Title 1. P1C1. Physical World and Measurement_With Solve.pdf ✅

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
†fŠZRMr I cwigvc  Mastery Practice Sheet 3 inch e„wó nq Zvn‡j H kn‡i cwZZ e„wói AvqZb dzU GK‡i (acre-ft) wbY©q K‡iv| [†`Iqv Av‡Q, 1 = acre.ft = (43.560 ft2 )ft = 43560 ft3 ] mgvavb: V = Al = 26  106  (3.28)2  1 6 ft3 = 46619733.33 ft3 = 1070.24 acre.ft (Ans.) 9|  Nb‡Z¡i †Kv‡bv ZijfwZ© miæ b‡ji ga ̈ w`‡q r e ̈vmv‡a©i †MvjK v cÖvšÍ‡e‡M (mg`aæwZ‡Z) Pjgvb| cÖvšÍ‡eM, v Gi gvb †Mvjv‡Ki e ̈vmva© (r), Zi‡ji NbZ¡ () Ges Zi‡ji mv›`aZv mnM () Gi wbf©i K‡i| gvÎv mgxKi‡Yi mvnv‡h ̈ cÖvšÍ‡e‡Mi GKwU ivwkgvjv wbY©q Ki| mgvavb: v   x r y  z GLv‡b, [p] = [ML–3 ] [v] = LT–1 [r] = L [] = ML–1T –1 [LT–1 ] = k[ML–3 ] x [L]y [ML–1T –1 ] z [LT–1 ] = k[M]x+z [L]–3x+y–z [T]–z z = 1 – 3x + y – z = 1 x + z = 0  x = –1, y = –1  cÖvšÍ‡e‡Mi ivwkgvjv, v = k –1 r –1  = k r †hLv‡b k GKwU mgvbycvwZK aaæeK| (Ans.) 10| c„w_exi e ̈vmva© 4000 gvBj, Gi cwiwa KZ? mgvavb: C = 2r = 2  3.1416  4000  1.609 km = 40.4  103 km (Ans.) 11| iscyi n‡Z XvKvi `~iZ¡ 402.3 km| GB `~iZ¡ gvB‡j cÖKvk K‡iv| mgvavb: lmile = 1.609 km  d = 402.3 km = 402.3 1.609 mile = 250 mile (Ans.) 12| †jvnvi †ÿ‡Î AvšÍtAvYweK `~iZ¡ 2.5  10–10 m| GB `~iZ¡ A ̈vs÷ag GK‡K cÖKvk K‡iv| mgvavb: We know, lÅ = 10–10 m d = 2.5  10–10 m = 2.5  10–10 10–10 Å = 2.5 Å (Ans.) 13| Puv‡`i fi 7.33  1022 kg| G‡K cvD‡Û cÖKvk K‡iv| mgvavb: We know, 1kg = 2.2046 cvDÛ M = 7.33  1022 kg = 16.16  1022 cvDÛ (Ans.) 14| †mvwWqvg evwZ †_‡K njy` e‡Y©i Av‡jvi Zi1⁄2‣`N© ̈ 5893 Å| nm GK‡K Gi gvb KZ? mgvavb: l Å = 0.1 nm GLv‡b, d = 5893 Å = 589.3 nm (Ans.) 15| Joule GK‡K cÖKvwkZ gvb‡K erg GK‡K cÖKvk K‡iv| mgvavb: 1 Joule = 1N.m = 1 kgms–2 .m = 1 kg.1m2 .1s–2 = (1000 g) . (100 cm)2 . 1s–2 = 107 gcm2 s –1 = 107 erg (Ans.) Avevi, 1erg = 1 gcm2 s –2 16| †Kv‡bv GKK c×wZ‡Z `~i‡Z¡i GKK n‡jv 1s G Av‡jvK †h `~iZ¡ AwZμg K‡i Zvi mgvb Ges mg‡qi GKK n‡jv c„w_ex m~‡h©i Pviw`‡K GKevi Nyi‡Z †h mgq jv‡M Zvi mgvb| GB c×wZ‡Z GKK †e‡Mi gvb‡K SI c×wZ‡Z cÖKvk K‡iv| mgvavb: GLv‡b, `~i‡Z¡i GKK = 3  108 m mg‡qi GKK = 365  86400 s = 31536000 s  †e‡Mi GKK = 3  108 31536000 = 9.513 ms–1 (Ans.) 17| GK ÒcvigvYweK fi GKKÓ Gi mgvb fi m¤ú~Y©iƒ‡c kw3‡Z iƒcvšÍwiZ n‡j Kx cwigvY kw3 wbM©Z n‡e? mgvavb: E = 1.66  10–27  (3  108 ) 2 J = 1.494  10–10 J = 933.75 Mev (Ans.) 18| y = a + bt + ct2 | GLv‡b y wgUv‡i t †m‡K‡Û cÖKvk Ki‡j b Gi GKK I gvÎv wbY©q K‡iv| mgvavb: GLv‡b, [b] [T] = [L]  [b] = [LT–1 ]  GKK = ms–1 (Ans.)
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|