Tiêu đề Statics Varsity Practice Sheet Solution.pdf ✅

08 w ̄’wZwe` ̈v Statics weMZ mv‡j DU-G Avmv cÖkœvejx 1. †Kv‡bv GKwU we›`y‡Z wμqviZ 2 N I 2 2 N ej؇qi jwä 2 5 N n‡j, Zv‡`i ga ̈eZ©x †KvYÑ [DU 21-22] 135 45 225 90 DËi: 45 e ̈vL ̈v: (2 5) 2 = 22 + (2 2) 2 + 2  2 2  2cos  20 = 4 + 8 + 8 2cos  cos = 1 2   = 45 2. †Kv‡bv we›`y‡Z 3P Ges 2P `yBwU e‡ji jwä R| cÖ_g ejwUi gvb wØ ̧Y Ki‡j jwäi gvbI wØ ̧Y nq| ej؇qi AšÍM©Z †KvY KZ n‡e? [DU 20-21, 14-15, 12-13, 12-13, 09-10; GST 23-24; JU 22-23]  3  6 2 3 5 6 DËi: 2 3 e ̈vL ̈v: R 2 = 9P2 + 4P2 + 12P2 cos = 13P2 + 12P2 cos Avevi, 4R2 = 36P2 + 4P2 + 24P2 cos  52P2 + 48P2 cos = 40P2 + 24P2 cos  24P2 cos = – 12P2  cos = – 1 2   = 2 3 3. †Kv‡bv GKwU we›`y‡Z wμqviZ  P I 2  P ej؇qi jwä 7  P n‡j, Zv‡`i ga ̈eZ©x †KvY KZ? [DU 19-20] 30 90 60 180 DËi: 60 e ̈vL ̈v: 7P2 = P2 + 4P2 + 2  P  2Pcos  4P2 cos = 2P2  cos = 1 2   = 60 4. `yBwU mgvb gv‡bi ej P Gi me©wb¤œ jwäi gvb KZ? [DU 18-19] 2P 0 P P 2 DËi: 0 e ̈vL ̈v: `yBwU mgvb gv‡bi ej ci ̄úi wecixZ w`‡K wμqv Ki‡j Zv‡`i jwäi gvb k~b ̈| 5. 2 N Ges 5 N gv‡bi `yBwU ej GKB †iLvq GKB w`‡K wμqviZ| Dnv‡`i me©vwaK jwä n‡e- [DU 17-18] 7 N 3 N 29 N 5 N DËi: 7 N e ̈vL ̈v: Rmax = P + Q = 2 + 5 = 7 N 6. GKwU mgevû wÎfz‡Ri evû·qi mgvšÍiv‡j GKBμ‡g mgwe›`y‡Z Kvh©iZ 6, 10, 14 GKK gv‡bi wZbwU †e‡Mi jwäi gvb n‡eÑ [DU 16-17] 4 3 units 7 3 units 10 3 units 15 3 units DËi: 4 3 units e ̈vL ̈v: GLv‡b, wμqviZ ej·qi g‡a ̈ mvaviY AšÍi n‡jv 4  wb‡Y©q jwä = 3  mvaviY AšÍi = 4 3 GKK 14 120 6 120 120 10 Shortcut: GKB we›`y‡Z wμqviZ wZbwU e‡ji g‡a ̈ mvaviY AšÍi d n‡j, jwä = 3  d 7. 8 N Ges 3 N gv‡bi `yBwU ej GKwU we›`y‡Z 60 †Kv‡Y GKwU e ̄‘‡Z wμqviZ| ej؇qi jwäi gvbÑ [DU 15-16; JnU 09-10] 73 N 97 N 55 N 11 N DËi: 97 N e ̈vL ̈v: R = P 2 + Q2 + 2PQcos = 8 2 + 32 + 2  8  3cos60 = 97 N
2  Higher Math 2nd Paper Chapter-8 8. †Kv‡bv we›`y‡Z P Ges 2P gv‡bi `yBwU ej wμqvkxj | cÖ_g ejwU‡K wØ ̧Y K‡i wØZxqwUi gvb 8 GKK e„w× Kiv n‡j, Zv‡`i jwäi w`K AcwiewZ©Z _v‡K| P Gi gvb- [DU 13-14] 1 2 4 8 DËi: 4 e ̈vL ̈v: P 2P = 2P 2P + 8  2P + 8 = 4P  P = 4 9. 3 GK‡Ki `yBwU mgvb ej 120 †Kv‡Y GK we›`y‡Z KvR K‡i| G‡`i jwäi gvb KZ? [DU 11-12; JU 21-22; JnU 13-14] 4 3 GKK 3 GKK 2 3 GKK 3 GKK DËi: 3 GKK e ̈vL ̈v: R = 2 3cos 120 2 = 3 GKK 10. 3P Ges 5P gv‡bi `yBwU ej ci ̄úi j¤^fv‡e wμqv K‡i| Zv‡`i jwäi gvb KZ? [DU 99-00, 97-98, 95-96; IU 17-18, 16-17, 15-16] 3P 2 2P 34P 43P DËi: 34P e ̈vL ̈v: R = 9P2 + 25P2 = 34P2 = 34P GKK weMZ mv‡j GST-G Avmv cÖkœvejx 1. wZbwU ej P, 3P, P mvg ̈ve ̄’vq _vK‡j cÖ_g `yBwU e‡ji ga ̈eZ©x †KvYÑ [GST 23-24; RU 19-20] 60 90 120 150 DËi: 150 e ̈vL ̈v: P 2 = P2 + 3P2 + 2  P  3Pcos  2 3cos = – 3  cos = – 3 2   = 150 2. †Kv‡bv we›`y‡Z 3P Ges 2P `yBwU e‡ji jwä R| cÖ_g ejwUi gvb wØ ̧Y Ki‡j jwäi gvbI wØ ̧Y nq| ej؇qi AšÍM©Z †KvY KZ n‡e? [GST 23-24; JU 22-23; DU 20-21, 14-15, 12-13, 12-13, 09-10]  3  6 2 3 5 6 DËi: 2 3 e ̈vL ̈v: R 2 = 9P2 + 4P2 + 12P2 cos = 13P2 + 12P2 cos Avevi, 4R2 = 36P2 + 4P2 + 24P2 cos  52P2 + 48P2 cos = 40P2 + 24P2 cos  24P2 cos = – 12P2  cos = – 1 2   = 2 3 3. wZbwU mgZjxq ej P, Q Ges R †Kv‡bv we›`y‡Z wμqv K‡i mvg ̈ve ̄’vq Av‡Q| hw` P Ges Q Gi gvb h_vμ‡g 5 3 N I 5 N Ges Zv‡`i ga ̈eZ©x †KvY  2 nq, Zvn‡j R, Q Gi m‡1⁄2 KZ †KvY •Zwi Ki‡e? [GST 22-23]  4  3 2 3 3 4 DËi: 2 3 e ̈vL ̈v: Q = 5 N P = 5 3 N   R S GLv‡b, P sin = Q sin(270 – ) = R 90  – cos sin = 5 5 3  – tan = 3   = 2 3 4. F Ges 2F gv‡bi mgwe›`y e‡ji wμqvw`K Ges GKB we›`y‡Z wμqviZ 2F Ges 2F + 2 gv‡bi ej؇qi jwäi wμqvw`K GKB n‡j F Gi gvb KZ GKK? [GST 21-22] 1 2 1 2 4 DËi: 1 e ̈vL ̈v: m`„k‡KvYx wÎfz‡Ri aviYv AbyhvqxÑ F 2F = 2F 2F + 2  2F + 2 = 4F  F = 1 F 2F 2F 2F+2 Shortcut: jwäi wμqvw`K GKB n‡j 1g †ÿ‡Î I 2q †ÿ‡Î ej؇qi Abycv‡Zi gvb mgvb n‡e|
w ̄’wZwe` ̈v  Varsity Practice Sheet Solution 3 weMZ mv‡j Agri-G Avmv cÖkœvejx 1. P gv‡bi `ywU e‡ji jwä P n‡j, ej؇qi ga ̈eZ©x †KvY †KvbwU? [Agri Guccho 21-22; CU 22-23] 0 45 135 120 DËi: 120 e ̈vL ̈v: R = P = 2Pcos 2  cos  2 = 1 2   2 = 60   = 120 2. ABC wÎfz‡Ri wZbwU kxl© we›`y‡Z 2, 2, p wZbwU m`„k mgvšÍivj ej wμqv Ki‡Q| G‡`i jwä wÎfz‡Ri fi‡K›`aMvgx n‡j, p Gi gvb- [Agri Guccho 21-22] 2 3 4 5 DËi: 2 e ̈vL ̈v: wÎfz‡Ri wZbwU kxl© we›`y‡Z mggv‡bi wZbwU ej wμqv Ki‡j G‡`i jwä wÎfzRwUi fi‡K›`aMvgx nq| 3. 3 N I 5 N gv‡bi `yBwU ej ci ̄úi j¤^| Zv‡`i jwäi gvb †KvbwU? [Agri Guccho 21-22] 5 N 6 N 34 N 31 N DËi: 34 N e ̈vL ̈v: R = 3 2 + 52 = 34 N 4. `ywU mgvb gv‡bi e‡ji jwä G‡`i ̧Yd‡ji eM©g~j n‡j, ej؇qi ga ̈eZ©x †KvY KZ? [Agri Guccho 19-20; IU 19-20] 0 30 60 120 DËi: 120 e ̈vL ̈v: R = P.P = 2Pcos 2  cos  2 = 1 2   2 = 60   = 120 weMZ mv‡j JU-G Avmv cÖkœvejx 1. †Kv‡bv we›`y‡Z 3P Ges 2P `yBwU e‡ji jwä R| cÖ_g ejwUi gvb wØ ̧Y Ki‡j jwäi gvbI wØ ̧Y nq| ej؇qi AšÍM©Z †KvY KZ n‡e? [JU 22-23; GST 23-24; DU 20-21, 14-15, 12-13, 12-13, 09-10]  3  6 2 3 5 6 DËi: 2 3 e ̈vL ̈v: R 2 = 9P2 + 4P2 + 12P2 cos = 13P2 + 12P2 cos Avevi, 4R2 = 36P2 + 4P2 + 24P2 cos  52P2 + 48P2 cos = 40P2 + 24P2 cos  24P2 cos = – 12P2  cos = – 1 2   = 2 3 2. 30 wgUvi •`N ̈© wewkó AB `‡Ði A cÖv‡šÍ 20 kg IRb I B cÖv‡šÍ P kg IRb Szjv‡bv Av‡Q| G‡`i jwä C we›`y‡Z wμqvkxj| AC Gi •`N ̈© 20 wgUvi n‡j, P ejwUi gvb KZ ? [JU 22-23] 20 N 30 N 40 N 50 N DËi: mwVK DËi †bB e ̈vL ̈v: 20  AC = P  BC  20  20 = P  10  P = 40 kg-wt  ej = (40  9.8) N  mwVK DËi †bB| P kg-wt B 20 A B C B 20 kg-wt 10 3. 2, 5 Ges 3 gv‡bi wZbwU ej †Kvb GK we›`y‡Z wμqviZ| Giv ci ̄úi fvimvg ̈ m„wó Ki‡j cÖ_g `yBwU e‡ji ga ̈eZ©x †KvYÑ [JU 21-22; RU 23-24, 21-22, 19-20; CU 13-14, 11-12, 03-04, 02-03] 30 45 60 90 DËi: 90 e ̈vL ̈v: 3 2 = 22 + 5 + 2  2  5cos  cos = 0   = 90 4. 3 GK‡Ki `yBwU mgvb ej 120 †Kv‡Y GK we›`y‡Z KvR K‡i| G‡`i jwäi gvb KZ? [JU 21-22; JnU 13-14; DU 11-12] 4 3 GKK 3 GKK 2 3 GKK 3 GKK DËi: 3 GKK e ̈vL ̈v: R = 2 3cos 120 2 = 3 GKK
4  Higher Math 2nd Paper Chapter-8 5. 60 †Kv‡Y wμqviZ 5 GKK gv‡bi `ywU mgvb e‡ji jwä KZ? [JU 21-22] 2 5 15 10 + 5 3 10 + 5 3 DËi: 15 e ̈vL ̈v: R = 2 5cos 60 2 = 2 5  3 2 = 15 GKK 6. `yBwU e‡ji jwä 40 N, hv ÿz`aZi e‡ji wμqv‡iLvi Dci j¤^| ÿz`aZi ejwU 30 N n‡j, e„nËg ej KZ? [JU 19-20] 60 N 70 N 50 N 85 N DËi: 50 N e ̈vL ̈v: Q Q 40 30 Q 2 = 402 + 302  Q = 50 N 7. ABC wÎfz‡Ri AB, AC evû Øviv `ywU e‡ji gvb I w`K m~wPZ n‡jv| Zv‡`i jwä wÎfzRwUi cwi‡K›`aMvgx n‡j wÎfzRwUÑ [JU 11-12] mg‡KvYx ev mgevû welg evû mg‡KvYx ev mgwØevû †Kv‡bvwUB bq DËi: mg‡KvYx ev mgwØevû e ̈vL ̈v: †Kv‡bv wÎfz‡Ri `yBwU evû Øviv ej I w`K m~wPZ n‡j Ges jwä wÎfzRwUi cwi‡K›`aMvgx n‡j, wÎfzRwU mg‡KvYx ev mgwØevû nq| weMZ mv‡j RU-G Avmv cÖkœvejx 1. 2, 5 Ges 3 gv‡bi wZbwU ej †Kvb GK we›`y‡Z wμqviZ| Giv ci ̄úi fvimvg ̈ m„wó Ki‡j cÖ_g `yBwU e‡ji ga ̈eZ©x †KvYÑ [RU 23-24, 21-22, 19-20; JU 21-22; CU 13-14, 11-12, 03-04, 02-03] 30 45 60 90 DËi: 90 e ̈vL ̈v: 3 2 = 22 + 5 + 2  2  5cos  cos = 0   = 90 2. mggv‡bi `yÕwU e‡ji jwäi eM© ej؇qi ̧Yd‡ji wZb ̧Y n‡j ej `ywUi ga ̈eZ©x †KvY KZ? [RU 23-24; CU 16-17]  3  2  4  6 DËi:  3 e ̈vL ̈v: R 2 = 3P 2  R = 3P = 2Pcos 2  cos  2 = 3 2   2 =  6   =  3 3. 2 3 †Kv‡Y Kvh©iZ 4 N I 1 N gv‡bi ej `yBwUi jwäi gvbÑ [RU 23-24] 21 N 19 N 13 N 23 N DËi: 13 N e ̈vL ̈v: R = 4 2 + 12 + 2  4  1 cos 2 3 = 17 – 4 = 13 N 4. †Kv‡bv we›`y‡Z 60 †Kv‡Y wμqviZ `yBwU mgvb ej‡K GKB we›`y‡Z wμqviZ 15 N e‡ji mvnv‡h ̈ fvimvg ̈ ivL‡j, e‡ji gvb †KvbwU? [RU 23-24] 5 2 N 5 3 N 2 2 N 3 2 N DËi: 5 3 N e ̈vL ̈v: 152 = P2 + P2 + 2  P  Pcos60  3P2 = 152  P 2 = 5  15  P = 75 = 5 3 N A_ev, P = Q n‡j, R = 2cos 2  15 = 2Pcos30  P = 15 3 = 5 3 N