Tiêu đề 1. P1C1. Physical World and Measurement_With Solve.pdf ✅

†fŠZRMr I cwigvc  Practice Content 1 †f.ZRMr I cwigvc Physical World and Measurement cÖ_g Aa ̈vq weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 1. ̄...z M‡Ri wcP 0.05 cm Ges e„ËvKvi † ̄..‡ji fvMmsL ̈v 50 n‡j, jwNô MY‡bi gvb KZ? [BUET Preli 22-23] 0.01 mm 0.02 mm 0.05 mm 0.1 mm DËi: 0.01 mm e ̈vL ̈v: jwNô MYb = wcP e„ËvKvi † ̄‹‡ji fvMmsL ̈v = 0.05 cm 50 = 0.001 cm = 0.01 mm 2. †Kvb †ù‡ivwgUv‡ii †h‡Kv‡bv `yB cv‡qi ga ̈eZ©x `~iZ¡ h_vμ‡g 51 mm, 47 mm, 52 mm| hw` GKwU mgZ‡jvËj †j‡Ýi eμZ‡ji D”PZv mgZj c„ô †_‡K 3.82 mm nq, Z‡e †jÝwUi eμZvi e ̈vma© KZ? [BUET Preli 21-22] 11.985 mm 110.986 mm 110.985 mm None DËi: 110.985 mm e ̈vL ̈v: R = d 2 6h + h 2 = 502 6  3.82 + 3.82 2  R = 110.985 mm     d = 51 + 47 + 52 3 = 50 mm 3. ej I kw3i gvÎv h_vμ‡gÑ [BUET 09-10] LT–2 and MLT–2 MLT–2 and ML2T –2 LT–2 and ML2T –2 MLT–2 and ML–2T –3 DËi: MLT–2 and ML2T –2 4. GKwU ̄øvBW K ̈vwjcv‡m©i cÖavb † ̄..‡ji ÿz`a N‡ii gvb 1 mm Ges fvwb©qvi † ̄..‡ji 10 Ni cÖavb † ̄..‡ji 9 N‡ii mgvb| GB † ̄..‡ji fvwb©qvi aaæeK KZ? [BUET 09-10] 0.01 cm 0.01 mm 0.05 cm 0.05 mm DËi: 0.01 cm e ̈vL ̈v: fvwb©qvi aaæeK = s n = 1 mm 10 = 0.1 mm = 0.01 cm weKí: cÖavb † ̄‹‡ji ÿz`aZg 1 Ni = 1 mm; fvwb©qvi † ̄‹‡ji ÿz`aZg 1 Ni = 9 mm 10 = 0.9 mm  fvwb©qvi aaæeK = 1 mm – 0.9 mm = 0.1 mm = 0.01 cm 5. GKRb QvÎ cixÿvMv‡i AwfKl©R Z¡i‡Yi gvb 9.88 m/sec2 wbY©q Kij| Aciw`‡K hLb †h 0.01 kg f‡ii †Kv‡bv evULvov‡K w ̄úas wbw3‡Z Szwj‡q w`j ZLb 0.0980 N ej †`Lv‡”Q| Zvi cixÿvjä AwfKl©R Z¡i‡Yi ÎæwU wbY©q K‡iv| [CKRUET 22-23] – 0.716% – 0.102% – 0.816% 0.102% – 0.916% DËi: – 0.816% e ̈vL ̈v: g g = 9.8 – 9.88 9.8  100% = – 0.816% 6. GKwU ̄øvBW K ̈vwjcv‡m©i cÖavb † ̄..‡ji 39 fvM fvwb©qvi † ̄..‡ji 40 fv‡Mi mgvb| cÖavb † ̄..‡ji GK fv‡Mi gvb 1.00 mm| fvwb©qvi aaæeK KZ? [KUET 06-07] 0.010 mm 0.020 mm 0.025 mm 0.100 mm DËi: 0.025 mm e ̈vL ̈v: fvwb©qvi † ̄‹‡ji 1 fv‡Mi gvb = 39 40 mm = 0.975 mm  fvwb©qvi aaæeK = (1 – 0.975)mm = 0.025 mm 7. e‡ji gvÎvi mgxKiY †KvbwU? [RUET 12-13] [MLT–2 ] [MLT] [MLT–1 ] [MLT–3 ] [MLT–4 ] DËi: [MLT–2 ] e ̈vL ̈v: ej = fi × Z¡iY = [M] × [LT–2 ] = [MLT–2 ] 8. U‡K©i gvÎv mgxKiY †KvbwU? [RUET 11-12] [ML2T 2 ] [ML–2T 2 ] [ML2T –2 ] [ML–2T –2 ] [MLT–2 ] DËi: [ML2T –2 ] e ̈vL ̈v: | |  = | r|  |F|  sin | |  = [L.MLT–2 ] = [ML2T –2 ] 9. wb‡Pi †KvbwU gvÎvMZ fv‡e w ̄ wZ ̄ vcK ̧Yvs‡Ki mgZyj ̈? [RUET 11-12] Stress Strain Surface tension Acceleration None DËi: Stress e ̈vL ̈v: w ̄’wZ ̄’vcK ̧YvsK = stress strain ; weK...wZi (strain) gvÎv †bB| 10. ZvcaviY ÿgZvi gvÎv mgxKiY †KvbwU? [CUET 13-14; RUET 12-13] ML2T 1  –2 ML2T –2  –1
2  Physics 1st Paper Chapter-1 ML–1T –2  –1 ML–2T –2  –1 ML–1T –1  –1 DËi: ML2T –2  –1 11. Length of a simple pendulum l = 100.0 0.5 cm, and time period T = (2.00  0.01) s. Determine the percentage of error in acceleration due to gravity ‘g’. [IUT 20-21]  1.5%  2.0%  1.05%  1.75% DËi:  1.5% e ̈vL ̈v: T = 2 L g  g = 4 2L T 2 g g × 100% = L L × 100% + 2 T T × 100% =      0.5 100 × 100% + 2 × 0.01 2 × 100% =  (0.5% + 1%) =  1.5% 12. U‡K©i gvÎv I e‡ji gvÎvi AbycvZ KZ? [BUTex 14-15] MLT–2 L ML2T –2 ML –1 DËi: L e ̈vL ̈v: || |F| = ML2T –2 MLT–2 = L 13. mv›`aZv ̧Yvs‡Ki gvÎvÑ [BUTex 11-12; KUET 06-07] [ML–2T –2 ] [ML–1T –3 ] [ML–1T –1 ] [M–2L 2T –1 ] DËi: [ML–1T –1 ] e ̈vL ̈v: F = A dv dy   = F A × dy dv = MLT–2 L 2 × L LT–1 = ML–1T –1 ACS Physics Department Gi g‡bvbxZ wjwLZ cÖkœmg~n 1| s = ut + 1 2 at2 Eqn wU gvÎv mgxKiY Øviv hvPvB Ki| mgvavb: GLv‡b, [S] = L [ut] = [LT–1 ]  [T] = [L] [at2 ] = [LT–2 ]  [T2 ] = [L]  Eqn wU ï×| (Ans.) 2| wbDU‡bi m~Î Abymv‡i M ̈vmxq gva ̈‡g k‡ãi †eM v = P D GLv‡b P = M ̈vmxq Pvc Ges D = NbZ¡| gvÎv we‡ePbvq mgxKiYwU mwVK wKbv hvPvB Ki| mgvavb: GLv‡b, [v] = LT–1 Avevi,    P D =    ML  –1T –2 ML–3 1 2 = [LT–1 ]  gvÎv we‡ePbvq mgxKiYwU mwVK| (Ans.) 3| gvÎv we‡ePbvq †`LvI †h, cv‡ki mgxKiYwU mwVK: Fs = 1 2 mv2 – 1 2 mu2 mgvavb: GLv‡b, [Fs] = [MLT–2 ]  [L] = [ML2T –2 ] [mv2 ] = [M]  [LT–1 ] 2 = [ML2T –2 ] [mu2 ] = [M]  [LT–1 ] 2 = [ML2T –2 ]  gvÎv we‡ePbvq mgxKiYwU mwVK| (Ans.) 4| 5 km †K ft G cÖKvk Ki| mgvavb: 5 km = 5000 m = 5000  3.28 ft = 16400 ft (Ans.) 5| AwfKl©R Z¡iY, g = 9.8 ms–2 n‡j F.P.S c×wZ‡Z g Gi gvb wbY©q Ki| mgvavb: g = 9.8 m 1 s2 = 9.8  3.28 ft 1 s2 = 32.1 ft s–1 (Ans.) 6| l = (4.00  0.05)cm Ges F = (4.00  0.02)N a‡i F = kl mgxKiY †_‡K k Gi gvb †ei Ki| mgvavb: F = kl  k = 1 N/cm k k = F F + l l  k k = 0.02 4 + 0.05 4  k = 0.0175 N/cm  k = (1  0.0175) N/cm (Ans.) 7| GKwU mvevb ey`ey` mij Qw›`Z ̄ú›`‡b `yj‡Q| ey`ey`wUi †`vjbKvj, T = PaD b S c mgxKiY Øviv cÖKvk Kiv hvq; †hLv‡b P nj Pvc, D nj NbZ¡ Ges S nj c„ôUvb, a, b I c Gi gvb wbY©q Ki| mgvavb: GLv‡b, T = PaD b S c [T] = [ML–1T –2 ] a  [ML–3 ] b  [MT–2 ] c  [T] = [M]a+b+c [L]– a – 3b [T]– 2a – 2c  a + b + c = 0 .... (i)  (i), (iii) I (iii) n‡Z,  – a – 3b = 0 .... (ii) a = – 3 2  – 2a – 2c = 1 ... (iii) b = 1 2 c = 1 (Ans.) 8| AvenvIqv cÖ‡K.kjxiv cÖvqkB cvwbi AvqZb‡K GKi dzU wn‡m‡e cÖKvk K‡ib| hw` 26 km2 †ÿÎ wewkó GKwU kn‡i 30 min G 2.0

4  Physics 1st Paper Chapter-1 19| †`LvI †h, KvR I U‡K©i gvÎv I GKK GKB| mgvavb: [KvR] = fi  Z¡iY  miY = [M]      L T 2  [L] = [ML2T –2 ] Avevi, [UK©] = Ae ̄’vb †f±i  ej = miY  fi  Z¡iY = [M]      L T 2  [L] = [ML2T –2 ]  KvR I U‡K©i gvÎv I GKK GKB| (†`Lv‡bv n‡jv) 20| †`LvI †h, L R Ges CR ivwk `ywUi GKK mg‡qi GKK| GLv‡b L Gi gvÎv [L2MT–2A –2 ], R Gi gvÎv [L2MT–3A –2 ], C Gi gvÎv [M–1L –2T 4A 2 ]| mgvavb:     L R =     L 2MT–2A –2 L 2MT–3A –2 = [T] = mgq [CR] = [M–1L –2T 4A 2 ]  [L2MT–3A –2 ] = [T] = mgq  L R I CR ivwk `ywUi GKK mg‡qi GKK| (†`Lv‡bv n‡jv) 21| AwfKl©R Z¡i‡Yi gvb 9.8 ms–2 | •`‡N© ̈i GKK wK‡jvwgUvi Ges mg‡qi GKK NÈv aiv n‡j AwfKl©R Z¡i‡Yi gvb KZ n‡e? mgvavb: g = 9.8 ms–2 =     9.8 1000 km     h 3600 –2 = 1.27008  105 kmh–1 (Ans.) 22| GKwU ej 15 kg f‡ii †Kv‡bv e ̄‧i Ici 1 wgwbU wμqv K‡i 4.6 kms–1 †eM Drcbœ K‡i| GB e‡ji gvb wbDU‡b cÖKvk K‡iv| mgvavb: F = 15  4.6  103 60 N = 1150 N (Ans.) 23| MwZ‡eM (v), mgq (T) Ges ej (F) †g.wjK ivwk| Nb‡Z¡i gvÎv wbY©q K‡iv| mgvavb:  = m V = ma Va = F l 3 .v.t–1  = NbZ¡ V = AvqZb v = †eM = F l 3 t 3.v.t2 = F v 3 .v.t2 = F v 4 .t2 = F v–4 t –2 (Ans.) 24| GK †gvj ev ̄Íe M ̈v‡mi †ÿ‡Î f ̈vb Wvi Iqvjm mgxKiY n‡jv:    P +  a V 2 (V – b) = RT, GLv‡b a I b aaæeK| a I b Gi SI GKK wbY©q K‡iv| mgvavb: GLv‡b, P = [ML–1T –2 ]  a V 2 = ML–1T –2  a = ML5T –2  a Gi GKK kgm5 s –2 (Ans.) Avevi, V I b Gi GKK GKB  b Gi GKK m 3 (Ans.) 25| MÖn m~‡h©i Pviw`‡K e„ËvKvi c‡_ Nyi‡Q| hw` ch©vqKvj (T), (i) K‡ÿi e ̈vmva© (r), (ii) m~‡h©i fi (M) Ges (iii) gnvKl©xq aaæeK (G) Gi Ici wbf©i K‡i Zvn‡j †`LvI †h, MÖn ̧‡jv †Kcjv‡ii Z...Zxq m~Î †g‡b P‡j| A_©vr †`LvI †h, T 2  r 3 mgvavb:  T = krx M yG z [T] = [L]x [M]y [M–1L 3T –2 ] z GLv‡b, – 2z = 1  z = – 1 2 x + 3z = 0  x = 3 2  T  r 3 2  T 2  r 3 (†`Lv‡bv n‡jv) 26| Qvcvi fz‡ji Kvi‡Y GKwU eB‡Z mij †`vjhy3 †Kv‡bv KYvi miY y Gi `ywU m~Î wjwce× Av‡Q| (K) y = a sin     2 T  t; (L) y = a sin vt| gvÎv we‡køl‡Yi gva ̈‡g †`LvI †Kvb m~ÎwU mwVK? mgvavb: (K) y = a sin     2 T  t GLv‡b, 2 T  t Gi gvÎv †bB|  [a] = [L]  [y] = [L]  GB mgxKiYwU mwVK| (L) y = a sin vt [vt] = [L] wKš‘ sin Gi Rb ̈ vt Gi gvÎv †bB KviY vt †KvY|