Q| AveZ©b Kvj 10s n‡j e¯‧wUi †K․wYK fi‡eM [RU. 09-10] A. 1.115kgmm2 s –1 B. 1.125kgm2 s –1 C. 1.256kgm2 s –1 D. 1.325kgm2 s –1 L = I = mr 22 T = 0.5 (2)2 2 10 = 1.256 kgm2 s –1

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 175 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 04 ANALYSIS OF CU QUESTION 01. †K․wYK fi‡eM L Gi gvb †KvbwU? [CU. 15-16] A. L = I B. L = 1 2 mv2 C. L = I 2 D. L = I E. L = 1 2 I 2 Ans D 02. GKwU PvKvi e¨vm 1m| GwU wgwb‡U 30 evi Nyi‡j Gi cÖv‡¯Íi •iwLK †eM ms–1 G KZ n‡e? [CU. 15-16] A. 30 B. C. 2 D. 60 E. 4 e¨vmva©, r = 0.5m, v = ω × r = 2 N t × r = 2 30 60 × 0.5 = 0.5 03. GKwU •e`y¨wZK cvLv cÖwZ wgwb‡U 600 evi Ny‡i| myBP eÜ Kivi ci 1000 cvK Ny‡i cvLvwU eÜ n‡j _vg‡Z KZ mgq jvM‡e? [CU. 15-16] A. 100 sec B. 150 sec C. 200 secD. 240 sec E. 120 sec θ = 2π × 1000, ωi= 2 N t = 2 600 60 = 100 ; ωf = 0; t = 2 i + f = 2 2000 100 = 200sec 04. †K․wYK fi‡eM I †K․wYK †e‡Mi g‡a¨ m¤úK© †KvbwU? [CU. 13-14] A. L = I 2 B. L= v C. L = r D. L = I E. L= r Ans D 05. GKwU KYv e„ËvKvi c‡_ cÖwZ †m‡K‡Û 5 evi AveZ©b K‡i| rads1 -G KYvwUi †K․wYK †eM KZ? [CU. 12-13] A. 0.314 B. 3.14 C. 31.4 D. 5 E. 10 †K․wYK †eM, = 2 N t = 2 5 1 = 31.4 rads–1 06. GKwU KYv e„ËvKvi c‡_ cÖwZ †m‡K‡Û 5 evi AveZ©b K‡i| rads1 G KYvwUi †K․wYK †eM KZ? [CU. 12-13] A. 0.314 B. 3.14 C. 31.4 D. 5 E. 10 31.4 1 2 2 3.14 5 t n 07. GKwU nvZNwoi †m‡K‡Ûi KuvUvi •`N©¨ 1.5cm n‡j Gi cÖv‡šÍi •iwLK †eM KZ?[CU. 08-09] A. 0.147 cm/s B. 0.157 cm/sC. 0.167 cm/s D. 0.177 cm/s •iwLK †eM, v = r = 2 N t r = 2 1 60 1.5 = 0.157cms–1 STEP 05 ANALYSIS OF DU-7 Clg QUESTION 01. GKwU e¯‧ 2 m e¨vmv‡a©i e„ËvKvi c‡_ 4.0 ms1 mg`ªæwZ‡Z Nyi‡Q| GKevi Ny‡i Avm‡Z e¯‧wUi KZ mgq jvM‡e? [DU-7Clg.A.19-20] A. ( 2 /2)s B. 2 s C. (1/)s D. (2/ 2 )s s = e„‡Ëi cwiwa = 2r = 2.2 = 4 2 s = vt t = 4 2 4 = 2 sec. STEP 06 ANALYSIS OF GST QUESTION 01. GKwU KYv 2.0 m e¨mv‡a©i e„ËvKvi c‡_ cÖwZ wgwb‡U 30 evi AveZ©b K‡i, Gi •iwLK †eM KZ? [AG-C. 2020-21] A. ms–1 B. 2ms–1 C. 4ms–1 D. 0.5ms–1 S B info •iwLK †eM = †K․wYK †eM e„ËvKvi c‡_i e¨vmva© v = r = 2N t = 2 30 60 2 = 2ms–1 02. GKwU •e`y¨wZK cvLvi myBP ÔAbÕ Ki‡j `kevi c~Y© N~Y©‡bi ci cvLvwUi †K․wYK †eM 20 rad/s nq| †K․wYK Z¡iY KZ? [AG-C. 2020-21] A. 1.83 rad/s2 B. 8.13 rad/s2 C. 3.18 rad/s2 D. 5.17 rad/s2 S C info 2 = 0 2 + 2 = 2 – 0 2 2 = 202 – 0 2 2 10 2 = 3.18 rads–2 PART A Analysis of General University Question JnU 01. †K․wYK fi‡eM Gi †¶‡Î †KvbwU mwVK? [JnU. 15-16] A. L = r p B. L = p r C. L = r .p D. L = p .r L = r p 02. nvZNwoi wgwb‡Ui KuvUvi †K․wYK †eM (angular velocity) KZ? [JnU. 16-17] A. 1800 m/s B. 1800 rad/s C. 600 rad/s D. 1200 rad/s †K․wYK †eM, = 2N t = 21 3600 = 1800 rads–1 KU 01. GKwU Nwoi N›Uvi KvUv †K․wYK †eM KZ? [KU. 13-14] A. 1.4510–4 rad s–1 B. 1.3510–4 rad s–1 C. 1.5510–4 rad s–1 D. 1.2510–3 rad s–1 = 2 T = 2 3.14 12 60 60 = 1.454410–4 rads–1 BRUR 01. 10kg f‡ii 0.5m PµMwZi e¨vmva© wewkó GKwU N~Y©x PvKvi wµqviZ ؇›Øi †gv‡g›U 10N-m n‡j †K․wYK Z¡iY KZ? [BRUR. 11-12] A. 4ms–2 B. 8ms–2 C. 4rads–2 D. 8rads–2 = I ev, = I = 10 Mk2 = 10 10 (0.5) 2 = 4rads–2 PART B Analysis of Science & Technology Question SUST 01. GKwU avZe †Mvj‡Ki fi 6g| GwU‡K 3m `xN© GKwU myZvi GK cÖv‡šÍ †eu‡a cÖwZ †m‡K‡Û 4 evi Nyiv‡bv n‡”Q| Gi †K․wYK fi‡eM KZ kgm2 s 1 ? [SUST. 16-17] A. 0.36 B. 0.46 C. 0.56 D. 1.36 L = I= Mr2 2N t = 610–3 3 2 2 4 1 = 1.36 kgm2 s –1 02. 1g f‡ii GKwU ey‡jU 1000 m/s †e‡M 1kg f‡ii GKwU Kv‡Vi UzKiv‡K AvNvZ Kij| ey‡jU we× Kv‡Vi UzKivwUi MwZ‡eM KZ n‡e? [SUST. 10-11] A. 0.999m/s B. 1.03 m/s C. 1.27 m/s D. 3.0m/s MV = mv ev, 1V = 1000. 1 1000 ev, V = 1 = .999m/s 03. •iwLK †eM v, †K․wYK †eM Ges Ae¯
vb †f±i r n‡j wb‡Pi †KvbwU mwVK?[SUST. 06-07] A. = r v B. v = r C. v = r D. r = v Ans B 04. GKwU e¯‧ 10m e¨vmv‡a©i GKwU e„Ë GKevi Nyi‡Z 10s mgq wb‡j e¯‧wUi †K․wYK †eM KZ n‡e? [SUST. 06-07] A. 1m/s B. 100m/s C. 0.628rad/s D. 6.28rad/s v = 2r t = 2 10 10 = 2 Avevi, v = r = v r = 2 10 = 5 = 0.628rad/s JUST 01. †Kv‡bv e¯‧i •iwLK MwZi Rb¨ †KvbwU mwVK? [JUST-C, 19-20] A. F = ma B. F 1 = – F 2 C. Aw`K ¸Yb D. †f±i ¸Yb Ans A 02. GKwU MÖv‡gv‡dvb †iKW© cÖwZ wgwb‡U 45 evi Ny‡i| Gi †K›`ª †_‡K 9 cm `~‡i †Kvb we›`yi `ªæwZ KZ? [JUST. 15-16] A. 0.22 ms-1 B. 0.32 ms-1 C. 0.42 ms-1 D. 0.52 ms–1 E. 0.62 ms–1 v = r = 2n t r = 2 45 0.09 60 = 0.427ms–1

176 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES MBSTU 01. `„p e¯‧i †ÿ‡ÎÑ [MBSTU-A, Set-2 19-20] A. L B. L 2 C. L I 2 D. L L = I L 02. e¨vmva© †f±i ( r ) •iwLK fi‡eM (p ) I †K․wYK fi‡eM (L )Gi g‡a¨ m¤úK© †KvbwU? [MBSTU. 14-15] A. L = 2P r B. L = r p C. p = r L D. p = L r Ans B 03. N~Y©biZ ‡Kv‡bv e¯‧ KYvi e¨vmva© ‡f±i I •iwLK fi‡e‡Mi ‡f±i ¸Ydj‡K Kx e‡j? [MBSTU. 14-15] A. UK© B. †K․wYK fi‡eM C. RoZvi åvgK D. PµMwZi e¨vmva© Ans B BSMRSTU 01. GKwU e¯‧ 0.5m e¨vmv‡a©i e„ËvKvi c‡_ cÖwZ wgwb‡U 60 evi AveZ©b K‡i| Gi •iwLK †eM KZ? [BSMRSTU-B, 19-20] A. 2 ms–1 B. 0.5 ms–1 C. 4 ms–1 D. ms–1 •iwLK †eM, v = r = 2n t r = 2 60 60 0.5 = 02. mvK©vm †Ljvq GKwU evBK 20 m/s †e‡M GKwU e„ËvKvi c‡_ Nyi‡Q| e„ËvKvi c‡_i e¨vmva© 200 m n‡j, evBKwUi †K․wYK †eM KZ wQj? [BSMRSTU-B, 19-20] A. 0.001 rad/s B. 0.01 rad/s C. 1 rad/s D. 0.1 rad/s •iwLK †eM, v = r = v r = 20 200 = 0.1 rad/s NSTU 01. GKwU KYv 4m e¨vmv‡a©i e„ËvKvi c‡_ cÖwZ †m‡K‡Û 1 evi AveZ©b K‡i, Gi •iwLK †eM KZ? [NSTU-A, 19-20] A. 2 ms1 B. 8 ms1 C. ms1 D. 4 ms1 V = r = 2 T × r = 2 × 4 = 8 ms–1 DU Technology 01. GKwU N~Y©biZ KYvi e¨vmva© †f±i r = (2 i + 2 j – k) m Ges cÖhy³ ej F = (6 i + 3 j – 3 k) N n‡j, U‡K©i gvb KZ? [DU-Tech. 2020-21] A. 36 B. 45 C. 49 D. 42 S B info = r F = i j k 2 2 –1 6 3 –3 = i (6 + 3) j(6 + 6) + k(6 – 12) = – 3 i – 6 k | | = (–3) 2 + (6) 2 = 45 02. GKwU KYv e„ËvKvi c‡_ cÖwZ wgwb‡U 180 evi AveZ©b Ki‡j KYvwUi K¤úv¼ KZ n‡e? [DU-Tech. 2020-21] A. 1 Hz B. 3 Hz C. 60 Hz D. 180 Hz S B info KYvwUi K¤úv¼ , f = 180 60 = 3 Hz STEP 07 ANALYSIS OF ENGINEERING & BUTex QUESTION CKRuet. Combind 01. e¨vmva© †f±i r = 2 i + 3 j + 2 k Ges ej †f±i F = 2 i + 3 j + 2 k n‡j e‡ji åvgK wbY©q Ki| [CKRUET. 2020-21] A. 2 i – 2 k B. 0 C. 2 i – 3 j – 2 k D. 2 i + 2 k E. None of them S A info = r × F = i 2 2 j 3 2 k 2 2 = i (6 – 4) – j (4 – 4) + k(4 – 6) = 2 i – 2 k RUET 01. 20 evi †Nvivi ci GKwU ‣e`y¨wZK cvLvi †K․wYK †eM 30rad/s n‡Z n«vm †c‡q 10rad/sec nq| †K․wYK g›`b n‡e- [RUET. 11-12] A. 3.18rad/sec2 B. 2.5rad/sec2 C. 2rad/sec2 D. None of them = 2 1 – 2 2 2 = (30) 2 – (10) 2 2 2 20 = 3.183rad/sec2 STEP 08 ANALYSIS OF MEDICAL & DENTAL QUESTION DAT 01. GKwU PvKvi e¨vm 1 wgUvi| GwU cÖwZ wgwb‡U 30 evi Nyi‡j Gi cÖv‡šÍi •iwLK †eM ms–1 G KZ n‡e? [DAT; 19-20] A. B. 2 C. 60 D. 30 •iwLK †eM, v = r = 2Nr t = 2Nd 2t = 2301 260 = 2 STEP 09 ANALYSIS OF HSC BOARD QUESTION 01. e„Ëxq MwZi †ÿ‡Î †K․wYK fi‡e‡Mi ivwk †KvbwU? [Xv. †ev. 2016] A. mr B. mr2 C. mr 2 D. m2 r Ans B 02. 0.01m ‣`‡N©¨i GKwU Nwoi wgwb‡Ui KvuUvi cÖvšÍxq we›`yi •iwLK †e‡Mi gvb KZ?[iv.‡ev. 2019] A. 1.5410-5ms-1 B. 1.6410-5ms-1 C. 1.7410-5ms-1 D. 1.8410-5ms-1 S C info V = r = 2 3600 × 0.01 = 1.74 × 10–5 ms–1 03. wb‡Pi †Kvb m¤úK©wU mwVK? [iv. †ev. 2016] A. L = r F B. L = F r C. L = r P D. L = P r Ans C 04. GKwU Nwoi †m‡KÛ, wgwbU I N›Uvi KuvUvi †K․wYK †e‡Mi AbycvZ- [iv. †ev. 2016] A. 720 t 60 t 1 B. 1 t 60 t 720 C. 1 t 12 t 720 D. 720 t 12 t 1 S D info = 2 T s : min : h = 2 60 : 2 3600 : 2 43200 = 720 : 12 : 1 05. e‡ji åvgK Gi mgxKiY- [P. †ev. 2019; Xv. †ev. 2015] A. = r F B. = I C. = d l dt D. me¸‡jv mwVK Ans D 06. UK© ( ) Gi Rb¨Ñ [P.‡ev. 2019] i. = I ii. = r F iii. = d L dt wb‡Pi †KvbwU mwVKÑ A. i I ii B. i I iii C. ii I iii D. i,iiIiii Ans D 07. Nwoi wgwb‡Ui KuvUvi •`N©¨ Kg‡j [P. †ev. 2017] A. •iwLK †eM ev‡o, †K․wYK †eM ev‡o B. •iwLK †eM K‡g, †K․wYK †eM K‡g C. •iwLK †eM w¯’i _v‡K, †K․wYK †eM ev‡o D. ‣iwLK †eM K‡g, †K․wYK †eM w¯’i _v‡K Ans D 08. c„w_exi N~Y©b bv _vK‡j c„w_exc„‡ôi †Kv‡bv ¯
v‡b e¯‧i IRb [P. †ev. 2017] A. e„w× cv‡e B. k~b¨ n‡e C. Amxg n‡e D. AcwiewZ©Z _vK‡e Ans A 09. WvBwfs-Gi jvd †`qvi mgq mvZviæiÑ [P. †ev. 2016] A. RoZvi åvgK aªyeK B. †K․wYK fi‡eM aªye C. †K․wYK Z¡iY aªye D. †K․wYK †eM aªye Ans B

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 177 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 10. Nwoi NÈvi KuvUvi †K․wYK †eM KZ? [P. †ev. 2016] A. /30 rad/s B. /30 rad/min C. /360 rad/min D. /720 rad/min Ans C 11. c„w_exi wbR A‡ÿ N~Y©‡bi Rb¨ ÒAvB‡dj UvIqv‡iiÓ †K․wYK †eM n‡e [w`. †ev. 2017] A. 1.99107 rad s1 B. 7.26105 deg s1 C. 4.167103 deg s1 D. 4.167103 rad s1 S B info = 2 T = 2 24 × 3000 rads–1 = 7.26 × 10–5 rads–1 12. e„ËvKvi c‡_ mgvb mg‡q mgvb †K․wYK `~iZ¡ AwZµgKvix †Kv‡bv KYvi •iwLK †e‡Mi [w`. †ev. 2017] A. ïay gv‡bi cwieZ©b n‡e B. aªyeK n‡e Ans C C. ïay w`‡Ki cwieZ©b n‡e D. gvb I w`K DfqB cwieZ©b n‡e 13. GKwU •e`y¨wZK cvLvi fi 10kg Ges †Kv‡bv Aÿ mv‡c‡ÿ PµMwZi e¨vmva© 2m| cvLvwU‡Z 3rads–2 †K․wYK Z¡iY m„wó Ki‡Z KZ gv‡bi UK© cÖ‡qvM Ki‡Z n‡e? [wm.‡ev 2021] A. 40Nm B. 20Nm C. 30Nm D. 120Nm S D info = I = mr 2 = 10 2 2 3 = 120Nm 14. GKwU †`qvj Nwoi wgwb‡Ui KuvUvi •`N©¨ 18cm. Gi cÖv‡šÍi •iwLK †eM KZ? [wm. †ev. 2016] A. 1.88104ms1 B. 3.14104ms1 C. 9.67103ms1 D. 0.58ms1 Ans B 15. hLb †Kv‡bv KYvi Ici cÖhy³ UK© k~b¨ ZLb wb‡Pi †Kvb ivwkwU aªæeK nq? [e. †ev. 2016] A. ej B. †K․wYK fi‡eM C. •iwLK fi‡eM D. e‡ji NvZ Ans B 16. wb‡Pi †KvbwU aªyeK n‡j †Kv‡bv KYvi Dci cÖhy³ UK© k~b¨ n‡e?[e. †ev. 2016] A. ej B. e‡ji NvZ C. •iwLK fi‡eM D. †K․wYK fi‡eM Ans D Concept 10 †KŠwYK MwZkw³ I †gvU kw³ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. Ek= 1 2 I 2 = L 2 2I, = 2N t = 2 T 02. evjwZ n‡Z co‡e bv, v = gr = g r 03. cvwb fwZ© evjwZi N~Y©b msL¨v (cvwb hLb bv c‡o), N = t 2 g r 04. A¶ mv‡c‡¶ Mwo‡q Pjv PvKv, wmwjÛvi ev †Mvj‡Ki †gvUkw³ i. Mwo‡q Pjv †MvjvKvi e¯‘i †gvU kw³ = mv2 ii. Mwo‡q Pjv †Mvj‡Ki †¶‡Î †gvU kw³ = 7mv2 10 iii. Mwo‡q Pjv wmwjÛvi/‡KvY‡Ki †gvU kw³ = 3 4 mv 2 iv. Lvov Ae¯’vq lm ‣`‡N©¨i `Û KZ n‡q c‡o †M‡j Zvi †K․wYK †eM = 3g L CONCEPTUAL MATH MEx 01 GKwU wmwjÛv‡ii fi 50kg Ges e¨vmva© 0.20m wmwjÛviwU A‡¶i mv‡c‡¶ Gi RoZvi åvgK 1.0kgm2 wmwjÛviwU hLb 2ms–1 †e‡M Avbyf~wgKfv‡e Mov‡Z _v‡K ZLb Zvi †gvU kw³ KZ n‡e? Ek = 3 4 mv2 = 3 4 50 2 2 = 150j MEx 02 2kg f‡ii GKwU cv_i‡K 12m `xN© GKwU m~Zvi mvnv‡h¨ †eu‡a Avbyf‚wgK Z‡j Nyiv‡bv n‡”Q| myZvwU m‡e©v”P 19.6N Uvb mn¨ Ki‡Z cv‡i| myZv bv wQ‡o cv_iwU‡K m‡e©v”P KZ `ªæwZ‡Z Nyiv‡bv †h‡Z cv‡i? N~Y©b †eM, v = gr = 9.8 12 = 10.84ms–1 MEx 03 GKwU evjwZ‡Z wKQz cvwb Av‡Q| evjwZwUi 2.45m e¨vmv‡a©i GKwU Dj¤^ e„ËvKvi c‡_ Nyiv‡bv n‡”Q wKš‧ cvwb co‡Q bv| Zvn‡j evjwZwUi †K․wYK †eM KZ? evjwZwUi †K․wYK †eM, = g r = 9.8 2.45 = 2rad/sec MEx 04 GKwU cvwb fwZ© evjwZ‡K 3.95m j¤^v `wo Øviv †eu‡a Dj¤^ Z‡j e„ËvKv‡i †Nviv‡bv n‡”Q| cÖwZ wgwb‡U KZevi †Nviv‡j evjwZ †_‡K †Kvb cvwb co‡e bv| evjwZi N~Y©b msL¨v, N = t 2 g r = 60 2 9.8 3.95 = 15 evi NOW START PRACTICE 01. 40kg fi I 0.20m e¨vmv‡a©i GKwU †ejbvKvi e¯‧ 2ms–1 †hv‡M Avbyf~wgK Z‡j Mov‡Z _v‡K e¯‧wUi †gvU MwZkw³ KZ? 02. 400kg fi I 1m e¨vmv‡a©i GKwU wb‡iU †MvjK 2ms–1 †e‡M Mwo‡q Pj‡Q| Gi †gvU MwZ kw³ KZ? 03. GKwU evjwZ‡Z wKQz cvwb Av‡Q| evjwZwUi 3.2m e¨vmv‡a©i GKwU Dj¤^ e„ËvKvi c‡_ Nyiv‡bv n‡”Q wKš‧ cvwb co‡Q bv| Zvn‡j evjwZwUi †K․wYK †eM KZ? 04. 2kg f‡ii GKwU cv_i‡K 10m `xN© GKwU m~Zvi mvnv‡h¨ †eu‡a Avbyf‚wgK Z‡j Nyiv‡bv n‡”Q| myZvwU m‡e©v”P 19.6N Uvb mn¨ Ki‡Z cv‡i| myZv bv wQ‡o cv_iwU‡K m‡e©v”P KZ `ªæwZ‡Z Nyiv‡bv †h‡Z cv‡i? NOW PRACTICE SOLVE : 01. †gvU MwZkw³ = 3 4 mv2 = 3 4 40 (2)2 = 120J 02. †gvU kw³, Ek = 7 10 mv 2 (wb‡iU †Mvj‡Ki RoZvi åvgK, I = 2 5 mv 2 = 7 10 400 (2)2 = 1120J 03. evjwZwUi †K․wYK †eM, = g r = 9.8 3.2 = 1.75 rad/sec 04. N~Y©b †eM, v = gr = 9.8 10 = 9.8ms–1

178 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. †KvbwU U‡K©i GKK- [DU. 09-10] A. Dyne/cm B. N-m C. N/m D. N/m.s Ans B 02. 24400J MwZkw³ wewkó GKwU PvKv cÖwZ wgwb‡U 602 evi †Nv‡i| PvKvwUi N~Y©b RoZv åvgK n‡e- [DU. 06-07] A. 40.5kg m2 B. 12.3 kg m 2 C. 10 kg m2 D. 406.7 kg m2 Ek = 1 2 I 2 ev I = 2Ek 2 = 2 24400 2 602 3.14 60 2 = 12.3 STEP 02 ANALYSIS OF JU QUESTION 01. GKwU Mvwoi PvKv 30 min G 2000 evi Ny‡i 10km c_ AwZµg K‡i| PvKvi cwiwa wbY©q Ki? [JU-A, Set-C. 20-21] A. 20 m B. 15 m C. 10m D. 5 m S D info = N t = 2000 30 60 = 1.11 rad/s–1 v = r = d t r = d t = 10000 1800 1.11 = 5.05 5 [mg‡e‡Mi †ÿ‡Î d = vt] 02. L 2 2I represents.................. of a particle [JU. 12-13] A. Rotational kinetic energy B. Potential energy C. torque D. Power L 2 2I = (I) 2 2I = 1 2 I 2 = Rotational kinetic energy STEP 03 ANALYSIS OF RU QUESTION 01. GKwU wmwjÛv‡ii fi 50kg Ges e¨vmva© 0.2m. wmwjÛviwUi A‡¶i mv‡c‡¶ Gi RoZvi åvgK 1kgm2 . wmwjÛviwUi hLb 4ms–1 †e‡M Avbyf~wgK fv‡e Mov‡Z _v‡K, ZLb Gi †gvU MwZkw³ n‡e? [RU. 12-13] A.150J B. 300J C. 450J D. 600J †gvU MwZkw³, E = 3 4 mv 2 = 3 4 50 (4)2 = 600J STEP 04 ANALYSIS OF CU QUESTION 01. GKwU Nwoi †m‡K‡Ûi KuvUvi †K․wYK †eM KZ? [CU-A, Set-4. 20-21] A. rad s–1 B. /3 rad s–1 C. /2 rad s–1 D. /30 rad s–1 S D info Nwoi †m‡K‡Ði KuvUvi ch©vqKvj 60s| myZivs †K․wYK †eM = 2 60 = 30 rads–1 02. †KvbwU N~Y©vqgvb e¯‧i MwZkw³? [CoU. A, 19-20, 15-16] A. K.E = 1 2 M 2 B. K.E = 1 2 I 2 C. K.E = 1 2 I 2 D. K.E = 1 2 M 2 N~Y©vqgvb e¯‘i MwZkw³, K.E = 1 2 I 2 = L 2 2I STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. GKwU wmwjÛv‡ii fi 50kg Ges e¨vmva© 0.2m| wmwjÛviwUi A‡¶i mv‡c‡¶ Gi RoZvi åvgK 1.0 kg-m 2 | wmwjÛviwU hLb 2m/s †e‡M Abyf~wgKfv‡e Mov‡Z _v‡K ZLb Zvi †gvU MwZkw³ KZ n‡e? [KU. 15-16] A. 150J B. 100J C. 200J D.180J †gvU MwZkw³ = 3 4 mv2 = 3 4 50 (2)2 = 150J PART B Analysis of Science & Technology Question JUST 01. N~Y©vqgvb e¯‧i MwZkw³ k, hw` = 1 nq Z‡e e¯‧i RoZvi åvgK n‡e- [JUST-B, 19-20] A. 2k B. 1 2 k C. k D. 4k K = 1 2 I 2 I = 2k 2 = 2K. BSMRSTU 01. 10 kg f‡ii cošÍ e¯‧i Z¡iY KZ, hLb evZv‡mi evav 78 N? [BSMRSTU-H, 19-20] A. 2.2 ms2 B. 2.5 ms2 C. 3.02 ms2 D. 22 ms 2 e¯‘i IRb, W = mg = 10 98 = 98 N evZv‡mi evav Fk = 78 N a = W – Fk m = 98 – 78 10 = 2ms–2 mwVK DËi Ack‡b †bB| wKš‧ KvQvKvwQ Ackb 2.2 ms–2 STEP 06 ANALYSIS OF MEDICAL & DENTAL QUESTION DAT 01. wb‡Pi e¯‧mg~‡ni g‡a¨ †KvbwUi MwZkw³ †ewk? [DAT; 19-20] A. fi 2M Ges †eM 3V B. fi 3M Ges †eM 2V C. fi M Ges †eM 4V D. fi 3M Ges †eM V MwZkw³, Ek = 1 2 MV2 , myZivs, GLv‡b- Ek = 1 2 2 (3)2 = 9 J Ek = 1 2 3 (2)2 = 6 J Ek = 1 2 1 (4)2 = 8 J Ek = 1 2 3 (1)2 = 1.5 J STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. Wvj fv½vi hvZvK‡j [Xv. †ev. 2016 i. Aÿ msjMœ KYvi †K․wYK †eM me‡P‡q †ewk ii. wKbv‡ii KYvi •iwLK †eM †ewk iii. cÖwZwU KYvi †Kv‡bv gyn~‡Z©i †K․wYK fi‡eM mgvb wb‡Pi †KvbwU mwVK? A. ii B. ii I iii C. i I iii D. i, ii I iii Ans A 02. 1 rps = ? [Xv. †ev. 2015] A. 2 rad s1 B. rad s1 C. 2 rad s1 D. 4 rad s1 Ans C

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 179 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 11 e¨vswKs msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. iv¯Ívi e¨vswKs, tan = v 2 rg 02. †ij jvB‡bi †ÿ‡Î e¨vswKs, tan = v 2 rg = h x ; [hLb LyeB †QvU] (GLv‡b, h = wfZ‡ii cvZ A‡cÿv evB‡ii cv‡Zi D”PZv, x= cvZ؇qi ga¨eZ©x `~iZ¡) CONCEPTUAL MATH MEx 01 GKwU iv¯Ívq 40m e¨vmv‡a©i euvK i‡q‡Q Ges iv¯ÍvwU 30 †Kv‡Y Xvjy| GKwU Mvwo m‡Ÿ©v”P KZ †e‡M euvK AwZµg Ki‡Z cv‡i? tan = v 2 rg v = rg tan v = 40 9.8 tan30 = 15.04ms–1 NOW START PRACTICE 01. GKwU †ijˇq Uªv‡Ki fi‡K›`ª †ijjvB‡bi †j‡ fj n‡Z 0.80m Dc‡i i‡q‡Q| †ijjvBb `ywUi `~iZ¡ 1.2m UªvKwU 50m e¨vmv‡a©i e¨vswKsnxb eµc‡_ m‡ev”P© KZ `yªwZ‡Z wbivc‡` evuK wb‡Z cvi‡e02. 75m e¨vmv‡a©i e„ËvKvi c‡_ †Kvb †gvUi mvB‡Kj Av‡ivnx KZ †e‡M Nyi‡j Dj¤^ Z‡ji mv‡_ 30° †Kv‡Y AvbZ _vK‡e wbY©q Ki? NOW PRACTICE SOLVE : 01. v 2 rg = h x v = hrg x = 0.8 50 9.8 1.2 = 18.07ms–1 02. tan = v 2 rg , v = tanrg = tan30 75 9.8 = 20.60ms–1 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF JU QUESTION 01. GKwU Mvwoi wbivc‡` euvK †bqvi kZ© n‡jv- [JU. 16-17] A. v (tanrg) 2 1 B. v (tanrg) C. v > (tanrg) D. v > (tanrg) 2 1 tan = v 2 rg ; v = (tanrg) 2 1 or, v (tan rg) 2 1 02. 100m e¨v‡mi e„ËvKvi c‡_ †Kvb gv‡Vi mvB‡Kj Av‡ivnx KZ †e‡M Nyi‡j Dj¤^ Z‡ji mv‡_ 450 †Kv‡Y AvbZ _vK‡eb? [JU. 11-12] A. 11.1359ms–1 B. 22.1359ms–1 C. 16.1359ms–1 D. 33.1359ms–1 v = tanrg = tan45 50 9.8 = 22.1359ms–1 03. †Kvb †gvUi mvB‡Kj Av‡ivnx r e¨vmv‡a©i e„ËvKvi c‡_ KZ †e‡M Nyi‡j wZwb Dj¤^ Z‡ji mv‡_ †Kv‡Y AvbZ _v‡Kb? [JU. 09-10] A. rg tan B. rg tan C. tan–1 rg D. rg (tan g) Ans B STEP 02 ANALYSIS OF RU QUESTION 01. hw` v = `ªæwZ, r = e¨vmva©, g = AwfKl©R Z¡iY nq, Zvn‡j wb‡¤œi †Kvb ivwkwU gvÎvnxb? [RU. 17-18] A. v 2 r g B. v 2 g r C. v2 gr D. v 2 rg v 2 rg = [LT –1 ] [L][LT–2 ] = [L 2 T –2 ] [L 2 T –2 ] = gvÎvnxb 02. 50m e¨vmv‡a©i iv¯Ívi euv‡K 9.4 ms–1 †e‡M GKwU mvB‡Kj Pvjv‡bvi mgq Av‡ivnxi bwZ †KvY KZ? [RU. 17-18] A. 1.1˚ B. 11˚ C. 88˚ D. 90˚ iv¯Ívi AvbwZ †KvY, = tan–1 v 2 rg = tan–1 9.4 2 50 9.8 = 11˚ 03. GKwU iv¯Ív 40m e¨vmv‡a©i evK wb‡q‡Q| H ¯
v‡b iv¯ÍvwU 4m PIov Ges Dnvi †fZ‡ii wKbviv n‡Z evwn‡ii wKbviv 0.8m DPzu| m‡e©v”P KZ †e‡M H ¯
v‡b †_‡K †bqv m¤¢e? [RU. 09-10] A.8.96ms–1 B. 9.92ms–1 C. 8.82ms–1 D. 8.97ms–1 v 2 rg = h x ev, v = rgh x = 40 9.8 0.8 4 = 8.85ms–1 STEP 03 ANALYSIS OF CU QUESTION 01. iv¯Ívi e¨vswKs wbf©i K‡i- [CU. 07-08] A. Mvwoi `ªæwZ I f‡ii Dci B. Mvwoi fi I iv¯Ívi euv‡Ki Dci C. ïay gvÎ Mvwoi f‡ii Dci| D. Mvwoi `ªæwZ I iv¯Ívi euv‡Ki e¨vmv‡a©i Dci Mvwoi `ªæwZ I iv¯Ívi euv‡Ki e¨vmv‡a©i Dci; tan v 2 rg STEP 04 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. †Kvb mvB‡Kj Av‡ivnx 100m e¨vmv‡a©i e„ËvKvi c‡_ 20m/s †e‡M Nyi‡Z †M‡j Dj¤^ Z‡ji mv‡_ KZ †Kv‡Y AvbZ _vK‡Z n‡e? [JnU. 08-09] A. 50° B. 48.2° C. 20.2° D. 22.2° tan = v 2 rg ev, tan = (20) 2 100 9.8 ev, = 22.2° KU 01. GKwU iv¯Ív 60m e¨vmva© euvK wb‡q‡Q| H ¯
v‡b iv¯ÍvwU 6m PIov Ges Zvi wfZ‡ii wKbviv n‡Z evB‡ii wKbviv 0.6m DPuy| m‡e©v”P KZ †e‡M H ¯
v‡b wbivc‡` evK †bqv hv‡e? [KU. 12-13] A. 9.2ms–1 B. 8.6ms–1 C. 5.5ms–1 D . 7.67ms–1 v 2 rg = h x ev, v = rgh x = 60 9.8 0.6 6 = 7.668ms–1 x h

180 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. GKwU iv¯Ív 50m e¨vmv‡a© euvK wb‡q‡Q| H ¯
v‡b iv¯ÍvwU 5m PIov Ges Gi †fZ‡ii wKbviv n‡Z evB‡ii wKbviv 0.5m DPuy| m‡e©v”P KZ †e‡M H ¯
v‡b wbivc` euvK †bqv m¤¢e? [KU. 10-11] A. 8 B. 7 C. 10 D. 9 v 2 rg = h x ev,= v 2 60 9.8 = 0.1 ev, v = 7ms–1 PART B Analysis of Science & Technology Question SUST 01. 150 m e¨vmva© wewkó GKwU euvKv c‡_ 70 km.h–1 †e‡M GKwU Mvwo wbivc‡` Pvjv‡Z n‡j c_wU‡K KZ wWwMÖ †Kv‡Y AvbZ ivL‡Z n‡e? [SUST-A, 19-20] A. 14.42 B. 10.42 C. 12.42 D. 8.42 E. 11.42 r = 150mv = 70 kmh1 = 70 1000 3600 ms1 = 19.44 ms1 tan = v 2 rg = tan1 v 2 rg = tan1 19.442 150 9.8 = 14.42 02. GKwU wgUvi‡MR †Uªb 200m e¨vmv‡a©i †ij jvB‡bi evu‡K Nyi‡Q| N›Uvq 50.4kms–1 †e‡M PjšÍ Mvwoi †¶‡Î `ywU †ijjvB‡bi D”PZvi cv_©K¨ KZ m ivL‡Z n‡e? [SUST. 15-16] A. 0.001 B. 0.01 C. 0.1 D. 0.15 E. 0.20 h = 0.1x = 0.1 1 ev, h = 0.1m wgUvi‡MR jvB‡b w¯úZ ؇qi ga¨eZ©x `~iZ¡ = 1m 03. Zzwg 50m e¨vmv‡a©i GKwU e„ËvKvi c‡_ N›Uvq 25 km †e‡M GKwU mvB‡Kj Pvjv‡j mvB‡KjwU Dj†¤^i mv‡_ KZUzKz †n‡j _vK‡e? [SUST. 12-13] A.3.81° B. 4.61° C. 5.61° D. 6.51° E. 6.81° tan = v 2 rg = (6.9444) 2 50 9.8 v = 25 3.6 = 6.9444 = 5.61 JUST 01. 200 m e¨vmva©wewkó GKwU euvKv c‡_ 50.4 kmh–1 †e‡M GKwU Mvwo Pj‡Q| iv¯ÍvwU 2m cÖ¯
 n‡j, evB‡ii cvk^© †fZ‡ii cvk^© A‡cÿv KZ m DuPz n‡Z n‡e? [JUST-A, Set-Ka 19-20] A. 0.1 B. 0.2 C. 0.3 D. 0.4 v = 50.4 kmh–1 = 50.4 × 1000 3600 = 14 ms–1 h x = v 2 rg h = v 2 x rg = 142 × 2 200×9.8 = 0.2m BSMRSTU 01. 50m e¨vmv‡a©i iv¯Ívi euv‡K 9.8 m/s †e‡M GKwU mvB‡Kj Pvjv‡bvi mgq Av‡ivnxi bwZ †KvY KZ? [PUST-A, 19-20] A. 1.1 B. 11 C. 88 D. 89 tan = v 2 rg = (9.8) 2 50 9.8 = 0.196 = 11 02. 75m e¨vmv‡a©i e„ËvKvi c‡_ †Kvb gUi mvB‡Kj Av‡ivnx KZ †e‡M Nyi‡j Dj¤^ Z‡ji mv‡_ 30 †Kv‡Y AvbZ _vK‡e? [BSMRSTU-B, 19-20] A. 19.5 ms–1 B. 21.9 ms–1 C. 20.6 ms–1 D. 29.3 ms–1 tan = v 2 rg v = tanrg = tan30 75 9.8 = 1 3 75 9.8 = 20.599 = 20.6 ms–1 STEP 05 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. GKRb mvB‡Kj Av‡ivnx N›Uvq 24 km †e‡M 30 m e¨vmv‡a©i GKwU e„ËvKvi c‡_ †gvo wb‡”Q| Zvu‡K Dj‡¤^i mv‡_ KZ †Kv‡Y †n‡j _vK‡Z n‡e? [KUET. 14-15] A. 836 B. 756 C. 856 D. 92 E. 841 tan = v 2 rg ; = tan–1 v 2 rg = tan–1 (6.67) 2 30 9.8 = 8.60o = 836 02. GKRb mvB‡Kj Av‡ivnx 20 †m‡K‡Û 70m e¨vmv‡a©i GKwU e„ËvKvi c‡_ †gvo wb‡”Q| Zv‡K Dj‡¤^i mv‡_ KZ †e‡M †n‡j _vK‡Z n‡e? [KUET. 12-13] A. 30 3 B. 35 12 © C. 36 22 © D. 35 45 © E. 35 v = s t = 2r t ; = tan–1 v 2 rg = tan–1 4 2 r 2 t 2 rg = tan–1 4 2 70 202 9.8 = 3512 STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. GKRb mvB‡Kj Av‡ivnx 500m e¨vmv‡a©i e„ËvKvi c‡_ 5ms–1 †e‡M Nyi‡Z †M‡j Dj¤^ Z‡ji mv‡_ KZ †Kv‡Y AvbZ _vK‡e? [iv.‡ev 2021] A. 0.27 B. 0.28 C. 0.29 D. 0.30 S C info = tan–1 v 2 rg = tan–1 5 2 500×9.8 = 0.29 wb‡Pi DÏxc‡Ki Av‡jv‡K 02 I 03bs cÖ‡kœi DËi `vI : GKwU iv¯Ívi euv‡Ki e¨vmva© 50m| iv¯Ívi cÖ¯’ 5m Ges evwn‡ii cÖvšÍ wfZ‡ii cÖvšÍ A‡cÿv 0.25m DuPz| 02. iv¯ÍvwUi cÖK…Z e¨vswKs †KvY KZ? [iv.‡ev 2021] A. 1.86 B. 2.86 C. 3.86 D. 5.86 S B info tan = sin = h n = sin–1 0.25 5 = 2.86 x h 03. iv¯Ívi e¨vswKs wbf©i K‡i- [iv.‡ev 2021] i. Mvwoi f‡ii Dci ii. euv‡Ki eµZvi Dci iii. Mvwoi †e‡Mi Dci wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans C wb‡Pi DÏxcKwUi Av‡jv‡K 04 I 05 bs cÖ‡kœi DËi `vI : GKwU c‡_i A I B `ywU ¯’v‡b h_vµ‡g 25m I 36m e¨vmv‡a©i euv‡Ki cÖ‡Z¨KwUi e¨vswKs †KvY 10(c_wUi cÖ¯’ 80cm) [P. †ev. 2015] 04. A ¯
v‡bi ev‡K wfZ‡ii cvk¦© n‡Z evB‡ii cvk¦© KZ DuPz n‡e? A. 2.17 cm B. 2.17 m C. 13.89 cm D. 13.89 m S C info tan = sin 0 [ ÿz`ªZg gv‡bi Rb¨] sin = h d h = dsin = 80 sin10˚ = 13.89 cm 05. euvK `ywU‡Z †Kv‡bv Mvwoi m‡e©v”P MwZ‡e‡Mi AbycvZ KZ? A. 5 : 6 B. 6 : 5 C. 25 : 36 D. 36:25 S A info v 2 1 r1 = v 2 2 r2 v 2 1 : v 2 2 = r1 : r2 v1 : v2 = r1 : r2 = 25 : 36 v1 : v2 = 5 : 6 wb‡Pi DÏxc‡Ki Av‡jv‡K 06 I 07bs cÖ‡kœi DËi `vI : A I B `ywU ¯’v‡b h_vµ‡g 25m I 36m e¨vmv‡a©i euv‡Ki iv¯Ívi cÖ‡Z¨KwUi e¨vswKs †KvY 2.5| (Dfq iv¯Ívi cÖ¯’ 5m)| 06. A ¯
v‡bi euv‡Ki iv¯Ívi †fZ‡ii cvk¦© A‡cÿv evB‡ii cvk¦© KZ DuPz n‡e? [w`.‡ev 2021] A. 0.109m B. 0.218m C. 2.498m D. 4.995m S B info tan = sin = h n h = x sin = 5 × sin 2.5 = 0.218 m x h 07. euv‡Ki iv¯Ív `ywU‡Z †Kv‡bv Mvwoi m‡e©v”P MwZ‡e‡Mi AbycvZ- [w`.‡ev 2021] A. 5 : 6 B. 16 : 15 C. 25: 48 D. 36: 65 S A info tan = v 2 rg v 2 r v r VA : VB = 5 : 6 08. iv¯Ívi euv‡Ki Xvj w`‡j- [wm.‡ev 2021] i. hvbevnb PjvP‡j AwaKZi wbivc` nq ii. †K›`ªgyLx ej cvIqv hvq iii. mvB‡Kj Av‡ivnx eµc‡_i †K‡›`ªi w`‡K †n‡j _v‡K wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans D

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 181 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES SAQ Short Answer Questions WRITTEN PART BAQ Broad Answer Questions 01. 2.0kg f‡ii GKwU e¯‧ w¯
i Ae¯
vq _vKv Av‡iKwU e¯‧i mv‡_ w¯
wZ¯
vcK msNl© NUv‡jv Ges msN‡l©i ci cÖ_g e¯‧wU Zvi Avw`‡e‡Mi GK-PZz_©vsk †eM wb‡q GKB w`‡K Pj‡Z _vKj| AvNvZcÖvß e¯‧wUi fi KZ? Solve m1 = 2kg, u2 = 0ms–1 ; v1 = u1 4 v1 = m1 – m2 m1 + m2 u1 1 4 = 2 – m2 2 + m2 ; m2 = 1.2 kg 02. GKwU 10N ej 2kg fi wewkó GKwU w¯
i e¯‧i Dci wµqv K‡i| hw` 5 †m‡KÛ ci e‡ji wµqv eÜ n‡q hvq Z‡e cÖ_g n‡Z 12 †m‡K‡Û e¯‧wU KZ `~iZ¡ AwZµg Ki‡e? Solve F = ma a = F m = 10 2 =5ms–2 s1 = 1 2 at2 = 1 2 5 5 2 = 125 2 m v = at = 5 5 = 25ms–1 s2 = vt = 25(12 – 5) = 175m s = s1 + s2 = 62.5 + 175 = 237.5m 03. 1000kg f‡ii GKwU Mvwoi PvKv I iv¯Ívi g‡a¨ w¯
wZ Nl©‡Yi mnM ev ¸Yv¼ s = 0.8 n‡j MvwowU m‡ev©”P KZ Xvjy iv¯Ívq wcQwj‡q bv c‡o †_‡g _vK‡Z cvi‡e? Solve g‡b Kwi, Xvjy iv¯ÍvwU Avbyf~wg‡Ki mv‡_ †Kv‡Y _vK‡j MvwowU wcQwj‡q bv c‡o †_‡K _vK‡e| s = Fs R = mg sin mg cos s = tan 0.8 = tan = 3839 mgsin mg mgcos Fs 04. 70kg f‡ii ev·‡K 500N Abyf‚wg‡Ki e‡j †g‡Si Dci w`‡q Uvbv n‡”Q| ev·wU hLb P‡j ZLb ev· I †g‡Si ga¨eZ©x Nl©Y mnM 0.50| ev‡·i Z¡iY wbY©q Ki| Solve GLv‡b, m = 70kg F = 500N a = ? Awfjw¤^K cÖwZwµqv, R = ev‡·i IRb = 70 9.8 = 686N Avgiv Rvwb, F fk = ma Avevi, k = fk R fk = k R = 0.5 686N = 343 N a = F – fk m = 500 – 343 70 = 2.24ms–2 05. GKwU 2.0 kg f‡ii e¯‧ 6.0 m/s `ªæwZ‡Z MwZkxj wQj| e¯‧wUi MwZi Kx gv‡bi Mo ej cÖ‡qvM Kiv n‡j, e¯‧wU 7.0 104 s Gi g‡a¨ †_‡g hv‡e? Solve hw` wb‡Y©q e‡ji Mo gvb F nq, Z‡e e¯‘i e‡ji NvZ, J = e¯‘i fi‡e‡Mi cwieZ©b, P ev, F t = mvf mvi ev, F (7.0 104 s) = 0 (2.0 kg) (6.0 m/s) ev, F = 12 kg. m/s 7.0 104 s = 1.71 104 kg. m/s = 1.71 104 N 1.71 104 N 06. GKwU Zvgvi †Mvj‡Ki fi 0.05 kg| GwU‡K 2m `xN© GKwU myZvi GK cÖv‡šÍ †eu‡a cÖwZ †m‡K‡Û 5 evi Nyiv‡bv n‡”Q| †MvjKwUi †K․wYK fi‡eM KZ? Solve Avgiv Rvwb, †K․wYK fi‡eM, L = I ................. (i) Avevi, I = Mr2 = (0.05 kg (2m)2 = 0.05 4 kg. m2 = 0.2 kg. m2 = 2 5 rad/sec = 31 .42 rad/sec GLv‡b, M = fvi = 0.05 kg r = N~Y©b-Aÿ n‡Z `~iZ¡ = 2m = †K․wYK †eM I = RoZvi åvgK =? (i) bs n‡Z cvB, L = 0.2 31.42 kg.m2 /sec 6.284 kg.m2 /sec 07. 3ms1 †e‡M 2kg f‡ii GKwU e¯‧ 0.5 kg f‡ii Ab¨ GKwU w¯
i e¯‧i m‡½ †mvRvmywR w¯
wZ¯
vcK msN‡l© wjß nq| msN‡l©i ci wØZxq e¯‧i †eM KZ n‡e? Solve Avgiv Rvwb, v2f = 2m1 m1 + m2 v1i + m2 m1 m1 + m2 v2i = 2 2 kg 2 kg + 0.5 kg 3ms1 + 0.5 kg 2 kg 2 kg + 0.5 kg 0 = 4.8 ms1 4.8 ms1 GLv‡b, GLv‡b e¯‘i fi, m1 = 2kg cÖ_g e¯‘i Avw`‡eM, v1i= 3 ms 1 wØZxq e¯Íyi fi, m2 = 0.5 kg wØZxq e¯‘i Avw`‡eM, v2i = 0 wØZxq e¯‘i †kl †eM, v2f = ? 08. 10kg f‡ii GKwU e¯‧i Dci KZ ej cÖ‡qvM Ki‡j e¯‧ Lvov (i) Dc‡ii w`‡K 1.2ms–2 (ii) wb‡Pi w`‡K 2.8ms–2 Z¡i‡Y MwZkxj n‡e| Solve (i) GLv‡b, m = 10kg a = 1.2ms–2 , g = 9.8ms–2 Avgiv Rvwb F – mg = ma F = m(g + a) F = 10(9.8 + 1.2) F = 10 11 = 110N (ii) GLv‡b, m = 10kg a = 2.8 ms–2 wb‡P w`‡K Avgiv Rvwb, a Z¡i‡Y wb¤œMvgx Ki‡Z mg – F = ma F = m(g – a) = 10(9.8 – 2.8) = 10 7 = 70N 09. 15000kg R¡vjvbxmn GKwU i‡K‡Ui fi 20000kg| i‡K‡Ui mv‡c‡ÿ 3000ms–1 `ªæwZ‡Z R¡vjvwb 200kgs–1 nv‡i cy‡o| i‡KUwU hw` Lvov Dc‡ii w`‡K MwZkxj n‡q _v‡K Z‡e, (i) i‡K‡Ui Dc‡ii w`‡K av°v (ii) wb‡ÿ‡ci mgq i‡K‡Ui Dci cÖhy³ jwä ej (iii) R¡vjvwb †kl nIqvi mgq m„ó cÖhy³ jwä ej (iv) R¡vjvwb †kl nIqvi gyn~‡Z© i‡K‡Ui †eM wbY©q Ki| Solve (i) i‡K‡Ui Dc‡ii w`‡K av°v F = Vr dm dt = 3000 200 = 6 105N (ii) wb‡ÿ‡ci mgq i‡K‡Ui Dci cÖhy³ jwä ej = EaŸ©gyLx av°vÑi‡K‡Ui IRb = Vr dm dt – mg 6 105 – 20000 9.8 = 4.04 105N (iii) R¡vjvwb †kl nIqvi gyn~‡Z© jwäej = Vr dm dt – mg m = i‡K‡Ui †gvU fi Ñ R¡vjvwbi fi = 2000 – 15000 = 5000kg jwä ej = Vr dm dt – mg = 6 105 – 5000 9.8 = 5.51 105N (iv) i‡K‡Ui †eM v = ? GLv‡b, mg¯Í R¡vjvwb cyo‡Z mgq jv‡M t = 15000 200 = 75s i‡K‡Ui †eM V = V0 + Vr ln m0 m – gt = 0 + 3000ln 20000 5000 – 9.8 75 = 3424ms–1

182 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 10. fvi‡K›`ªMvgx Ges Z‡ji mwnZ j¤^ eivei Aÿ mv‡c‡ÿ GKwU AvqZvKvi cv‡Zi RoZvi åvgK 5kgm2 | cvZwUi cÖ¯
 1m Ges fi 12kg n‡j •`N¨© KZ? Solve GLv‡b, RoZvi åvgK I = 5kgm2 ; cÖ¯’, b = 1m; M = 12kg Avgiv Rvwb, fvi‡K›`ªMvgx I Z‡ji mwnZ j¤^ eivei Aÿ mv‡c‡ÿ AvqZvKvi cv‡Zi RoZvi åvgK, I = M 12(l 2 + b2 ) 5 = 12 12 (l 2 + 12 ) 5 = l 2 + 1 l = 2m (Ans.) 11. GKwU •e`y¨wZK cvLv wgwb‡U 1200 evi Ny‡i| myBP eÜ Kivi 3 wgwbU ci cvLvwU eÜ n‡q †Mj cvLvwUi †K․wYK Z¡iY KZ? †_‡g hvIqvi Av‡M cvLvwU KZevi Nyi‡e? Solve GLv‡b, 0 = 1200 rev min–1 = 1200 2 60 rads–1 = 40 rad s–1 t = 3 min = 180s = ? ‡K․wYK miY, – 0 = ? Avgiv Rvwb, f = 0 + t 0 = 40 + + 180 = – 0.7rads –2 Avevi, = 0 + 0 + f 2 t – 0 = 40 + 0 2 180 = 3600 rad = 3600 2 rev = 1800 rev 12. †ev‡ii nvB‡Wªv‡Rb cigvYy g‡W‡j GKwU B‡jKUªb GKwU †cÖvU‡bi Pviw`‡K 5.2 –11m e¨vmv‡a©i GKwU e„ËvKvi c‡_ 2.18 106ms–1 †e‡M cÖ`wÿY K‡i| B‡jKUª‡bi fi 9.1 10–31kg n‡j †K›`ªgyLx e‡ji gvb KZ? Solve GLv‡b, r = 5.2 10–11m = 2.18 106 ms–1 m = 9.1 10–31 kg Avgiv Rvwb, †K›`ªgyLx ej F = mv2 r = 9.110–31(2.18106 ) 2 5.2 = 8.32 10–8N 13. GKwU iv¯Ív 60m e¨vmv‡a© euvK wb‡q‡Q| H ¯
v‡b iv¯ÍvwU 6m PIov Ges Gi wfZ‡ii wKbviv n‡Z evB‡ii wKbviv 0.6m DPzu| m‡ev©”P KZ †e‡M H ¯
v‡b wbivc` euvK †bIqv m¤¢e? Solve h x = sin = tan = v 2 rg v 2 rg = h x v = hrg x = 0.6 60 9.8 6 ms–1 = 7.66ms–1 14. 9.1 10–31 kg fi wewkó GKwU B‡jKUªb hw` wbDwK¬qvm‡K †K›`ª K‡i 0.53 10–10m e¨vmv‡a©i Kÿc‡_ Nyi‡Z _v‡K, Z‡e Zvi †K․wYK †eM †ei Ki| [cøv‡¼i aªæeK = 6.63 10–34Js] Solve L = nh 2 I = nh 2 = nh 2I = h 2mr 2 = 6.6310–34 29.110–31(0.5310–10) 2 = 4.1 1016 rads–1 [r = 0.53 10–10m Gi Rb¨ n = 1 aiv hvq] 15. gnvwe‡k¦i †h PviwU †g․wjK ej Av‡Q †m¸‡jv wjL| [RU-C, Set-1. 19-20] Solve gnvwe‡k^i PviwU †g․wjK ej: 1. gnvKl© ej, 2. Zwor-Pz¤^Kxq ej, 3. mej wbDwK¬q ej, 4. `ye©j wbDwK¬q ej| 16. NÈvq 60 gvBj †e‡M Pjgvb GKwU Mvwoi BwÄb nVvr eÜ n‡q hvq| MvwowU _vg‡Z 5 wgwbU mgq jvM‡j Gi Z¡iY KZ? [JnU-A, Set-2. 19-20] Solve GLv‡b, u = 60 22 15 = 88ft/s; a = v – u t = –88 5 60 = –88 3 102 = –29.33 10–2 = –0.2933 ft/s2 17. 4000 kg f‡ii GKwU wjdU 240 kg f‡ii GKwU ev· enb Ki‡Q| hLb wjd‡Ui Zv‡ii (Supporting cable) Dci EaŸ©gyLx Uvb 48000 N nq ZLb EaŸ©gyLx Z¡iY KZ? wjdUwU w¯
i Ae¯
vb †_‡K 3s mg‡q KZ D”PZvq DV‡e? [BUTEX; 19-20] Solve T –mg = ma 48000 – (4000 + 240) 9.8 = (4000 + 240) a a = 1.52075 ms–2 h = 1 2 at2 [u = 0 ms–1 ] = 1 2 (1.52075) 3 2 = 6.843375m mg T [T > mg] 18. 50 ms1 mg‡e‡M PjšÍ 2500 kg f‡ii GKwU Mvwo g›`‡bi d‡j 2500 m `~iZ¡ AwZµg Kivi ci †_‡g †Mj| MvwowU _vgv‡bvi Rb¨ cÖ`Ë ej Ges _vgvi mgq wbY©q Ki| [DU.A 19-20] Solve †`Iqv Av‡Q, u = 50 ms1 m = 2500 kg ; s = 2500 m ; v = 0 v 2 = u 2 + 2as ; v2 u 2 = 2as a = v 2 u 2 2s = 0 2 502 2 2500 ms2 = 0.5 ms2 a = 0.5 ms2 g›`b a = 0.5 ms2 cÖ`Ë ej, F = ma = (2500 0.5) N = 1250 N ; v = u + at ; v u = at t = v u a = 0 50 0.5 s = 100 s 19. mgvb •`N©¨ I r = 0.5cm e¨vmv‡a©i `ywU B¯úvZ Zv‡ii mvnv‡h¨ 45 kg f‡ii GKwU UªvwdK jvBU Szjv‡bv Av‡Q| hw` Zvi `ywU Abyf~wg‡Ki mv‡_ 15o †KvY •Zwi K‡i, Zvn‡j UªvwdK jvB‡Ui IR‡bi Rb¨ Zvi `ywUi •`N©¨ weK…wZi cwigvY KZ n‡e? †`Iqv Av‡Q, B¯úv‡Zi Bqs-Gi ¸Yv¼ 2 1011Nm–2 | [BUET; 19-20] Solve ; jvwgi Dccv`¨ Abymv‡i, W sin 150o = T1 sin(90o + 15o ) = T2 sin(90o +15o ) T2 T1 15o 15o W Solve T1 = W sin 150o cos 15o = 851.9465 N; T2 = T1 = 851.9465 N Avgiv Rvwb, Y = T/A l/L l L = T/A Y = 851.9465 (510–3 ) 2 21011 l L = 5.4427 10–5 20. 150 g f‡ii GKwU wµ‡KU ej 12m/s †e‡M MwZkxj n‡q GKwU e¨vU Øviv AvNvZ Kivi d‡j ejwU 20 m/s †e‡M wd‡i Av‡m| ejwUi Dci wµqviZ e‡ji AvNv‡Zi mgqKvj 0.01 s, ejwUi Dci e¨v‡Ui Mo ej wbY©q Ki| [BUET; 19-20] Solve Avgiv Rvwb, m (v – u) = Ft F = m(v–u) t = 0.15 (12 + 20) 0.01 N = 480 N

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 183 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 21. GKwU •e`y¨wZK cvLv wgwb‡U 1500 evi Ny‡i| myBP eÜ Kivi 4 wgwbU ci cvLvwU eÜ n‡q hvq| cvLvwUi †K․wYK Z¡iY KZ? †_‡g hvevi Av‡M cvLvwU KZevi Nyi‡e, Zv wbY©q Ki? 5+5=10 [KUET; 19-20] Solve 0 = 2N t = 2 1500 60 = 50 rad s–1 = 0 rads–1 , t = 4min = 240 sec = – 0 t = 0 – 50 240 = – 5 24 rads–2 Avevi, †_‡g hvIqvi c~‡e© Nyb©b msL¨v N n‡j, N = 2 = 1 2 0 + 2 t = 1 2 50 + 0 2 240 = 3000 rev †_‡g hvevi c~‡e© cvLvwU 3000 evi Nyi‡e| 22. 0.56 kg fi wewkó GKwU wgUvi †¯‥‡ji 30 cm wPwýZ `v‡Mi j¤^ A‡ÿi mv‡c‡ÿ wgUvi †¯‥jwUi N~Y©b RoZv wbiƒcY Ki| †¯‥jwU‡K cvZjv iW wn‡m‡e we‡ePbv Ki| [BUET; 19-20] Solve RoZvi åvg‡Ki mgvšÍivj Aÿ Dccv`¨ n‡Z cvB, N~Y©b RoZv = IG + Mh2 = MI2 12 + Mh2 = 0.561 2 12 + 0.56 (0.3)2 kg – m 2 = 0.097 kg – m 2 23. nvB‡Wªv‡Rb cigvYyi B‡jKUªb wbDwK¬qvm‡K †K›`ª K‡i 5.2 10–11 e¨vmv‡a©i e„ËvKvi Kÿc‡_ 2.20 106 ms–1 †e‡M Nyi‡Q| B‡jKUª‡bi †K›`ªgyLx Z¡iY KZ? [NSTU-A, 19-20] Solve †K›`ªgyLx Z¡iY, a = v 2 r = (2.20 106 ) 2 5.2 10–11 = 9.30 1022 ms–2 STEP 3 mKj cvV¨eB‡qi NCTB QUESTIONS ANALYSIS cÖkœ e¨vLvmn mgvavb cÖkœ 01. me‡P‡q kw³kvjx ej n‡”Q- [Bm&nvK, Avwgi] A. gnvKl© ej B. Zwor Pz¤^Kxq ej C. mej wbDwK¬q ej D. `ye©j wbDwK¬q ej Ans C 02. wPÎwU wbDU‡bi †Kvb m~Î cÖKvk K‡i? A B R F A. 1g m~Î B. 2q m~Î C. 3q m~Î D. 2q I 3q m~Î Ans C 03. wbDU‡bi MwZi m~Î †_‡K Avgiv Rvb‡Z cvwi- [Bm&nvK, Avwgi] i. e¯‘i Ici ej wµv Ki‡j e¯‘wU Z¡iY wb‡q Pj‡Z _v‡K ii. e¯‘i Z¡iY cÖhy³ e‡ji e‡M©i mgvbycvwZK iii. e‡ji AwfgyLB Z¡i‡Yi AwfgyL wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans B 04. fi‡e‡Mi †¶‡Î wb‡Pi mwVK weeiY- [Bm&nvK, Avwgi] (i) fi‡eM = fimiY/†eM (ii) fi‡e‡Mi GKwU †f±i ivwk (iii) fi‡e‡Mi gvÎv mgxKiY [MLT1 ] wb‡Pi †KvbwU mwVK? A. i B. i I ii C. ii I iii D. i,ii I iii Ans C 05. wbDU‡bi MwZi wØZxq m~‡Î ma KF, GLv‡b- [Bm&nvK, Avwgi] i. K GKwU mgvbycvwZK aªæeK ii. K Gi gvb ivwk¸wji GK‡Ki Dci wbf©i K‡i iii. K-Gi gvb S.I c×wZ‡Z 1 wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i,ii I iii Ans D 06. GKwU e¯‧ mvg¨ve¯
vq _vK‡e ev w¯
i _vK‡e hw` Gi- [Bm&nvK, Avwgi] i. †eM k~b¨ nq ii. miY k~b¨ nq iii. Z¡iY k~b¨ nq wb‡Pi †KvbwU mwVK? A. i B. i I ii C. iii D. i, ii I iii Ans B 07. N~Y©biZ †Kv‡bv e¯‧KYvi e¨vmva© †f±i I •iwLK fi‡e‡Mi †f±i ¸Ydj‡K Kx e‡j? [Bm&nvK, Avwgi] A. UK© B. †K․wYK fi‡eM C. RoZvi åvgK D. PµMwZi e¨vmva© Ans B 08. wP‡Î AB e¯‧wU O-†K †K›`ª K‡i XY A‡ÿi PZzw`©‡K Nyi‡Z cv‡i| wb‡Pi †Kvb DËwU fzj? [Bm&nvK, Avwgi] A. d F B. d F C. r F D. a ˆrFsin d F ˆ dFsin Or, r F ˆ rFsin 09. †RU BwÄb ev i‡K‡Ui Kvh©bxwZi wfwË n‡jv- [Bm&nvK, Avwgi] A. f‡ii wbZ¨Zv m~Î B. kw³i wbZ¨Zv m~Î C. •iwLK fi‡e‡Mi wbZ¨Zv m~Î D. †K․wYK fi‡e‡Mi wbZ¨Zv m~Î Ans C 10. GKwU †gwkbMvb †_‡K cÖwZ †m‡K‡Û n msL¨K ey‡jU wbM©Z nq| cÖwZwU ey‡j‡Ui fi m kg Ges Gi †eM v ms–1 n‡j †gwkbMv‡bi Ici cÖhy³ ej (N GK‡K) n‡jv- [Bm&nvK, Avwgi] A. mnv B. mn v C. mn D. mv n Ans A 11. 73 m DuPz n‡Z GKwU ej †d‡j †`qv n‡jv| wb‡P evwji g‡a¨ 1 m Xz‡K †_‡g †Mj| G‡ÿ‡Î- [Bm&nvK, Avwgi] A. †KejgvÎ fi‡eM msiwÿZ _vK‡e B. †KejgvÎ MwZkw³ msiwÿZ _vK‡e C. fi‡eM I MwZkw³ DfqB msiwÿZ _vK‡e D. fi‡eM I MwZkw³ †Kv‡bvUvB msiwÿZ _vK‡e bv Ans A 12. mylg MwZ‡eM PjšÍ GKwU e¯‧i Ici- [Bm&nvK, Avwgi] A. GKwU †bU ej wµqv K‡i B. GKwU †bU k~b¨ ej wµqv K‡i C. GKwU mylg Z¡iY wµqv K‡i D. Dc‡ii †Kv‡bvwUB bq Ans B wb‡Pi DÏxcKwUi Av‡jv‡K 13 I 14 bs cÖ‡kœi DËi `vI: GKwU cv_i A I B `ywU ¯
v‡b h_vµ‡g 25 m I 36 m e¨vmv‡a©i euv‡Ki cÖ‡Z¨KwUi e¨vswKs †KvY 10° (cv_iwUi cÖ¯
 80 cm)| [Bm&nvK, Avwgi] 13. A ¯
v‡bi euv‡K †fZ‡ii cvk¦© n‡Z evB‡ii cvk¦© KZ DuPz n‡e? A. 2.17 cm B. 2.17 m C. 13.89 cm D. 13.89 m S C info tan = sin = d h h = dsin = 0.8 × sin10° = 0.1389 m = 13.89 cm 14. euvK `ywU‡Z †Kv‡bv Mvwoi m‡e©v”P MwZ‡e‡Mi AbycvZ KZ? [Bm&nvK, Avwgi] A. 5 : 6 B. 6 : 5 C. 25 : 36 D. 36 : 25 S A info tan = v 2 rg v 2 r v r v1 : v2 = 5 : 6

184 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 15. †KvbwU †K․wYK fi‡e‡Mi msiÿY m~Î? [Bm&nvK, Avwgi] A. L = aªæeK B. P = aªæeK C. = aªæeK D. F = aªæeK Ans A wb‡Pi DwÏcK c‡o 16 I 17 bs cÖ‡kœi DËi `vI: c„w_ex †_‡K Puv‡`i `~iZ¡ 3.84 × 105 km Ges Puv` c„w_ex‡K e„ËvKvi Kÿc‡_ 27.3 w`‡b GKevi cÖ`wÿY K‡i| 16. Puv‡`i †K․wYK `ªæwZ- [Bm&nvK, Avwgi] A. 2.85 rad s–1 B. 2.66 × 10–6 rad s–1 C. 2.66 × 105 rad s–1 D. 2.85 × 10–6 rad s–1 Ans B 17. Puv‡`i •iwLK †eM- [Bm&nvK, Avwgi] A. 2.044 km s–1 B. 1.55 km s–1 C. 1.022 km s–1 D. 5 × 103 km s–1 Ans C 18. †K›`ªgywL ej- [Bm&nvK, Avwgi] i. GKwU Kvh©nxb ej ii. GwU e¯‘i MwZc‡_i w`‡K wµqv K‡i iii. Gi gvb mv2 r wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans B 19. GKwU Mvwoi wbivc‡` euvK †bIqvi kZ© n‡jv- [Bm&nvK, Avwgi] A. (tan rg) 1 2 B. (tan rg) C. > (tan rg) D. > (tan rg) 1 2 Ans A 20. •iwLK Z¡iY I †K․wYK Z¡i‡Yi m¤úK©- [Bm&nvK, Avwgi] A. a = r B. a = r C. a = r2 D. a = r Ans D 21. AB A‡ÿi mv‡c‡ÿ PvKwZi RoZvi åvgK KZ? [Bm&nvK, Avwgi] A. 1 2 Mr2 B. 1 4 Mr2 C. Mr2 D. 3 2 Mr2 Ans B 22. GKwU wjdU a Z¡i‡Y wb‡P bvg‡Q| m f‡ii GKRb e¨w³ wjd‡Ui †g‡S‡Z KZ ej cÖ‡qvM Ki‡e? [Bm&nvK, Avwgi] A. ma – mg B. ma C. m (g – a) D. m (g + a) Ans C 23. wb‡Pi †Kvb †jLwPÎwU MwZkxj KYvi MwZkw³ I fi‡eM P-Gi cwieZ©b wb‡`©k K‡i? [Bm&nvK, Avwgi] A. Ek 1 P B. Ek 1 P C. Ek 1 P D. Ek 1 P Ans D 24. GK e¨w³ †cø‡bi g‡a¨ GKwU w¯úªs Zzjv h‡š¿i mvnv‡h¨ GKwU e¯‧i IRb cwigvc Ki‡Qb| A¨v‡iv‡cøbwU w¯
i Ae¯
v †_‡K Lvov Ic‡ii w`‡K µgvMZ DV‡Z _vK‡j w¯úªs Zzjv h‡š¿i cv‡Vi Kxiƒc cwieZ©b n‡e? [Bm&nvK, Avwgi] A. µgk e„w× cv‡e B. µgk n«vm cv‡e C. cÖ_‡g e„w× cv‡e Ges Zvici n«vm cv‡e D. AcwiewZ©Z _vK‡e Ans C 25. †Kv‡bv e¯‧i RoZvi åvgK wbf©i K‡i Gi- [Zcb, AvwRR] A. fi Ges N~Y©b A‡ÿi Dci B. AvqZ‡bi Dci C. †K․wYK †e‡Mi Dci D. †K․wYK fi‡e‡Mi Dci Ans A 26. †Kvb `„p e¯‧i PµMwZi e¨vmva© †KvbwU? [Zcb, AvwRR] A. K= 1 M B. K= M 1 C. K = 1 M D. K = M 1 Ans C 27. GKwU PvKvi RoZvi åvgK 2kgm2 | PvKvwU wgwb‡U 30 evi Nyi‡Q| Gi †K․wYK fi‡eM KZ? [Zcb, AvwRR] A. B. 2 C. 3 D. 4 2 60 2 30 2 t 2 n L I I 28. 2 kg f‡ii GKwU e¯‧‡K 3 m `xN© GKwU myZvi GK cÖv‡šÍ †eu‡a 4 rads–1 †K․wYK †e‡M Nyiv‡bv n‡”Q| myZvi Dci Uvb n‡e- [Zcb, AvwRR] A. 50 N B. 48 N C. 100 N D. 96 N S D info T = mw2 r = 2 × 42 × 3 = 96 N 29. 2kg f‡ii GKwU e¯‧ 4m `xN© GKwU myZvi GK cÖv‡šÍ †eu‡a 2ms-1 †e‡M Nyiv‡bv n‡”Q| myZvi Dci Uvb n‡e- [Zcb, AvwRR] A. 10 N B. 20 N C. 5 N D. 4 N F = ma 5N 2 2 2 5 2s v m 2 2 30. †KvbwU RoZvi åvgK msµvšÍ mgvšÍivj Aÿ Dccv`¨? [Zcb, AvwRR] A. Iz = Ix + Iy B. I = Ig + MK2 C. I = Ig + MK D. I = Ig + Mh2 Ans D 31. w¯
i Ae¯
vq _vKv GKwU e¯‧ we‡ùvwiZ n‡q m1 I m2 f‡ii `yBwU e¯‧‡Z cwiYZ n‡q h_vµ‡g v1 I v2 †e‡M wecixZ w`‡K Pjgvb| v1 v2 Gi AbycvZ KZ? [Zcb, AvwRR, Xv. we. 2019-20] A. m1 m2 B. m2 m1 C. – m2 m1 D. m2 m1 S B info m1v1 – m2v2 = 0 [∵ wecixZ w`‡K] m1v1 = m2v2 v1 v2 = m2 m1 32. wµqv I cÖwZwµqv ej- [Zdv¾j, gwnDwÏb] (i) GKwU e¯‘ Dci wµqv K‡i (ii) `ywU wfbœ e¯‘i Dci wµqv K‡i (iii) ci¯ú‡ii wecixZgyLx (iv) G‡`i †hvMdj k~b¨ wb‡Pi †KvbwU mwVK? A. i B. i I ii C. i I iii D. ii,iii I iv Ans D 33. RoZv wb‡¤œi †Kvb wel‡qi Dci wbf©i K‡i? [Zdv¾j, gwnDwÏb] A. e¯‘i AvqZb B. e¯‘i †eM C. e¯‘i fi D. c„ô Uvb Ans C 34. W IR‡bi GKwU †QvU ej‡K GKwU nvjKv myZv Øviv Szjv‡bv n‡jv| hLb F e‡j Aybyf~wgK fv‡e cÖevwnZ evZvm ejwUi Dci wµqv Kij ZLb ejwU wPÎ Abyhvqx Dj‡¤^i mv‡_ †KvY K‡i Ae¯
vb Kij| ● F W †Kvb mgxKiYwU , F I W Gi g‡a¨ m¤úK© wb‡`©k K‡i? A. cos = F W B. sin = F W C. tan = F W D. tan = W F Ans C 35. wPÎ Abyhvqx GKwU eøK GKwU AvbZ Z‡ji Dci w¯
i Ae¯
v‡b Av‡Q| [Zdv¾j, gwnDwÏb] R f W wb‡Pi †Kvb †Rvov ï× DËi? A. f = W cos ; f = W sin B. R = W; f = R C. R= W cos ; f = W sin D. W sin = W cos ; R = W cos e¯‘i IRb w †K `ywU Dcvs‡k wefvwRZ Kiv hvq| w cos cÖwZwµqv ej R Gi wecixZ w`‡K Ges Wsin Nl©b ej f –Gi wecixZ w`‡K wµqvkxj| e¯‘wU w¯’i Ae¯’v‡b Av‡Q| wsin = f Ges wcos = R

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 185 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 36. e‡ji wµqv eÜ n‡q †M‡j e¯‧wUi- [Zdv¾j, gwnDwÏb] A. mylg Z¡i‡Y P‡j B. †eM n«vm cvq C. mylg †e‡M P‡j D. †eM k~b¨ nq Ans C wb‡Pi wPÎ j¶¨ Ki Ges 37 I 38 bs cÖ‡kœi DËi `vI: [†gvK‡Q` m¨vi] 5kg f‡ii e¯‘ 5ms–1 mgy`ªywZ‡Z 5m e¨vmv‡a©i e„Ë c‡_ Nyi‡Q| ● ● ● ● ● r = 5 m V= 5ms –1 c~e© DËi cwðg `wÿY `w¶Y 37. †K›`ªgyLx e‡ji gvb KZ n‡e? [†gvK‡Q` m¨vi] A. 20 N B. 25 N C. 30 N D. 35 N F = mv2 r = 5(5) 2 5 = 25 N 38. †K›`ªgyLx e‡ji w`K †Kvb w`‡K? A. DËi B. `w¶Y C. c~e© D. cwðg Ans A 39. †h e‡ji Kvi‡Y e¯‧ e„ËvKvi c‡_ Ny‡i †mB ej hw` cÖZ¨vnvi K‡i †bqv nq Zvn‡j e¯‧wU MwZ RoZvi Kvi‡Y- [†gvK‡Q` m¨vi] (i) H gyn~‡Z© †e‡Mi w`K eivei Qz‡U hv‡e (ii) e„Ëc‡_i ¯úk©K eivei Qz‡U hv‡e (iii) e¨vmva© eivei †K‡›`ªi w`‡K Qz‡U hv‡e wb‡Pi †KvbwU mwVK? A. i B. i I ii C. ii I iii D. iii` Ans B 40. hLb †Kvb KYvi Dci cÖhy³ UK© k~b¨ nq ZLb wb‡Pi †Kvb ivwkwU aªæeK nq? [†gvK‡Q` m¨vi] A. ej B. †K․wYK fi‡eM C. •iwLK fi‡eM D. e‡ji NvZ Ans B 41. `yÕwU mgvb f‡ii e¯‧i g‡a¨ w¯
wZ¯
vcK msNl© NU‡j G‡`i †eM- [†gvK‡Q` m¨vi] (i) GKB _vK‡e (ii) cwiewZ©Z n‡e (iii) wewbgq Ki‡e wb‡Pi †KvbwU mwVK? A. i B. ii C. iii D. ii I iii e¯‘ `ywUi fi mgvb n‡j A_©vr m1 = m2 n‡j V1f = V2i Ges V2f = V1i| AZGe mgvb f‡ii `ywU e¯‘i g‡a¨ msNl© NU‡j GKwU e¯‘ AciwUi †eM cÖvß n‡e| A_©vr e¯‘Øq †eM wewbgq K‡i| 42. wbDU‡bi MwZi Z…Zxq m~Îvbymv‡i wµqv I cÖwZwµqvi ga¨Kvi †Kv‡Yi cwigvY KZ? [cÖvgvwYK m¨vi] A. 0° B. 90° C. 180° D. 360° Ans C 43. `ywU GKB f‡ii gv‡e©j mij‡iLv eivei GKB w`‡K Pjgvb Ae¯
vq msN‡l© wjß n‡j G‡`i †eM- [igv m¨vi] i. GKB _vK‡e ii. cwiewZ©Z n‡e iii. wewbgq Ki‡e wb‡Pi †KvbwU mwVK? A. i B. ii C. iii D. ii I iii Ans C 44. GKRb G¨v_‡jU js Rv¤ú †`qvi c~‡e© wKQzUv c_ †`Šovq- Gi KviY- [igv m¨vi] A. w¯’wZRoZv e„w×i Rb¨ B. MwZRoZv e„w×i Rb¨ C. wµqvej e„w×i Rb¨ D. cÖwZwµqv ej e„w×i Rb¨ kixi‡K MwZ RoZvi cÖfv‡e †i‡L †ekx AMÖmi nevi Rb¨ GKRb G¨v_‡jU js-Rv¤ú †`Iqvi Av‡M wKQzUv c_ †`․ovq| 45. 4N ej GKwU e¯‧i Dci 1 sec e¨vcx wµqv Ki‡j fi‡e‡Mi cwieZ©b KZ?[BKivg m¨vi] A. 2kg ms–1 B. 4kg ms–1 C. 8kg ms–1 D. 16kg ms–1 P = F t = 4 1 = 4kg ms–1 46. ؇›Øi (Couple) †¶‡Î mZ¨- [BKivg m¨vi] i. `ywU mggv‡bi wKš‘ wecixZgyLx ej †Kv‡bv `„p e¯‘i Dci `ywU wfbœ we›`y‡Z wµqv K‡i| ii. Ø›Ø GKwU e¯‘‡K N~Y©vqgvb Ki‡Z cv‡i iii. Ø›Ø cvwbi U¨vc Lyj‡Z mvnvh¨ Ki‡Z cv‡i wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans D 47. MvQ †_‡K 2.0 kg f‡ii GKwU KvuvVvj †mvRv wb‡Pi w`‡K covi mgq hw` 5.5 m/s2 Z¡iY nq, Zvn‡j evZv‡mi evav KZ wbDUb? [kvgmyi m¨vi] A. 4.3N B. 6.8 N C. 8.6 N D. 5.5N F = m(g – a) = 2 (9.8 – 5.5) = 8.6N 48. 0.150kg f‡ii GKwU cv_‡ii LÛ‡K 0.75m j¤^v GKwU myZvi GKcÖv‡šÍ †eu‡a e„ËvKvi c‡_ cÖwZ wgwb‡U 90 evi Nyiv‡j myZvi Dci Uvb wbY©q Ki| [kvgmyi m¨vi] A. 9.99N B. 9.90N C. 9.95N D. 9.98N F = m 2 r = 0.15 4 2 902 602 0.75 = 9.99N 49. 20 evi Nyievi ci GKwU •e`y¨wZK cvLvi †K․wYK †eM 30rad/sec n‡Z n«vm †c‡q 10 rad/sec nq| †K․wYK g›`b n‡e- [kvgmyi m¨vi] A. 3.18 rad/sec2 B. 2.5 rad/sec2 C. 6.36 rad/sec2 D. 5 rad/sec2 = 2 – o 2 2 = 302 – 102 2 2 20 = 3.18 rads–2 STEP 4 MCQ CONCEPT TEST WRITTEN cÖkœ 01. mgvb f‡ii `ywU e¯‧i g‡a¨ w¯
wZ¯
vcK msNl© n‡j wb‡Pi †KvbwU mwZ¨? (1g e¯‧i Avw` I †kl †eM u1 I v1 Ges 2q e¯‧i Avw` I †kl †eM u2 I v2) A. u1 = v2 B. u1 = v1 C. u1 = u2 D. u2 = v2 02. GKwU PvKvi fi 6 kg Ges †Kv‡bv Aÿ mv‡c‡ÿ PµMwZi e¨vmva© 30 cm| PvKwU‡Z 3 rad s–2 Z¡iY m„wó Ki‡Z KZ gv‡bi UK© cÖ‡qvM Ki‡Z n‡e? A. 1.62 Nm B. 1.8 Nm C. 16.2 Nm D. 18 m 03. 10 kg fiwewkó GKwU e¯‧i Dci 50 N ej 1 min a‡i wµqv K‡i| †e‡Mi cwieZ©b KZ n‡e? A. 50 ms–1 B. 3000 ms–1 C. 300 ms–1 D. 500 ms–1 04. GKwU 588 N IR‡bi e¯‧‡K 0.70 ms–2 Z¡iY w`‡Z Gi Dci KZ ej cÖ‡qvM Ki‡Z n‡e? A. 40 N B. 42 N C. 50 N D. 52 N 05. 10 kg f‡ii †Kv‡bv e¯‧i Ici cÖhy³ ej 20 N I Nl©Y ej 5 N n‡j e¯‧wUi Z¡iY n‡eA. 2 ms–2 B. 1.5 ms–2 C. 1.25 ms–2 D. 1.08 ms–2 06. `ywU MÖv‡gv‡dvb †iK‡W©i cÖ‡Z¨KwUi †K›`ª n‡Z 0.5m `~‡ii •iwLK †eM h_vµ‡g 35ms–1 Ges 20ms–1 | Mªv‡gv‡dvb؇qi ‣iwLK †e‡Mi AbycvZ †ei Ki? A. 7:4 B. 4:7 C. 7:3 D. 2:7 07. GKwU fvix PvKvi †K›`ªMvgx A‡¶i mv‡c‡¶ Gi RoZvi åvgK 100 kgm2 | PvKvwU‡Z 2.5 rads–2 †K․wYK Z¡iY w`‡Z KZ U‡K©i cÖ‡qvRb n‡e? A. 100 N-m B. 1 N-m C. 50 N-m D. 10 N-m 08. GKwU 7.0 kg f‡ii e¯‧ GKwU wjd‡Ui †g‡Si Dci w¯
i Ae¯
vq Av‡Q| wjd‡Ui DaŸ©Mvgx Z¡iY 2 m/s2 n‡j e¯‧i Dci †g‡S KZ©„K ej KZ? A. 68.6 N B. 54.6 N C. 82.6 N D. 0.0 N 09. 4kg I 6kg f‡ii `yBwU e¯Í h_vµ‡g 10m/s Ges 5m/s †e‡M GKB w`‡K MwZkxj| ci¯úi av°v LvIqvi ci e¯Í `yBwU hy³ Ae¯
vq Pj‡Z _vK‡j, hy³ e¯‧i †eM KZ? A. 10m/s B. 7m/s C. 6m/s D. 4m/s 10. f~wgi mv‡_ me©wbgœ 30 †Kv‡Y AvbZ Z‡j ¯
vwcZ e¯‧ wcQ‡j †b‡g hvq| Nl©Yv¼ KZ? A. sin 300 B. tan 300 C. cos 300 D. 1

186 cÖ_g cÎ wbDUwbqvb ejwe`¨v ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 11. c„w_ex N~Y©‡bi Rb¨ welye †iLvq Aew¯
Z †Kvb e¯‧i †K›`ªvwfgyLx Z¡iY KZ? [c„w_exi e¨vmva© R = 6400 km] A. 9.8 m/sec2 B. 4.9 m/sec2 C. 0.068 m/sec2 D. 0.034 m/sec2 12. 50ft •`‡N©¨i GKwU myZvi GK cÖv‡šÍ GKwU KYv 50ft/s mg`ªæwZ‡Z N~Y©vqgvb| KYvwUi Dci †K›`ªgyLx Z¡iY KZ? A. 200ft/s2 B. 50ft/s2 C. 10ft/s2 D. 25ft/s2 13. Nl©Y ej 5N n‡j 5kg f‡ii GKwU e¯‧‡K 5ms-2 Z¡i‡Y Ki‡Z ej cÖ‡qvMA. 15N B. 20N C. 25N D. 30N 14. GKwU Mvav 40N ej Øviv 8kg f‡ii GKwU ev· f‚wgi mv‡_ 37 †Kv‡Y Uvb‡Q| ev‡·i Z¡iY KZ? A. 4ms–2 B. 6ms–2 C. 8ms–2 D. 10ms–2 15. 20 ms–1 †e‡M AvMZ 0.2 kg f‡ii GKwU ej GKRb †L‡jvqvo a‡i 0.2 s mg‡q _vwg‡q w`j| †L‡jvqvo KZ…©K cÖhy³ Mo ej KZ? A. –10 N B. 10 N C. 20 N D. –20 N 16. mej wbDwK¬q e‡ji gvb hw` 1 aiv nq, Z‡e `ye©j wbDwK¬q ej, ZworPz¤^Kxq ej Ges gnvKl© e‡ji Av‡cw¶K mejZvi gvb h_vµ‡g cÖvqA. 10–2 , 10–12 I 10 –39 B. 10–39, 10–12 I 10–2 C. 10–12, 10–2 I 10–39 D. †KvbwUB bq 17. gnvKv‡k Aew¯
Z GKwU kvUj gnvKvk hv‡bi fi 3 103 kg Ges R¡vjvwbi fi 50 kg| R¡vjvwb 5 kg s1 nv‡i e¨eüZ n‡j Ges 150 m s1 mylg `ªæwZ‡Z wbM©Z n‡j kvUj hv‡bi Dci av°v n‡eA. 945 N B. 850 N C. 750 N D. 650 N 18. GKwU PvKvi fi 10 kg Ges PµMwZi e¨vmva© 0.5 m Gi RoZv åvgK †KvbwU? A. 1.5 kg m2 B. 2.5 kg m2 C. 2.5 N D. 0 19. 60 kg f‡ii GKwU e¯‧i Ici KZ ej cÖ‡qvM Ki‡j 1 wgwbU ci Gi †eM 10 ms–1 n‡e? A. 20 N B. 05 N C. 40 N D. 10 N 20. ej I e‡ji wµqvKv‡ji ¸Ydj‡K wK e‡j A. e‡ji NvZ B. åvgK C. NvZ ej D. kw³ OMR SHEET 07. A B C D 14. A B C D 01. A B C D 08. A B C D 15. A B C D 02. A B C D 09. A B C D 16. A B C D 03. A B C D 10. A B C D 17. A B C D 04. A B C D 11. A B C D 18. A B C D 05. A B C D 12. A B C D 19. A B C D 06. A B C D 13. A B C D 20. A B C D WRITTEN PART 21. 1 cvDÛ fi wewkó GKwU nvZzwo 3 Bw `yi n‡Z GKwU †c‡i‡Ki gv_vq AvNvZ K‡i Ges Dnv cÖvq 10 1 †m‡KÛ mgq MwZkxj wQ‡jv| †c‡i‡Ki gv_vi Dci cÖhy³ e‡ji gvb KZ? DËi:....................................................................................... 22. GKwU PvKv 4 wgwbU 10 †m‡KÛ mg‡q 200 evi Ny‡i 500 wgUvi c_ †h‡Z cv‡i| PvKvwUi cwiwa¯
 GKwU KYvi ‣iwLK †eM n‡e? DËi:....................................................................................... 23. w¯
i Ae¯
v †_‡K 1000 ‡KwR IR‡bi GKwU Mvox 10 †m‡KÛ mg‡q 20wg/‡m †eM AR©b K‡i| MvoxwU‡Z KZ wbDUb ej cÖ‡qvM Kiv n‡qwQj? DËi:....................................................................................... 24. 100 cvDÛ IR‡bi GKwU Mvwo‡K 45 cvDÛ IR‡bi GKwU Avbyf‚wgK e‡j †g‡Si Dci w`‡q †U‡b †bqv n‡”Q| w¯
ive¯
v †_‡K hvÎv ïiæ K‡i 10s G MvwowU 80 ft †M‡j †g‡Si Nl©Yv¼ KZ? DËi:....................................................................................... 25. GKRb evjK GKwU mvB‡K‡j P‡o 20m e¨vmv‡a©i GKwU e„ËvKvi c‡_ 3 min G 5 evi AveZ©b K‡i| mvB‡Kj mn evj‡Ki fi 25kg n‡j, Zvi †K›`ªvgyLx ej KZ? DËi:....................................................................................... 26. 10 kg f‡ii GKwU e¯‧ 4 ms1 †e‡M DËi w`‡K Pj‡Q| 5 kg f‡ii Aci GKwU e¯‧ 2ms1 †e‡M `wÿY w`‡K Pj‡Q| †Kvb GK mgq e¯‧ `ywUi g‡a¨ msN‡l©i d‡j wgwjZ e¯‧wU KZ †e‡M Pj‡e? DËi:....................................................................................... 27. 10N ej 2kg f‡ii GKwU w¯
i e¯‧i Dci wµqv K‡i| 4 sec ci e‡ji wµqv eÜ n‡q †M‡j cÖ_g n‡Z 8sec G e¯‧wU KZ `~iZ¡ AwZµg Ki‡e? DËi:....................................................................................... 28. 10,000kg R¡vjvbxmn GKwU i‡K‡Ui fi 15000kg| R¡vjvbx hw` 200 kg/s nv‡i cy‡o i‡K‡Ui mv‡c‡¶ 2000m/s †e‡M wbM©Z nq, Zvn‡j i‡K‡Ui Dci cÖhy³ av°v ev _ªvó KZ? DËi:....................................................................................... 29. fi †K›`ªMvgx Ges Z‡ji mwnZ j¤^ eivei A¶ mv‡c‡¶ GKwU AvqZvKvi cv‡Zi RoZvi åvgK 5 kgm2 | cvZwUi cÖ¯
 1m Ges fi 12 kg n‡j •`N©¨ KZ? DËi:....................................................................................... 30. GKwU •e`y¨wZK cvLvi myBP Ab Ki‡j `kevi c~Y© N~Y©‡bi ci †K․wbK †eM 20 rad/sec nq| †K․wbK Z¡iY KZ? DËi:....................................................................................... MCQ ANSWER ANALYSIS WRITTEN cÖkœ bs DËi e¨vL¨v 01 A mgvb f‡ii e¯‘ w¯’wZ¯’vcK msNl© n‡j ci¯úi †eM wewbgq K‡i| v1 = m1 – m2 m1 + m2 × u1 = 2m2 m1 + m2 × u2; m1 = m2 n‡j, v1 = m1 – m1 m1 + m1 × u1 + 2m1 2m1 × u2 ; v1 = u2 Ges v2 = u1 02 A = I = mr2 = 6 × (0.3)2 × 3 = 1.62 Nm 03 C F = ma = m v – u t v – u = Ft m = 300 ms–1 04 B W = mg m = W g = 588 9.8 = 60 F = ma = 60 × 0.70 = 42 N 05 B a = F – Fk m = 20 – 5 10 = 1.5 ms–2 06 A 1 2 = v1 v2 = 35 20 = 7 : 4

ASPECT PHYSICS cÖ_g cÎ wbDUwbqvb ejwe`¨v 187 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES cÖkœ bs DËi e¨vL¨v 07 A = I = 10 10 = 100 N-m 08 C T = F = m (g + a) = 7(9.8 + 2) = 82.6 N 09 B cÖ_g e¯‘i fi m1 = 4kg, wØZxq e¯‘i fi m2 = 6kg; cÖ_g e¯‘i †eM v1 = 10ms–1 , wØZxq e¯‘i †eM v2 = 5ms–1 v = m1v1 + m2v2 m1 + m2 ev, 4 10 + 6 5 4 + 6 = 70 10 = 7ms–1 10 B s = tans ev, s = tan300 11 A g = GM R 2 = 6.67 6 1024 (6.4 106 ) 2 10–11 = 9.8m/s2 12 B a = v 2 r = 502 50 = 50ft/s2 13 D F = 5 + ma = 5 + 5 5 = 30 N 14 A Fcosθ = mg , a = Fcos m = 40cos37 8 = 4 ms–2 15 C †_‡g hvIqvq, v = 0 ; †L‡jvqvo KZ…©K cÖhy³ Mo ej, F = ma = m u t = 0.2 20 0.2 = 20 N. 16 C `ye©j wbDwK¬q ej, ZworPz¤^Kxq ej I gnvKl©xq ej h_vµ‡g mej wbDwK¬q e‡jj 10–12 10–2 , 10–39 ¸Y| 17 C gnvk~‡b¨ g = 0 e‡j, i‡K‡Ui Dci wµqviZ ej, F = vr dm dt = 150 5 = 750 N 18 B I = mr2 = 10 (0.5)2 = 2.5 kgm2 19 D F= ma = m. v – u t = 60 10 60 = 10N 20 A e‡ji NvZ = F t = fi‡e‡Mi cwieZ©b| 21 GLv‡b, t = 0.1s, m = 1 lb, g =32 fts–2 , h = 0.25 ft, v = 2gh = 4ms1 , t 0.16m 2 u v s GLb, Fs = mgh s mgh F = 40 poundal 22 cwiwa 2r = 500 200 = 2.5 m, v = r = 2N t r = 2r t N = 2.5 200 250 = 2 ms1 23 F = ma; a = vu t = 20 0 10 = 2ms2 F = 1000 2 = 2000N 24 GLv‡b P = 45 32 = 1440 poundel, m = 100 lb, u = 0, t = 10s, s = 80ft, a = 2s/t2 = 1.6fts2 , R = mg = 100 32 = 3200 Rvwb, F = ma = 100 1.6 = 160, Fs =1440160 = 1280, s = Fs /R=1280/3200 = 0.4 25 F=m 2 r=m 2N T 2 r=25 2 5 180 2 20 = 15.23N 26 O m2 = 5kg; u2 = –2ms–1 O m1 = 10kg; u1 = 4ms–1 m1u1 + m2u2 = (m1 + m2)V 10 4 + 5 (–2) = (10 + 5) V; V = 2ms1 27 a = F m = 5ms2 S1 = 1 2 5 4 2 = 40m; [†h‡nZz u = 0] v = at = 5 4 = 20ms-1 S2 = vt = 4 20 = 80m S = S1 + S2 = 40 + 80 = 120m ASPECT SPECIAL: S = F t1 m t1 2 + t2 = 10 4 2 4 2 + 4 = 120 m 28 F = dm dt . vr = 200 × 2000 = 4 × 105N 29 2 2 b 12 M I ev, 2 2 1 12 12 5 l ev, 5 1 2 l ev, 5 1 2 l ev, l 2 m 30 1 = 0, 2 = 20 rad/sec N = 10 evi = 2N = 2.10 =20 = 2 20 (20) 0 2 2 2 2 1 2 2 2 3.18rad /sec 10

188 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES SURVEY TABLE  Kx coe? †Kb coe? †Kv_v n‡Z coe? KZUzKz coe?  TOPICS MAGNETIC DECISION [hv co‡e] MAKING DECISION [†h Kvi‡Y co‡e] VVI For This Year DU JU RU CU GST Engr. HSC Written MCQ THEORY CONCEPT-01 KvR msµvšÍ Z_¨vejx 70% 70% 65% 50% 40% 10% 55% - CONCEPT-02 kw³ Ges kw³i iƒcvšÍi 60% 40% 45% 55% 30% 05% 40% - CONCEPT-03 msiÿYkxj ej I AmsiÿYkxj e‡ji cv_©K¨ I •ewkó¨ 55% 75% 55% 60% 40% 08% 30% - CONCEPT-04 ÿgZv msµvšÍ Z_¨vejx 30% 40% 50% 55% 50% 10% 35% - MvwYwZK cÖ‡qvM CONCEPT-01 KvR msµvšÍ MvwYwZK cÖ‡qvM 65% 75% 80% 75% 80% 50% 20% CONCEPT-02 kw³ msµvšÍ MvwYwZK cÖ‡qvM 70% 70% 75% 70% 75% 60% 70% CONCEPT-03 MwZkw³ Ges wefe kw³ msµvšÍ MvwYwZK cÖ‡qvM 60% 70% 75% 70% 80% 45% 65% CONCEPT-04 fi‡eM msµvšÍ MvwYwZK cÖ‡qvM 75% 85% 80% 85% 70% 65% 70% CONCEPT-05 ¶gZv msµvšÍ MvwYwZK cÖ‡qvM 80% 70% 75% 70% 85% 45% 90% CONCEPT-06 Kzqv msµvšÍ MvwYwZK cÖ‡qvM 60% 55% 65% 60% 60% 75% 80% DU. = Dhaka University, JU. = Jahangirnagar University, RU. = Rajshahi University, CU = Chittagong University, GST = General, Science & Technology, Engr. = Engineering. STEP 1 mvRv‡bv me Z_¨ THEORY fwZ© †ivMxi c_¨ Concept 1 KvR msµvšÍ Z_¨vejx ✅ KvR = ej miY = F. S ev W = Fscos ✅ KvR: †¯‥jvi ivwk ✅ GKK: J, eV, Kwh, Nm, Kgm2 s –2 , erg, Poundalft (lbft2 s –2 ) ✅ 1 Joule =107 erg ✅ 1erg = 10–7 J ✅ 1eV = 1.6 10–19 J ✅ Kv‡Ri gvÎv mgxKiY [ML2T –2 ] ✅ Ryj I K¨vjwii g‡a¨ m¤úK©: 1 Cal = 4.2 Joule; 1 J = 0.24 Cal ✅ KvR I kw³i K‡qKwU AcÖPwjZ GKK: GK‡Ki bvg wK‡mi GKK MvwYwZK dgy©jv AvM© wmwRGm (CGS) c×wZ‡Z Kv‡Ri cig GKK| 1erg = 1 dyne 1 cm 1 erg = 10–7 Joule 1 J = 107 erg dzU cvDÛvj GdwcGm (FPS) c×wZ‡Z Kv‡Ri cig GKK| 1 ft – poundal = 4.2 105 erg MÖvg-†mw›UwgUvi wmwRGm (CGS) c×wZ‡Z Kv‡Ri AwfKlx©q GKK| 1 gm – cm = 980 erg dzU-cvDÛ GdwcGm (FPS) c×wZ‡Z Kv‡Ri AwfKlx©q GKK| 1 ft = 1b = 1.356 Joule wK‡jvMÖvg-wgUvi Gg‡KGm (MKS) c×wZ‡Z Kv‡Ri AwfKlx©q GKK| 1 kg – m = 9.8 Joule ◈ AvaywbK c`v_©weÁv‡b KvR ev kw³i cwigv‡ci Rb¨ B‡jKUªb †fvë (eV) bvgK GKwU GKK e¨envi Kiv nq| 1eV = 1.610–19 J ✅ e‡ji Øviv KvR: msÁv hw` ej cÖ‡qv‡Mi d‡j e‡ji w`‡K e‡ji cÖ‡qvM we›`yi miY N‡U ev e‡ji w`‡K mi‡Yi abvZ¥K Dcvsk _v‡K Z‡e H mi‡Yi Rb¨ K…Z KvR‡K e‡ji Øviv KvR e‡j| kZ©: [0 < 90] •ewkó¨ w¯’wZkw³ n«vm cvq MwZkw³ e„w× cvq Z¡i‡Yi m„wó nq D`vniY GKwU e¯‘‡K Qv‡`i Dci n‡Z wb‡P †d‡j †`Iqv| GKwU w¯’i dzUej‡K ej cÖ‡qvM Ki‡j hw` dzUejwU e‡ji w`‡K m‡i hvq| evB‡ii Drm n‡Z gnvKlx©q e‡ji wecix‡Z KvR| `yBwU wecixZ PvR© ev Avavb AvKwl©Z n‡j| ✅ e‡ji weiæ‡× KvR: msÁv hw` ej cÖ‡qv‡Mi d‡j e‡ji wecixZ w`‡K e‡ji cÖ‡qvM we›`yi miY N‡U ev e‡ji wecixZ w`‡K mi‡Yi FYvZ¥K Dcvsk _v‡K Z‡e H mi‡Yi Rb¨ K…Z KvR‡K e‡ji weiæ‡× KvR e‡j| [90 < 180] •ewkó¨ w¯’wZkw³ e„w× cvq; MwZkw³ n«vm cvq; g›`‡bi m„wó nq; D`vniY †g‡S n‡Z e¯‘ †Uwe‡j †Zvjv; mg‡e‡M MwZkxj e¯‘ †eªK K‡l _vgvb; GKwU eB‡K †g‡S †_‡K Avjgvwi‡Z ivLv| KvR, kw³ I ÿgZv cÖ_g WORK, ENERGY AND POWER cÎ 05 Aa¨vq c„ôv 188