ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 239 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 JUST 01. hw` †Kvb cvnv‡oi kx‡l© Ges Lwbi Mfx‡i mij †`vj‡Ki †`vjbKvj mgvb nq, Zvn‡j cvnv‡oi D”PZv I Lwbi MfxiZvi AbycvZ n‡e- [MBSTU-A, Set-2 19-20] A. 3:4 B. 4:3 C. 1:2 D. 2:1 cvnv‡oi kx‡l© Ges Lwbi Mfx‡i mij †`vj‡Ki †`vjbKvj mgvb nIqv gv‡b `yB RvqMv‡ZB g Gi gvb mgvb| 2h = d h d = 1 2 h : d = 1 : 2 02. c„w_ex‡K 6.4 106 m e¨vmv‡a©i Ges 55 gcc–1 Nb‡Z¡i †MvjK ai‡j Gi c„‡ô AwfKl© Z¡iY KZ n‡e? [JUST-C, 19-20] A. 9.83 ms–2 B. 8.93 ms–2 C. 8 ms–2 D. 9 ms–2 ρ = 5.5 g/cc = 5.5×03 kgm–3 [1g /cc=1000 kgm–3 ] g = 4 3 Rρ G = 4 3 × × 6.4×106 × 5.5 × 103 × 6.673 × 10–11 = 9.83 ms–2 ASPECT SPECIAL: c„w_exi g Gi gvb Rvbv _vK‡j AsK Ki‡Z n‡ebv| 03. c„w_ex m~‡h©i Pviw`‡K cwiågY Ki‡Q| c„w_exi ch©vqKvj cÖvq 365 w`b Ges c„w_exi †K›`ª †_‡K m~‡h©i `~iZ¡ m 11 1.510 n‡j m~‡h© fi KZ? [JSTU: 14-15] A. 3×1030kg B. 2×1030kg C. 5×1030kg D. 2×1015kg E. 3×1015kg 2 2 3 GT 4π r m 11 2 3 2 11 6.673 10 365 24 60 60 4 1.5 10 = 2 1030kg BSMRSTU 01. c„w_ex c„ô †_‡K 1000 wK.wg. wfZ‡i AwfKl©xq Z¡i‡Yi gvb KZ? c„w_exi e¨vmva© 6.4 106 cm? [BSMRSTU-B, 19-20] A. 9.8ms–2 B. 32 ms–2 C. 9.4 ms–2 D. 8.34 ms–2 d MfxiZvq AwfKl©R Z¡iY, gd ge = 1 – d R = 1 – 5 32 = 27 32 gd = 27 32 9.8 = 8.3ms–2 STEP 05 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. GKwU K…ò Mne‡ii NUbv w`M‡šÍ 6.9km Dnvi NbZ¡ KZ? [G = 6.67 10–11Nm2 kg–2 ] [KUET. 18-19] A. 4.6 1018km.m–3 B. 4.66 1018km.m–3 C. 5.1 1021g.cm–3 D. 3.38 1018kgm–3 E. 4.2 1021g/cm3 R = 2GM c 2 M = 4.653 1030kg = M V = M 4 3 R 3 = 3.38 1018kgm–3 STEP 06 ANALYSIS OF MEDICAL & DENTAL QUESTION AFMC 01. welyexq I †giæ A‡ji e¨vmv‡a©i `~iZ¡ KZ? [AFMC. 2020-21] A. 11km B. 22km C. 44km D. 33km S C info c„w_ex m¤ú~Y© †MvjvKvi bq| DËi-`wÿY wKQzUv Pvcv Ges wbiÿxq A‡j wKQzUv ùxZ, A_©vr c„w_exi AvK…wZ Dc‡MvjK c„w_exi †giæe¨vmv‡a©i †P‡q wbiÿxq- e¨vmva© cÖvq 22km †ewk| f~-c„‡ôi AwfKl©R Z¡i‡Yi gvb c„w_exi †K›`ª †_‡K `~i‡Z¡i e‡M©i e¨¯ÍvbycvwZK e‡j †giæ A‡j g gvb m‡e©v”P Ges wbiÿxq A‡j me©wb¤œ nq| KviY Ab¨ †h †Kvb ¯’v‡b g gvb GB `yBwUi cÖvwšÍK gv‡bi g‡a¨ _v‡K| STEP 07 ANALYSIS OF HSC BOARD QUESTION cÖ`Ë DÏxcKwU †`‡L cÖ‡kœi DËi `vI : [Xv. †ev. 2016] 01. 50kg f‡ii GKwU e¯‘‡K A n‡Z C †Z wb‡q †M‡j Gi IRb n‡eÑ [Xv. †ev. 2016] A. 490 N B. 272.2N C. 245N D. 217.8N S D info gc = R R + h 2 × gA = R R + R 2 2 × 9.8 = R 3R 2 2 × 9.8 = 4.36 ms –2 W = mg = 50 × 4.36 = 217.8 N 02. c„w_exc„‡ô Aÿvs‡ki Rb¨ g Gi mgxKiY n‡eÑ[c„w_exi fi I e¨vmva© h_vµ‡g M I R] [w`.‡ev. 2019] A. g= GM R 2 – 2R cos2 B. g= GM R 2 – 2R cos C. g= GM R 2 – R cos2 D. g= GM R 2 – R cos Ans A 03. †giy A‡cÿv welyexq A‡j AwfKl©R Z¡iY KZUv Kg? [Kz. †ev. 2017] A. 2R B. R C. R cos D. 2Rcos Ans D 04. c„w_ex c„ô, c„w_ex c„ô †_‡K h D”PZvq I c„w_ex c„ô n‡Z h MfxiZvq AwfKl©R Z¡iY h_vµ‡g g, gh I gbh n‡jÑ [mKj †evW© 2018] A. gbh < gh < g B. gh < gbh < g C. gh > gbh > g D. gh > g > gnh Ans B
240 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 3 f~-c„ô n‡Z h-D”PZv msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. g = GM (R + h) 2 02. h D”PZvq AwfKl©R Z¡iY, gh = R R + h 2 g ; gh = 1 2h R g [when, h << R] [ R 2 D”PZvq g = 0] ASPECT SPECIAL: ◈ D”PZv, h = g gh – 1 R ◈ h D”PZvq g f~c„‡ôi x% n‡j, R x x h 10 ◈ h D”PZvq AwfKl©R Z¡i‡Yi gvb n 1 n‡j h = ( n –1) R Model Example01 c„w_ex‡K 6400km e¨vmv‡a©i GKwU †MvjK ai‡j f~-c„ô n‡Z KZ D”PZvq AwfKl©xq Z¡i‡bi gvb f~-c„‡ôi AwfKl©xq Z¡i‡bi gvb 1 4 Ask n‡e? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] gm = Gi 1 4 = g 4 gh g = 1 4 Avgiv Rvwb, f~c„‡ô g = GM R 2 ...........................(1) f~-c„ô n‡Z h D”PZvq gh = GM (R+h) 2 .................(2) (2) (1) gh g = GM (R+h) 2 GM R 2 gh g = GM (R+h) 2 R 2 GM gh g = R 2 (R+h) 2 1 4 = R 2 (R+h) 2 1 2 = R R+h 1 2 = R R+h 2R = R + h h = R h = 6400km gh = 1 – 2h R g [hLb h<<R] AwfKl©R Z¡i‡Yi gvb f~-c„‡ôi AwfKl©R Z¡i‡Yi 1 n Ask n‡j| D³ ¯’v‡bi D”PZv, h = ( n – 1)R = ( 4 – 1)R = 6400km CONCEPTUAL MATH MEx 01 c„w_exi e¨vmva© 6400 km n‡j c„w_ex c„‡ôi 6400 km DuPz‡Z ‘g’ Gi gvb KZ n‡e? 2 R h R e g h g gh = ge R R + R 2 = R 2R 2 ge = 9.8 4 = 2.45 ms2 NOW START PRACTICE 01. c„w_ex c„ô n‡Z KZ D”PZvq AwfKl©xq Z¡i‡bi gvb c„w_ex c„‡ôi Z¡i‡bi gv‡bi kZKiv 81 fvM| c„w_exi e¨vmva© = 6.38 106m 02. c„w_ex‡K 6400km e¨vmv‡a©i GKwU †MvjK ai‡j f~-c„ô n‡Z KZ D”PZvq AwfKl©xq Z¡i‡bi gvb f~-c„‡ôi AwfKl©xq Z¡i‡bi gvb 1 64 Ask| NOW PRACTICE SOLVE : 01. D”PZv, x (10 - x )R h h = (10 – 81)R 81 9 6 6.38 10 9 R h m 5 7.110 02. AwfKl©R Z¡i‡bi gvb f~-c„‡ói AwfKl©R Z¡i‡bi 1 n Ask n‡j| D³ ¯’v‡bi D”PZv, h n 1R, D”PZv h 64 1R = (8 – 1)R = 7R= 76.4103 km = 4.48104 km
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 241 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. c„w_exi e¨vmva© 4000 gvBj n‡j AvbygvwbK KZ D”PZvq ga¨vKl©Y ej c„w_ex c„‡ôi ga¨vKl©Y e‡ji 1% n‡e? [DU. 07-08] A. 400 miles B. 4,000 miles C. 40,000 miles D. 36,000 miles 1% ev 1 100 n‡j, h = ( n – 1)R = ( 100 – 1)R = 9R = 36000 miles 02. c„w_exi e¨vmva© 6400 km n‡j c„w_ex c„‡ôi 6400 km DuPz‡Z 'g' Gi gvb KZ n‡e? [DU. 01-02; BU. 11-12; BRU. 13-14] A. 4.90 m/s2 B. 2.45 m/s2 C. 0 D. 9.8 m/s2 2 6400 6400 6400 e g h g 2 2.45 /sec 4 9.8 m or, R g g h h e 1 ev, 6400 = 1 6400 h e g g h ev, 1+1= h e g g ev, 4 = h e g g ev, 2 2.45 4 9.8 g ms h 03. GKwU K…wÎg DcMÖn c„w_exi Pviw`‡K f~-c„ô n‡Z 900 km Dc‡i †_‡K e„ËvKvi c‡_ Nyi‡Q| c„w_exi e¨vmva© 6400 km Ges f~-c„‡ô ga¨vKl©Y RwbZ Z¡iY 9.81 m/sec2 n‡j DcMÖnwUi †eM KZ? [DU. 00-01] A. 20.75 km/s B. 10 km/s C. 7.42 km/s D. 19.65 km/s DcMÖnwUi †eM, v = R h gR2 = (6400 900) 1000 9.81 (6400 1000) 2 = 7.42103ms–1 = 7.42 kms–1 STEP 02 ANALYSIS OF RU QUESTION 01. f~-c„‡ô AwfKl©R Z¡iY 10 ms–2 Ges GKwU Lwbi Zj‡`‡k 5 ms–2 | c„w_exi e¨vmva© 6106 m n‡j Lwbi MfxiZv n‡e- [RU. 15-16] A. 2.6 104 m B. 3 106 m C. 4.5 106 m D. †KvbwUB bq R d g g e d 1 ev, R d 1 10 5 ev, R d 1 2 1 ev, d R = 1 2 ev, 6 6 10 3 10 2 6 2 R d m STEP 03 ANALYSIS OF GST QUESTION PART A Analysis of General University Question BU 01. f~-c„ô †_‡K c„w_exi e¨v‡mi mgvb D”PZvq GKwU we›`y‡Z g Gi gvb KZ? [BU. 12-13, NUST. 13-14] A. 9.8m/s2 B. 4.9m/s2 C. 2.5m/s2 D. 1.1m/s2 gh ge = R R + h 2 = R R + 2R 2 = 1 3 2 gh = 9.8 9 = 1.088 = 1.1 PART B Analysis of Science & Technology Question SUST 01. f~c„‡ôi 200km D‡a© AwfKl©R Z¡iY KZ m/s2 ? (c„w_exi e¨vmva© 6400km Ges f~c„‡ô g Gi gvb 9.8m/s2 ) [SUST. 16-17] A. 9.19 B. 8.85 C. 7.75 D. 6.92 E. 5.89 gh ge = R R + h 2 ev, gh 9.8 = 6.4 6.4 + 0.2 2 ev, gh= 6.4 6.6 2 9.8= 9.21 STEP 04 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. c„w_exi e¨vmva© 6.4 106m Ges c„‡ô AwfKl©R Z¡iY 9.8ms–2 | f~-c„ô †_‡K 6.4 107m D”PZvq AwfKl©R Z¡i‡Yi gvb KZ? [KUET. 18-19] A.–186.2ms–2 B.– 9.8 ms–2 C.0.081 ms–2 D. 8.05 ms–2 g g = R 2 (R + h) 2 = g = 0.081ms–2 02. f~-c„ó n‡Z 1000km D‡aŸ© AwfKl©R Z¡iY KZ m/s2 ? [c„w_exi e¨vmva© = 6400km] [KUET. 16-17] A. 3.8 ms2 B. 7.33 ms2 C. 8.1 ms2 D.9.8 ms2 gh = R 2 (R + h) 2g gh = R 2 (R + h) 2g = (6.4106 ) 2 (6.4106 +1106 ) 29.8=7.33ms2 03. f~-c„‡ô KZ DPz‡Z †M‡j †mLvbKvi AwfKl©R Z¡i‡Yi gvb 25% n‡e| (c„w_exi e¨vmva© = 6.4106m) [KUET. 10-11] A.100km B. 25km C. 640km D. 6400km E. 64000km 25% ev, 1 4 n‡j h = ( n – 1)R = ( 4 – 1)R = 6400km STEP 05 ANALYSIS OF HSC BOARD QUESTION 01. c„w_exi e¨vmva© ‘R’ Ges c„w_ex‡Z AwfKl©R Z¡iY ‘g’| c„w_exc„ô n‡Z ‘h’ D”PZvq AwfKl©R Z¡iY KZ? [Xv. †ev. 2015] A. g(R – h) R B. gR2 (R + h) 2 C. gR (R + h) 2 D. g(R – h) 2 R 2 S B info g = GM R 2 ; g = GM (R + h) 2 = gR2 (R + h) 2 02. R c„w_exi e¨vmva© n‡j f~-c„ô n‡Z KZ D”PZvq g-Gi gvb k~b¨ n‡e? [P. †ev. 2016] A. R B. 2R C. R 2 D. 4R S C info g1 = 1 – h R g = R 2 = 1 – h R g 1 2 = 1 – h R h R = 1 2 h = R 2 03. c„w_exi N~Y©b eÜ n‡j welye †iLvi g Gi gvbÑ [w`. †ev. 2017] A. e„w× cv‡e B. n«vm cv‡e C. GKB _vK‡e D. k~b¨ n‡e Ans A 04. g-Gi gvb me©vwaK †Kv_vq? [w`. †ev. 2015] A. †giæ B. welye C. f~-‡K‡›`ª D. cvnv‡oi P~ovq Ans A
242 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 4 f~-c„ô n‡Z d-MfxiZv msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. d MfxiZvq AwfKl©R Z¡iY, gd= 1 d R g [R MfxiZvq g = 0] ASPECT SPECIAL: d MfxiZvq AwfKl©R Z¡i‡Yi gvb n 1 n‡j d = n n 1 ×g Model Example01 f~-c„ô n‡Z KZ Af¨šÍ‡i AwfKl©R Z¡i‡bi gvb 1 4 Ask| General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] Avgiv Rvwb, gh = g 1 – d R d = 1 – gh ge R = 1 1 4 R = 4 1 4 6400 103 = 4.8 106m d = n 1 n R = 4 1 4 R = 3 4 R = 3 4 6.4106 =4.8 106m CONCEPTUAL MATH MEx 01 c„w_ex c„ô n‡Z KZ Mfx‡i †M‡j g Gi gvb f~-c„‡ôi gv‡bi A‡a©K n‡eMfxiZv d = n 1 n R = 2 1 2 R = R 2 = 6.4106 2 = 3.2106 NOW START PRACTICE 01. f~-c„ô n‡Z KZ Af¨šÍ‡i AwfKl©R Z¡i‡bi gvb 1 5 Ask? NOW PRACTICE SOLVE : 01. d = n 1 n R = 5 1 5 R = 4 5 R = 4 5 6.4 106 = 5.12 106m REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF CU QUESTION 01. c„w_ex c„ô n‡Z KZ Mfx‡i †M‡j g Gi gvb f~-c„‡ôi gv‡bi A‡a©K n‡e- [CU. 07-08] A. 2.0 106 m B. 3.5 106 m C. 3.2 106 m D. 3 106 m MfxiZv d = n 1 n R = 2 1 2 R = R 2 = 6.4106 2 = 3.2106 STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question SUST 01. c„w_ex c„‡ô g Gi gvb 9.8m/s2 c„w_exi †K‡›`ª g Gi gvb k~b¨| c„w_exi c„ô †_‡K e¨vmv‡a©i A‡a©K MfxiZvq g Gi gvb KZ? [SUST. 07-08, 06-07, 09-10; RU. 12-13] A. 9.8m/s2 B. 4.9m/s2 C. 3.2m/s2 D. 2.5m/s2 R d g g e d 1 ev, d ge R R g 2 1 ev, 9.8 2 1 1 gd = 9.8 2 1 = 4.9ms-2 STEP 03 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. f~c„‡ôi KZ Mfx‡i AwfKl©Y Z¡i‡Yi gvb f~c„‡ôi gv‡bi GK PZ©z_vsk n‡e? [c„w_exi e¨vmva© = 6.4 103 km) [BUET: 02-03; KUET. 17-18] A. 8.4 103 km B. 4.8 103 km C. 4.0 103 km D. 5.2 103 km E. 6.8 103 km d = n – 1 n R = 3 4 R = 4.8 103 km KUET 01. c„w_ex c„ô †_‡K 1000km wfZ‡i AwfKl©xq Z¡i‡Yi gvb KZ? c„w_exi e¨vmva© 6.4106m. [KUET. 09-10] A. 9.8ms-2 B. 32ms-2 C. 9.41ms-2 D. 8.5ms-2 E. 8.34ms-2 e d g g R d ge gd 1 = 6400 1000 1 ev, gd = 8.268 = 8.33 STEP 04 ANALYSIS OF HSC BOARD QUESTION 01. f~-c„ô †_‡K KZ MfxiZvq AwfKl©R Z¡i‡Yi gvb f~-c„‡ôi AwfKl©R Z¡i‡Yi GK-Z…Zxqvsk n‡e? (R = c„w_exi e¨vmva©) [Xv. †ev. 2017] A. R 4 B. R 3 C. R 2 D. 2R 3 Ans D S D info ASPECT SPECIAL: MfxiZv d = n 1 n R = 3 1 3 R = 2R 3 02. c„w_exi e¨vmva© (R)-Gi Zzjbvq KZ cwieZ©b n‡j, AwfKl©R Z¡i‡Yi gvb f~-c„‡ôi AwfKl©R Z¡i‡Yi A‡a©K n‡e? [P. †ev. 2016] A. R 2 B. R 4 C. R 8 D. R 16 S A info AwfKl©R Z¡i‡Yi gvb f~-c„ô AwfKl© Z¡i‡Yi 1 x Ask n‡j: f~c„ô n‡Z D”PZv n‡j, h = ( x – 1) × R f~c„ô n‡Z Mfxi n‡j, h = x – 1 x × R
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 243 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES wP‡Îi wb‡`©kbvi Av‡jv‡K 03 I 04 bs cÖ‡kœi DËi `vI: wP‡Î m f‡ii GKwU e¯‘ c„w_exi myo½ w`‡q Pj‡Q| 03. e¯‘wUi MwZi aibÑ [e. †ev. 2016] A. •iwLK B. eµ C. †`vjb D. N~Y©b Ans A 04. A we›`y‡Z AwfKl© Z¡i‡Yi gvbÑ [e. †ev. 2016] A. 2.94ms2 B. 3.00ms2 C. 6.86ms2 D. 9.8ms2 S C info gA = 1 – d R g = 1 – R – 0.7R R × 9.8 = 6.86 ms–2 Concept 5 wewfbœ MÖ‡n g-Gi gv‡bi Zzjbv msµvšÍ MvwYwZK cÖ‡qvM FORMULA 0.1 gm = (e¨vmv‡a©i ¸Y) 2 f‡ii ¸Y × ge [AbycvZ ev fMœvsk bv n‡j] 02. gm = f‡ii ¸Y (e¨vmv‡a©i ¸Y) 2 × ge [AbycvZ fMœvsk n‡j] Model Example01 P‡›`ªi fi c„w_exi f‡ii 1 80 fvM| P‡›`ªi e¨vmva© c„w_exi e¨vmv‡a©i 1 4 fvM| c„w_exi c„‡ô AwfKl©xq Z¡i‡Yi gvb 9.8ms –2 n‡j, P›`ª c„‡ô AwfKl©xq Z¡iY KZ? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] GLv‡b, Mm = Me 1 80 = Me 80 ; Rm = Re 1 4 = Re 4 P›`ª c„‡ô, gm = GMm Rm 2 .....................(1) c„w_ex c„‡ô, ge = GMe Re 2 ....................(2) (1) (2) K‡i, gm ge = Mm Rm 2 Re 2 Me gm ge = Mm Me Re 2 Rm 2 gm ge = Me 80 Me Re 2 Re 4 gm = 1 80 (4)2 9.8 gm = 1.96ms–2 gm = f‡ii ¸Y (e¨vmv‡a©i ¸Y) 2 ge [AbycvZ fMœvsk n‡j] = 1 80 1 4 2 9.8 = 1.96ms–2 NOW START PRACTICE 01. g½j MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i 0.532 ¸Y Ges fi 0.11 ¸Y| f~-c„‡ô AwfKl©R Z¡i‡bi gvb 9.8 ms–2 g½‡ji c„‡ô AwfKl©b Z¡i‡bi gvb †ei Ki? NOW PRACTICE SOLVE : 01. f‡ii ¸Y (e¨vmv‡a©i ¸Y) 2 ge [AbycvZ fMœvsk n‡j]= 0.11 (0.532) 2 9.8 = 3.8ms2 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF JU QUESTION 01. †Kvb MÖ‡ni fi I e¨vmva© h_vµ‡g c„w_exi fi I e¨vmv‡a©i wظb n‡j IB MÖ‡ni c„‡ô AwfKl©R Z¡iY n‡e c„w_ex c„‡ói AwfKl©R Z¡i‡YiÑ [JU-H, Set-A. 20-21] A. wظY B. mgvb C. A‡a©K D. GK PZz_v©sk S C info Avgiv Rvwb, g1 = GM1 R1 2 ; g2 = GM2 R2 2 g2 g1 = M2 M1 R1 R2 2 = 2M M R 2R 2 g = g 2 M 2 = 2M1 M2 M1 = 2 R2 = 2R1 R1 R2 = 1 2 STEP 02 ANALYSIS OF RU QUESTION 01. c„w_ex I Puv‡`i c„‡ô AwfKl©xq Z¡i‡Yi AbycvZ KZ? (c„w_exi fi = 81 P‡›`ªi fi, c„w_exi e¨vmva© = 4 P‡›`ªi e¨vmva©) [RU. 15-16] A. 81:4 B. 81:6 C. 81: 10 D. 81:16 AwfKlx©q Z¡iY, g = 2 R GM ; g 2 R M m e g g = m e M M 2 e 2 m R R = 1 81 16 1 = 81:16 02. c„w_exc„‡ô I P›`ªc„‡ô `yBwU †m‡KÛ †`vj‡Ki •`‡N©¨i AbycvZ 81:16| c„w_exc„‡ô g Gi gvb 9.81ms–2 n‡j P›`ªc„‡ô g Gi gvb- [RU. 13-14] A. 1.91ms–2 B. 1.93ms–2 C. 1.94ms–2 D. 1.92ms–2 e m e m g g L L ev, 9.81 81 16 e e m m g L L g ev, gm =1.93ms–2 03. GKwU e¯‘i IRb c„w_ex‡Z 56.84N I P‡›`ª 9.8N P›`ª A‡c¶v c„w_ex‡Z AwfKl©R Z¡iY KZ ¸Y? [RU. 09-10] A. 5.9 B. 5.7 C. 6.0 D. 5.8 5.8 9.8 56.84 c m c m W W g g
244 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 04. c„w_exi e¨vmva© A‡a©K n‡j AwfKl©R Z¡i‡Yi gvb n‡e- [RU. 09-10] A. 9.8ms-2 B. 4.9ms-2 C. 39.2ms-2 D. 19.6ms-2 2 c e e c R R g g ev, 9.8 1 2 2 gc = 39.2ms-2 05. c„w_ex c„‡ô GKwU †jv‡Ki fi 60kg| P›`ª c„‡ô †jvKwUi fi KZ? (c„w_exi fi = 81 P‡›`ªi fi, c„w_exi e¨vmva© = 4 P‡›`ªi e¨vmva©) [RU. 06-07] A.12.85 kg B.13.85 kg C. 60 kg D. 60kg-wt ¯’vb‡f‡` g cwiewZ©Z nq, ¯’vb‡f‡` m cwiewZ©Z nq bv| STEP 03 ANALYSIS OF CU QUESTION 01. c„w_exi AwfKlx©q Z¡iY g=9.8 ms2 Ges e¨vmva© R = 6400 km n‡j, GKwU e¯‘i gyw³ †eM KZ n‡e? [CU. 11-12, 13-14] A. 10.2 kms1 B. 11.5 kms1 C. 12.0 kms1 D. 11.2 kms1 E. 11.2 kms2 c„w_exi gyw³‡eM, ve = 2gR = 3 2 9.8 6400 10 =11.2103 ms–1 = 11.2 kms–1 02. `ywU MÖ‡ni NbZ¡ mylg Ges mgvb wKš‘ cÖ_gwUi fi I e¨vmva© wظY cÖ_g MÖ‡ni Dcwifv‡Mi Ges wØZxq MÖ‡ni Dcwifv‡Mi 'g' Gi AbycvZ †KvbwU? [CU. 11-12] A. 2 : 1 B. 1:2 C. 4:1 D. 8:1 E. 1:4 1: 2 2 1 1 2 2 2 1 2 2 1 2 1 R R m m g g 03. GKwU MÖ‡ni fi I e¨vmva© DfqB h_vµ‡g c„w_exi fi I e¨vmv‡a©i wظY| f~-c„‡ô g = 9.8 m/sec2 n‡j H MÖ‡ni c„‡ô g Gi gvb KZ? [CU. 07-08] A. 9.8 m/sec2 B. 19.6 m/sec2 C. 4.9 m/sec2 D. 2.45 m/sec2 E. 8.4 m/sec2 2 m R e R e M m M e g m g 2 2 1 1 2 ev, 2 1 9.8 m g STEP 04 ANALYSIS OF GST QUESTION 01. A I B MÖn؇qi ¯Íi h_vµ‡g M I 2M, Ges e¨vmva© h_vµ‡g R I 2R n‡j Zv‡`i AwfKl©R Z¡i‡bi AbycvZ gA : gB KZ? [GST-A. 20-21] A. 1:1 B. 1:2 C. 2:1 D. 4:1 S C info gA gB = GMA rA 2 GMB rB 2 = MA MB rB 2 rA 2 = M 2M (2R) 2 R 2 = 1 2 4 = 2 1 PART A Analysis of General University Question BRUR 01. c„w_exi mv‡c‡ÿ gyw³‡eM vE Ges Puv‡`i mv‡c‡ÿ gyw³‡eM vM n‡j, wb‡Pi †KvbwU mwVK? [BRUR-F, Set-2, 19-20] A. vE > vM B. vE < vM C. vE = 1 vM D. vE = vM c„w_exi AwfKl©R Z¡iY Puv‡`i AwfKl©R Z¡iY A‡cÿv †ekx, gE > gM. ZvB vE > vM A_©vr, ve = 2gR ve g PART B Analysis of Science & Technology Question SUST 01. c„w_ex c„‡ô g Gi gvb 9.8m/s2 c„w_ex c„ô †_‡K c„w_exi e¨vmv‡a©i mg cwigvb D”PZvq g Gi gvb KZ m/s2 ? [SUST. 08-09] A. 9.8 B. 4.9 C. 2.45 D. 0 2 R R R g g e h ev, 2 2 1 9.8 h g ev, gh = 9.8 4 ev, gh = 2.45 ms–2 STEP 05 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. c„w_exi fi P‡›`ªi f‡ii 80 ¸Y Ges Zv‡`i e¨vmva© h_vµ‡g 12800 km Ges 3200 km | P›`ª c„‡ôi AwfKl©R Z¡i‡Yi gvb KZ? [KUET. 14-15] A. 163 cm/s2 B. 1.7 m/s2 C. 196 cm/s2 D. 1.9 m/s2 E. 1.64 m/s2 gm (c„w_exi f‡ii ¸Y) (c„w_exi e¨vmv‡a©¨i ¸Y) 2 ge = 1 80 1 4 2 × ge = e 16 g 80 1 = 1.96 5 9.8 ms2 = 196cms–2 02. Pvu‡`i e¨vmva© c„w_exi e¨vmv‡a©i 1 4 th Ges fi 1 80 th f~-c„‡ô AwfKl©R Z¡i‡Yi gvb 9.8m/s2 Puv‡`i AwfKl©R Z¡i‡Yi gvb †ei Ki| [KUET. 07-08] A. 3.8m/s2 B. 9.8m/s2 C. 0.196m/s2 D. 1.96cm/s2 E. 1.96m/s2 gm (c„w_exi f‡ii ¸Y) (c„w_exi e¨vmv‡a©¨i ¸Y) 2 ge = 1 80 1 4 2 × ge ev, 16 9.8 80 1 gm =1.96m/s2 STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. c„w_exc„‡ô †Kv‡bv e¯‘i fi 60kg n‡j Pvu‡` H e¯‘i fi KZ? Pvu‡`i AwfKl©R Z¡iY c„w_exi 1 6 ¸Y| [e.‡ev. 2019] A. 10kg B. 20kg C. 60kg D. 360kg Ans C
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 245 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 6 wewfbœ MÖ‡n IR‡bi Zyjbv msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. Wm = (e¨vmva© ¸Y) 2 f‡ii ¸Y We 02. F = Fh F –1 × Fe 03. W = Wh W –1 × We Model Example01 f~-c„‡ôi †Kvb e¯‘i IRb 100N Puv‡` Gi IRb KZ? c„w_exi fi I e¨vmva© h_vµ‡g Puv‡`i fi I e¨vmv‡a©i 81 Ges 4 ¸Y| General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] Avgiv Rvwb, IRb W = mg ‡`Iqv Av‡Q, f~-c„‡ô We = mge--------------(1) Me = 81 Mm Puv‡`i c„‡ô Wm = mgm---------(2) Re = 4Rm (2) (1) Wm We = gm ge -------------------(3) Avevi f~c„‡ô, ge = GMe Re 2 -----(4) Ges P‡›`ªi c„‡ô, gm = GMm Rm 2 ----(5) (5) (4) gm ge = GMm Rm 2 Re 2 GMe gm ge = Mm Me Re 2 Rm 2 gm ge = Mm 81Mm (4Rm) 2 R 2 m gm ge = 16 81 (3)bs mgxKiY †_‡K, Wm We = 16 81 Wm= 16 81 We = 20N Wm = (e¨vmva© ¸Y) 2 f‡ii ¸Y We = (4) 2 81 100 = 20N NOW START PRACTICE 01. f~-c„‡ô †Kvb †jv‡Ki IRb 648N n‡j wZwb Puv‡` wM‡q KZUzKz IRb nviv‡eb? c„w_exi fi I e¨vmva© h_vµ‡g Puv‡`i fi I e¨vmv‡a©i 81 Ges 4 ¸Y| NOW PRACTICE SOLVE : 01. Wm = (e¨vmva© ¸Y) 2 f‡ii ¸Y We = (4) 2 81 648 = 128; nviv‡bv IRb = We – Wm = 648 – 128 = 520N REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF RU QUESTION 01. c„w_ex‡Z GKwU e¯‘i IRb 196N n‡j Gi fi n‡e- [RU. 17-18] A. 0.2 kg B. 2 kg C. 20 kg D. 200 kg kg g w w mg m e e e e 20 9.8 196 02. †Kvb e¯‘i fi 100kg n‡j Dnvi IRb KZ? [RU. 09-10] A. 10 B. 98 C. 100 D. 980 w = mg = 1009.8 = 980N STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. 2kg f‡ii †Kvb e¯‘i Puv‡` IRb KZ wbDUb n‡e? [KU. 17-18] A. 1.6 B. 3.2 C. 9.8 D. 19.6 e m w w = 6 1 e m e m g g mg mg w mg N w w e e e m 2 9.8 19.6 6 Puv‡` IRb, wm= 6 1 19.6 = 3.2 N 02. f~-c„‡ô †Kvb †jv‡Ki IRb, 648N n‡j wZwb Puv‡` wM‡q KZUzKz IRb nviv‡eb| c„w_exi fi I e¨vmva© h_vµ‡g Puv‡`i fi I e¨vmva© 81 ¸Y I 4 ¸Y| [KU. 12-13; BU. 14-15] A. 520N B. 225N C. 250N D. 252N e m W W (e¨vmv‡a©i ¸b) 2 f‡ii ¸b ev, 81 4 648 2 Wm ev, Wm=128N nvivb IRb = 648 – 128 = 520N BU 01. P‡›`ªi fi m c„w_exi fi M-Gi 1/80 fvM I P‡›`ªi e¨vmva© r c„w_exi e¨vmva© R-Gi 1/4 fvM| P›`ªc„‡ô AwfKl©R Z¡i‡Yi gvb KZ? [RU. 15-16] A. 1 5 ge B. 1 6 ge C. 1 15ge D. 1 16 ge e m g g = f‡ii ¸Y (e¨vmv‡a©i ¸Y) 2 2 4 1 80 1 e m g g ev, gm = 16 80 ge = 1 5 ge 02. f~-c„‡ô GKwU e¯‘i IRb 90 N| c„w_exi e¨vmva© R n‡j c„w_exi †K›`ª †_‡K 3R `~i‡Z¡ e¯‘wUi IRb KZ ? [BU. 13-14] A. 10N B. 30N C. 90N D. 270N 2 c e e c R R w w ev, 2 3 1 90 wc ev, Wc = 10N
246 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 PART B Analysis of Science & Technology Question SUST 01. c„w_ex‡Z †Zvgvi IRb 50kg-wt g½j MÖ‡n KZ? g½j MÖ‡ni fi c„w_exi f‡ii 1 10 Ges g½j MÖ‡ni e¨vm c„w_exi e¨vmv‡a©i A‡a©K| [SUST. 11-12] A. 20kg-wt B. 25kg-wt C. 30kg-wt D. 40kg-wt E. 100kg-wt e m w w f‡ii ¸b (e¨vmv‡a©i ¸b) 2 ev, 10 4 2 1 10 1 2 e m w w ev, 10 4 50 wm ev, w kg wt m 20 02. P›`ªc„‡ô †Kvb e¯‘i IRb c„w_ex c„‡ôi IR‡bi Qqfv‡Mi GKfvM| P›`ª Ges c„w_exi Mo NbZ¡ mgvb a‡i wb‡j P‡›`ªi e¨vmva© KZ? [SUST. 10-11] A. 100km B. 500km C. 1000km D. 2450km e m e m g g w w c m R R (mylg Nb‡Z¡i †¶‡Î) ev, 6 6000 1 Rm ev, Rm 1000km 03. c„w_ex c„‡ô GKwU e¯‘i IRb 10 N n‡j c„w_exi †K‡›`ª Zvi IRb KZ? [SUST. 06-07] A. 10 kg B.10 N C. 0 N D. 0 kg c„w_exi †K‡›`ª, g = 0 IRb, W = mg = 0 N 04. GKwU wjdU 1 m/s2 Z¡i‡b 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 W Wh = g g + a Wh W = g + a g Wh W = 9.8 + 1 9.8 Wh W = 10.8 9.8 Wh W = 1.101 1 wh = 1.101 × w × 100% = 110.1% AvbygvwbK IRb e„w×: (110w 100w)% = 10% DU Technology 01. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظY wKš‘ fi A‡a©K| H MÖ‡ni c„‡ôi AwfKl©R Z¡iY KZ? [DU-Tech. 2020-21] A. 1.0g B. 0.5g C. 0.25g D. 0.125g S D info Avgiv Rvwb, c„w_exi AwfKl©R Z¡iY, ge = g = GM R 2 ge gp = GMe Re 2 Rp 2 GMp = Me Me 2 (2Re) 2 Re 2 ge gp = 8 1 gp = g 8 = 0.125 g STEP 03 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. GKwU wjdU 15ms-1 MwZ‡Z Dc‡i DV‡Q| 60kg f‡ii GKRb †jvK wjd‡U Ae¯’vb Ki‡j wjd‡Ui Dci Zvi cÖZxqgvb IRb n‡e-[BUET: 10-11] A. 588N B. 900N C. 750N D. 800N F = mg = 609.8 = 588N KUET 01. c„w_ex‡Z GKwU e¯‘i IRb 180kg| g½j MÖ‡ni fi c„w_exi f‡ii 1 9 Ges e¨vmva© 1 2 n‡j, g½jMÖ‡ni e¯‘wUi IRb KZ? [KUET. 15-16] A. 100kg wt B. 180kg wt C. 80kg wt D.1620kg wt E. 20kg wt Wm We = GMm R 2 m GMe R 2 e = Mm Re Re Rm 2 = 1 9 4 Wm = 180 4 9 = 80kg wt STEP 04 ANALYSIS OF HSC BOARD QUESTION 01. c„w_ex‡Z †Kv‡bv e¯‘i IRb 20N n‡j Puv‡` KZ? [wm.‡ev. 2019] A. 100N B. 39.2N C. 20N D. 3.33N S D info Fm = Fe 6 = 20 6 = 3.33N Concept 7 mylg NbZ¡ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. M = gR2 G = 4R 2 3 02. g = 4 3 RG CONCEPTUAL MATH MEx 01 mylg Nb‡Z¡i `ywU MÖ‡ni e¨vmv‡a©i AbycvZ 2:3 n‡j, MÖn `ywUi c„‡ôi Z¡i‡Yi AbycvZ KZ? g = 4 3 RG g Ges R mgvbycvwZK m¤ú‡K© we`¨gvb| g1 : g2 = 2 : 3 MEx 02 c„w_exi e¨vmva© R = 6.4 106 , G = 6.710–11 Gm. AvB GKK Ges AwfKl©xq Z¡iY g= 9.81ms –2 n‡j, Mo c„w_exi Mo NbZ¡ KZ? = 3g 4RG = 3g 43.1416(6.4106 )(6.710–11) = 5.5 103 kgm-3 NOW START PRACTICE 01. mylg Nb‡Z¡i `ywU MÖ‡ni e¨vmv‡a©i AbycvZ 4:5 n‡j, MÖn `ywUi c„‡ôi Z¡i‡Yi AbycvZ KZ? 02. c„w_ex c„ô †_‡K 300km †fZ‡i AwfKl©xq Z¡i‡Yi gvb wbY©q Ki| c„w_exi e¨vmva© 6.4 103 km, NbZ¡ 5.5 103 kgm–3 | 03. f~-c„‡ô gva¨vKl©YRwbZ Z¡iY g Gi gvb 9.8 ms-2 | c„w_exi mgvb AvK…wZi wKš‘ c„w_exi wظY Nb‡Z¡i Aci GKwU MÖ‡n g Gi gvb KZ? NOW PRACTICE SOLVE : 01. g2: g1 = 5 : 4 [g ] 02. g = 3 4 (R – D) G = 9.4 ms–2 03. g = 4 3 G R gp gE = p E ev, gp 9.8 = 2 1 ev, gp = 19.6ms–2
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 247 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. c„w_exc„‡ô gnvKlx©q †¶Î cÖvej¨ g KvíwbK GKwU MÖ‡ni NbZ¡ hw` c„w_exi Nb‡Z¡i mgvb nq Ges e¨vmva© hw` wظY nq Z‡e GB MÖ‡ni c„‡ô gnvKlx©q †¶Î cÖvej¨ KZ? [DU. 12-13] A. g B. 2g C. 4g D. 8g mylg Nb‡Z¡i †¶‡Î, P E P E R R E E ev, 2 1 EP g ev, EP = 2g 02. `ywU MÖ‡ni NbZ¡, mylg Ges mgvb, wKš‘ cÖ_gwUi e¨vmva© wØZxqwUi wظY| cÖ_g MÖ‡ni Dcwifv‡Mi Ges wØZxq MÖ‡ni Dcwifv‡Mi 'g' Gi AbycvZ n‡jv : [DU. 03-04] A. 2 : 1 B. 1 : 2 C. 4 : 1 D. 8 : 1 AwfKlx©q Z¡iY, g = 3 4 RG; MÖn `ywUi NbZ¡ mylg| g αR ; 1 2 R R g g 1 2 1 2 = 2:1 STEP 02 ANALYSIS OF RU QUESTION 01. c„w_ex‡K 6.4106m e¨vmv‡a©i Ges 55g/cc NY‡Z¡i †MvjK ai c„‡ô AwfKl©R Z¡iY n‡e- [RU. 11-12] A. 9.83ms-2 B. 9.0ms-2 C. 9.9ms-2 D. 8.9ms-2 g GR 3 4 11 6 3 6.67 10 6.4 10 5.5 10 3 4 = 9.83 Concept 8 gnvKl©xq cÖvej¨, gnvKl©xq wefe msµvšÍ FORMULA 01. cÖvej¨, E = GM d 2 ; wefe, V = GM R [Amx‡g gnvKl©xq wef‡ei gvb m‡e©v”P, GB m‡e©v”P gvb n‡jv k~b¨ (0)] 02. E = g = GM R 2 = 3 4 RG or, E g R [mylg Nb‡Z¡i †ÿ‡Î] CONCEPTUAL MATH MEx 01 R e¨vmv‡a©i c„w_exi c„‡ô AwfKl© wefe V n‡j c„ô n‡Z R D”PZvq wef‡ei gvb KZ? c„‡ô wefe, V = – GM R ; R D”PZvq wef‡ei gvb, V = – GM R + R = – GM 2R = V 2 MEx 02 c„w_ex‡K 5.5103 kgm–3 Mo Nb‡Z¡i •Zwi 6.4106 m e¨vmv‡a©i GKwU †MvjK wn‡m‡e we‡ePbv K‡i Gi c„‡ô wefe wbY©q Ki| Avgiv Rvwb, V = – R GM wKš‘ GLv‡b c„w_exi fi, M = 3 4 R 3 V = – 3R 4G R 3 = – 3 4 GR2 = – 3 4 3.146.710–11(6.4106 ) 2 5.5103 = – 6.32107 Jkg–1 MEx 03 c„w_exc„‡ô gnvKl©xq †¶Î cÖvej¨ E| KvíwbK GKwU MÖ‡ni NbZ¡ hw` c„w_exi Nb‡Z¡i mgvb nq Ges e¨vmva© hw` A‡a©K nq Z‡e GB MÖ‡ni c„‡ô gnvKl©xq †¶Î cÖvej¨ KZ? gnvKl©xq cÖvej¨, E = g g = 3 4 RG; MÖn `ywUi NbZ¡ mgvb| E g R; E 2 1 E R R E 1 1 2 2 MEx 04 M1 f‡ii GKwU e¯‘ n‡Z r1 `~i‡Z¡ cÖvej¨, M2 f‡ii e¯‘ n‡Z r2 `~i‡Z¡ cÖve‡j¨i A‡a©K| M1 I M2 Gi AbycvZ 1:10 n‡j r1 I r2 Gi AbycvZ KZ? gnvKl©xq cÖvej¨, E = 2 R GM 1 2 E E = 1 2 M M 2 2 1 R R 2 1 R R = 2 1 1 2 M M E E = 10 1 2 = 0.447 NOW START PRACTICE 01. c„w_exc„‡ô gnvKlx©q †¶Î cÖvej¨ g. KvíwbK GKwU MÖ‡ni NbZ¡ hw` c„w_exi Nb‡Z¡i mgvb nq Ges e¨vmva© hw` wظY nq Z‡e GB MÖ‡ni c„‡ô gnvKlx©q †¶Î cÖvej¨ KZ? NOW PRACTICE SOLVE : 01. gnvKl©xq cÖvej¨, E = g g = 3 4 RG; MÖn `ywUi NbZ¡ mgvb| E g R; E 2E R R E 1 1 2 2 =2g
248 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 RU QUESTION 01. f~-c~‡ôi †Kvb we›`y‡Z 10kg f‡ii e¯‘i AwfKl©R Z¡i‡Yi gvb 9.8ms2 n‡j H we›`y‡Z gnvKl©xq cÖve‡j¨i gvb n‡e- [RU. 12-13] A. 980O/Kg B. 98N/Kg C. 9.80N/Kg D. 0.98N/Kg AwfKl©R Z¡iY = gnvKlx©q cÖve‡j¨i msL¨vMZ gvb mgvb = 9.80 STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question BSMRSTU 01. R e¨vmv‡a©i c„w_exi c„‡ô AwfKl© wefe V n‡j c„ô n‡Z R D”PZvq wef‡ei gvb KZ? [BSMRSTU-H, 19-20] A. V/4 B. V/2 C. V D. 2V c„‡ô wefe, V = – GM R R D”PZvq wef‡ei gvb, V = – GM R + R = – GM 2R = V 2 NSTU 01. gnvKl©xq wefe I wefe kw³ Gi m‡e©v”P gvb k~b¨ nqÑ [NSTU-A, 19-20] A. Amx‡g B. †K‡›`ª C. fi‡K‡›`ª D. evqygÛ‡j Ans A STEP 03 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. GK e¨w³i IRb c„w_ex c„‡ô 785N Ges g½j c„‡ô 298N AwfKl©xq †¶‡Îi ZxeªZv KZ? [BUET: 09-10] A. 2.63/kg B. 6.09N/kg C. 3.72N/kg D. 9.8lb/kg E = F m = 298 785 9.8 = 298 9.8 785 = 3.72N/kg STEP 04 ANALYSIS OF HSC BOARD QUESTION wb‡Pi DÏxc‡Ki Av‡jv‡K 01 I 02 bs cÖ‡kœi DËi `vI : [iv. †ev. 2016] B I C we›`y‡Z h_vµ‡g m1 = 1 I m2 = 2kg f‡ii `ywU e¯‘ Av‡Q| AB = AC = 1m Ges BC = 2m Ges BD = CD 01. D we›`y‡Z gnvKlx©q †ÿÎ cÖve‡j¨i gvb A. G B. 2G C. 3G D. 4G S A info BD = CD = BC 2 = 2m 2 = 1m EB = Gm1 BD2 = G × 1 1 2 = G EC = Gm2 CD2 = G × 2 1 2 = 2G E = Ec – Ea = 2G – G = G A B C EB D EC 02. A I D we›`y‡Z gnvKlx©q wef‡ei AbycvZ A. 1 t 1 B. 1 t 2 C. 2 t 1 D. 3 t 1 S A info †h‡nZz AB + AC = BC †m‡nZz wÎfzRwU KvíwbK| A I D cÖK…Zc‡ÿ GKB we›`y, myZivs G‡`i gnvKlx©q wef‡ei AbycvZ n‡e 1:1| 03. 2 kg f‡ii †Kv‡bv e¯‘ n‡Z 2 m `~‡i †Kv‡bv we›`yi gnvKl©xq wefe KZ? [G = 6.673 10–11 Nm2 kg–2 ] [P. †ev. 2015] A. – 6.673 10–11 Jkg–1 B. – 3.3365 10–11 Jkg–1 C. 6.673 10–11 Jkg–1 D. 3.3365 10–11 Jkg–1 S A info V = – GMe r = – 6.673 × 10–11 × 2 2 = – 6.673×10–11 Jkg–1 wb‡Pi wPÎ †_‡K 04 I 05 bs cÖ‡kœi DËi `vI : [w`. †ev. 2016] c„w_exi fi Me = 61024kg 04. B I D Kÿc‡_i `~iZ¡ KZ? A. 1.02103 m B. 1.92106 m C. 1.02105 m D. 1.02106 m S B info – G Me rD = – 41 × 106 rD = 9.765 × 106 m – G Me rB = – 51 × 106 rB = 7.85 × 106 m rD – rB = 1.92 × 106 m 05. mKj Kÿc‡_i †KŠwYK †eM mylg n‡j DÏxc‡Ki †Kvb Kÿc‡_ †Kvb e¯‘i †K›`ªgyLx ej me©vwaK n‡e? A. F B. A C. B D. D Ans D 06. gnvKl©xq cÖvej¨ I gnvKl©xq wef‡ei g‡a¨ m¤úK© n‡jvÑ [h. †ev. 2016] A. E = dV dr B. F = Vr C. E = – dV dr D. E = V r Ans C 07. gnvKlx©q wef‡ei †ÿ‡Î- [wm. †ev. 2017] A. V = – GM r B. Gi GKK Jkg–1 C. GwU GKwU †f±i ivwk D. A I B DfqB Ans D
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 249 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 9 gyw³‡eM msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. gyw³‡eM, Ve = 2gR = 2GM R MÖ‡ni fi MÖ‡ni e¨vmva© 02. vp ve = gpRp geRe 03. vm ve = gmRm geRe CONCEPTUAL MATH MEx 01 g½j MÖ‡ni e¨vm 6000 km. Ges c„‡ô AwfKlx©q Z¡iY 3.8ms2 g½j MÖ‡n gyw³ †eM KZ? e¨vm d = 6000km, e¨vmva©: r = 3000km ve = 2gR = 2 3.8 3 106 = 4.77kms-1 NOW START PRACTICE 01. c„w_exi AwfKl©xq Z¡iY g = 9.8 ms–2 Ges e¨vmva© R = 6400 km n‡j, c„w_exi gyw³‡eM KZ n‡e? 02. ey‡ai fi I e¨vmva© h_vµ‡g 31026 kg Ges 8108 m n‡j, ey‡ai gyw³ †eM wbY©q Ki| 03. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظY| D³ MÖ‡ni c„‡ô AwfKl©R Z¡iY c„w_exi AwfKl©R Z¡i‡Yi AvU ¸Y| D³ MÖ‡n gyw³ †eM c„w_exi Zzjbvq KZ ¸Y? NOW PRACTICE SOLVE : 01. c„w_exi gyw³‡eM, ve = 2gR = 29.86.4106 =11.2103 ms–1 = 11.2 kms–1 02. ey‡ai gyw³ †eM, v = R 2GM = 2 1 8 11 26 8 10 2 6.67 10 3 10 = 7.1103 ms–1 = 7.1 kms–1 03. gyw³ †eM, v = 2gR vp = e p e p R R g g ve = 82ve = 4 ve REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. me©wb¤œ KZ †e‡M f~-c„ô n‡Z m f‡ii GKwU e¯‘‡K Dc‡ii w`‡K wb‡¶c Ki‡j Zv Avi KL‡bv wd‡i Avm‡e bv? [DU. 15-16] A. (2gR) B. (2) gR C. gR D. 2(gR) hw` Dr‡¶cY †eM gyw³ †eM A‡c¶v †ewk nq, Zv Avi c„w_ex‡Z wd‡i Av‡m bv| gyw³‡eM, v = 2gR ;mgxKi‡Y m bv _vKvq Avgiv ej‡Z cvwi †h, gyw³ †eM e¯‘i f‡ii Dci wbf©i K‡i bv| e¯‘ †QvU ev eo hvB †nvK bv †Kb, gyw³ †eM GKB n‡e| 02. f~-c„ô n‡Z Aí D”PZvq Ges f~-c„‡ôi mgvšÍiv‡j GKwU b‡fvhvb KZ `ªæwZ‡Z Pj‡j hvÎxiv IRbnxbZv Abyfe Ki‡e? c„w_exi e¨vmva© = 6400 km Ges g = 9.8 ms-2 [DU. 06-07] A. 7.9 km s-1 B. 7.1 km s-1 C. 3.5 km s-1 D.3.1 km s-1 v gR 6 9.86.410 7.9km/sec 03. g½j MÖ‡ni e¨vmva© 3.4x106 m Ges ga¨vKl©Y RwbZ Z¡iY 3.7 m/sec2 n‡j g½j MÖ‡n †Kvb e¯‘i gyw³‡eM KZ? [DU. 00-01] A. 12.58 km/s B. 3.55 km/s C. 5.02 km/s D. 11.20 km/s. gR m v 2 6 23.43.710 5.02km/sec STEP 02 ANALYSIS OF JU QUESTION 01. c„w_exc„‡ô AwfKl©R Z¡iY hw` GKB _v‡K Ges gyw³‡eM 0.5% evov‡Z c„w_exi e¨vmv‡a©i kZKiv cwieZ©b Ki‡Z n‡e- [JU-A, Set-D. 19-20] A. 1% e„w× B. 2% n«vm C. 2.5% e„w× D. 3% n«vm gyw³‡eM, V = 2gR ; V R gyw³‡eM 0.5% evov‡Z e¨vmva© = 1% e„w× Ki‡Z n‡e| GLv‡b, v1 = v, v2 = 1.005v R2 = v2 v1 2 R1 = 1.005v v 2 R1 = 1.01 R1 e„w× = (1.01 – 1) 100% = 1% 02. c„w_exi AwfKl©xq Z¡iY g = 9.8ms-2Ges e¨vmva© R = 6400km| GKwU e¯‘i gyw³ †eM KZ? [JU. 16-17; RU. 16-17] A. 11.2 km s-1 B. 12.2 km s-1 C. 16.2 km s-1 D. 25 km s-1 Ans A STEP 03 ANALYSIS OF RU QUESTION 01. aiv hvK, AwfKl©R Z¡i‡Yi gvb 10 ms–2 | f~-c„ô n‡Z 5m Dci †_‡K GKwU e¯Íy‡K wb‡P co‡Z w`‡j f~wg ¯úk© Kivi gyn~‡Z© Zvi †eM KZ ms1 ? [RU. 18-19] A. 5 B. 9.8 C. 10 D. 15 v 2 = u 2 + 2gh v = 2gh = 10 Note: f~wg ¯úk© Kiv gyn~‡Z© †eM PvB‡jB, v = 2gh 02. GKwU MÖ‡ni e¨vm 4108 m Ges fi 2.21028 kg| D³ MÖ‡n gyw³ †eM KZ? [RU. 17-18] A. 12.14 km/s B. 122.5 km/s C. 130.6 km/s D. 110.6 km/s e¨vmva©, R=(4108 /2)= 2108 m gyw³ †eM, v = R 2GM = 2 1 8 11 28 2 10 2 6.67 10 2.2 10 = 121.12ms–1
250 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. c„w_exi e¨vmva© GKwU MÖ‡ni e¨vmv‡a©i wظY| wKš‘ MÖ‡ni c„‡ôi AwfKl©R Z¡iY c„w_exi AwfKl©R Z¡iY AvU¸Y| D³ MÖ‡ni gyw³‡eM c„w_exi gyw³‡e‡Mi- [RU. 15-16] A. `yB¸Y B. Pvi¸Y C. wZb¸Y D. †KvbwUB bq vp ve e e p p g R g R Ue vp 4 2 1 2 8 1 ev, Vp = 2Ve 04. g½j MÖ‡ni e¨vm I c„‡ô AwfKl©xq Z¡iY h_vµ‡g 6000km I 3.8ms-2 n‡j gyw³‡eM n‡e- [RU. 12-13] A. 47.7kms-1 B. 7.7kms-1 C. 477kms-1 D. 4.77kms-1 e¨vm d = 6000km, e¨vmva©: r 3000km Vc 2gR = 2 3.8 3 = 4.77kms-1 05. c„w_exi e¨vmva© 6400km. 500kg f‡ii GKwU e¯‘i gyw³‡eM wbY©q Ki? [RU. 10-11] A. 11.2 kms–1 B. 32 kms–1 C. 11200 kms–1 D. ‡KvbwUB bq ve 2gR gyw³ †e‡M e¯‘i f‡ii Dci wbf©ikxj bq| STEP 04 ANALYSIS OF CU QUESTION 01. f~-c„‡ô e¯‘i gyw³‡eM 11.2 wK.wg/ †m. n‡j c„w_exi †P‡q 8 ¸Y fvix Ges wظY e¨v‡mi GKwU MÖ‡ni c„‡ô gyw³‡eM KZ? [CU. 16-17] A. 5.6 wK. wg./‡m. B. 11.2 wK. wg./‡m. C. 22.4 wK. wg./‡m. D. 44.8 wK. wg./‡m. E. 179.2 wK. wg./‡m. Ve= 2GM R Vp Ve = Mp Me Re Rp Vp Ve = 8 1 2 vP = 4 ve = 2Ve = 11.2 2 = 22.4 kms1 02. g½j MÖ‡ni e¨vm 6000m Ges c„‡ô AwfKlx©q Z¡ib 2 3.8 ms g½j MÖ‡n gyw³ †eM KZ? [CU. 10-11] A. 4.77 kms–1 B. 5.77 kms–1 C. 4.88 kms–1 D. 5.88 kms–1 Ve 2gR 2Rg 6 3.8 = 4.77kms–1 STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظb| D³ MÖ‡ni c„‡ô AwfKl©R Z¡ib c„w_exi AwfKl©R Z¡i‡bi 8 ¸b| D³ MÖ‡ni gyw³ †eM c„w_exi Zzjbvq KZ ¸b? [JnU. 08-09] A. 2 ¸b B. 2 ¸b C. 8 ¸b D. 4 ¸b MÖ‡ni gyw³ †eM, ve = 28 = 4 PART B Analysis of Science & Technology Question SUST 01. c„w_exi †K›`ª †_‡K c„w_ex‡K wN‡i N~Y©vqgvb GKwU DcMÖ‡ni `~iZ¡ wظb Kiv n‡j Zvi MwZ‡eM- [SUST. 10-11] A. 2¸b evo‡e B. 2 ¸Y Kg‡e C. 2 ¸b evo‡e D. 2 ¸b Kg‡e| 2 1 1 2 R R v v ev, 2 2 1 v v ev, v2 = 2 v JUST 01. c„w_ex‡Z gyw³‡eM Ve; GKwU i‡K‡Ui †eM (V) KZ n‡j GwU Puv‡`i gZ AvPiY Ki‡e? [JUST-B, 19-20] A. V = Ve /2 B. V = Ve / 2 C. V = Ve /4 D. V = Ve Dr‡ÿcY †eM, v = Ve 2 ev 7.88 kms–1 n‡j Zv e„ËvKvi c‡_ c„w_ex‡K cÖ`wÿY Ki‡e Ges Puv‡`i gZ DcMÖ‡n cwiYZ n‡e| MBSTU 01. e„n¯úwZi fi I e¨vmva© h_vµ‡g 1.9 10 27 kg Ges 7 107 m n‡j Gi gyw³‡eM n‡e- [MBSTU-B, Set-2 19-20] A. 8 104 ms1 B. 7 104 ms1 C. 6.02 104 ms1 D. 60 104 ms1 Ve = 2GM R = 2 6.673 10–11 1.9 1027 7 107 = 6.02 104 ms–1 MÖ‡ni bvg gyw³ †e‡Mi gvb c„w_ex 11.2kms–1 /7miles-1 /25000mileh-1 g½j MÖn 5.1kms-1 [Zcb m¨vi] /4.77kms-1 [Avgxi m¨vi] eya 7.1kms-1 e„n¯úwZ 6.02104ms-1 ASPECT SPECIAL: wewfbœ MÖ‡ni gyw³‡eM gyL¯’ ivL‡j A¼ Kivi cÖ‡qvRb n‡e bv| KviY GKwU wbw`©ó MÖ‡ni gyw³‡eM wbw`©ó| 02. f~-c„ô n‡Z 45° †Kv‡Y Dc‡ii w`‡K wbw¶ß †Kvb e¯‘i gyw³‡eM KZ n‡e? [MBSTU. 15-16] A. 11.2 2 Km s1 B. 11.2Km s 1 C. 11.2/ 2 Km s 1 D. 2 Km s 1 v = 2g sin R= 2 × 9.8 × sin 45 × 6400 = 11.2 2 kms 1 BSMRSTU 01. c„w_ex †_‡K V Avw`‡e‡M Ges f~-c„‡ôi mv‡_ 30 †Kv‡Y GKwU i‡KU‡K wb‡ÿc Kiv nj| b~b¨Zg †eM KZ n‡j i‡KUwU c„w_exi AwfKl© ej‡K AwZµg Ki‡Z cvi‡e? [BSMRSTU-C, 19-20] A. 11.2 kms1 B. 15.56 kms1 C. 22.4 kms1 D. 5.6 kms1 c„w_exi gyw³‡e‡Mi mgvb †eM w`‡q wb‡ÿc Ki‡j Zv c„w_exi AwfKl© ej AwZµg Ki‡Z cvi‡e| 02. †Kvb e¯‘i gyw³‡e‡Mi gvb A‡cÿvK…Z fvix e¯‘i †P‡q- [BSMRSTU-C, 19-20] A. †ekx n‡e B. Kg n‡e C. mgvb n‡e D. wbw`©ó bq gyw³‡eM e¯‘i f‡ii Dci wbf©i K‡i bv| GwU MÖ‡ni f‡ii Dci wbf©ikxj Ges GKwU wbw`©ó MÖ‡ni gyw³‡eM wbw`©ó| STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. c„w_ex †_‡K u Avw`‡eM Ges f~-c„‡ôi mv‡_ 300 †Kv‡Y GKwU i‡KU wb‡¶c Kiv nj| b~¨bZg †eM KZ n‡j i‡KUwU c„w_exi AwfKl©R ej‡K AwZµg Ki‡e? [BUET. 11-12] A. 11.2ms–1 B. 1 3 22.4kms1 C. 22.4kms–1 D. 5.6kms1 S B info vsin = 2gR ev, v = 2gR sin = 2gR sin30 = 22.4kms1 KUET 01. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظY| D³ MÖ‡ni AwfKl©R Z¡iY c„w_exi AwfKl©R Z¡i‡Yi AvU¸Y| D³ MÖ‡ni gyw³‡eM c„w_exi gyw³ †e‡Mi Zzjbvq KZ¸Y Zv wbY©q Ki? [KUET. 04-09] A. 2 ¸Y B. 4 ¸Y C. 8 ¸Y D. 10 ¸Y E. 16 ¸Y MÖ‡ni gyw³ †e‡Mi ¸Y = Rn gn = 28 = 4 ¸Y
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 251 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES CUET 01. c„w_exi AwfKl©xq Z¡iY 980cm/sec2 Ges GKwU e¯‘i gyw³ †eM 11.2 km/ sec c„w_exi e¨vmva© KZ? [CUET. 10-11] A. 6400km B. 640km C. 64000km D. None Ans A STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. g½j MÖ‡ni e¨vm 6000 km, Gi c„‡ô g-Gi gvb 3.8ms–2 n‡j g½j MÖn †_‡K †Kv‡bv e¯‘i gyw³ †eM KZ? [MAT. 18-19] A.9.7 kms–1 B. 4.77 kms–1 C. 3.77 kms–1 D. 11.2 kms–1 Ve = 2gR = 2 3.8 6000 = 4.77kms–1 [we. `ª.: wewfbœ MÖ‡n gyw³‡e‡Mi gvb wbw`©ó ZvB gyL¯Í ivLv fvj|] DAT 01. Puv‡`i gyw³ †eM KZ? [DAT. 18-19] A. 5 km/sec B. 11.2 km/sec C. 2.4 km/sec D. 4.3 km/sec GK`g mwVK DËi Ack‡b †`qv †bB| KvQvKvwQ DËi 2.4kms–1 STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. gyw³‡eM- [Xv. †ev. 2019] A. e¯‘i f‡ii Dci wbf©i K‡i B. Gi gvb c„w_ex c„‡ô 11.2 kms–1 C. AwfKl©R Z¡i‡Yi Ici wbf©i K‡i D. B I C DfqB S D info Ve = 2gR = 11.2 kms–1 wb‡Pi DÏxc‡Ki Av‡jv‡K 02 I 03 bs cÖ‡kœi DËi `vI: [Kz. †ev. 2015] †Lvjv gv‡V iwdK GKwU e¯Íy‡K we‡kl hvwš¿K e¨e¯’vq Dc‡i wb‡ÿc Kivi †Póv Ki‡Q| eÜz iwng Zuv‡K mZK© K‡i e‡j †ewk †Rv‡i wb‡ÿc Ki‡j e¯ÍywU Avi c„w_ex‡Z wd‡i Avm‡e bv| c„w_exi e¨vmva© = 64106 m Ges g = 9.78 ms-2 02. c„w_ex‡Z gyw³ †eM KZ? A. 11.19 ms-1 B. 11.19 kms-1 C. 11.20 kms-1 D. 11.20 ms-1 S B info Ve = 2gR = 2×9.78 × 6.4 × 106 = 11.19 kms-1 03. wK Kvi‡Y eÜz iwn‡gi Avk¼vwU mZ¨ n‡e? A. wb‡ÿc gyn~‡Z© e¯ÍywUi Dci jwä ej abvZ¥K n‡j B. e¯ÍywUi MwZkw³ K…Z Kv‡Ri mgvb n‡j C. wb‡ÿc gyn~‡Z© e¯ÍywUi Dci jwä ej k~b¨ n‡j D. e¯ÍywUi MwZkw³ cÖ‡qvRbxq K…Z Kv‡Ri Kg n‡j Ans B 04. g½jMÖ‡ni c„‡ô g = 3.8 ms2 Ges Gi e¨vmva© 3103 km. g½jc„‡ô gyw³‡eM KZ n‡e? [wm. †ev. 2017] A. 4.0 kms1 B. 4.8 kms1 C. 7.8 kms1 D. 11.0 kms1 S B info ve = 2gR = 2 × 3.8 × 3 × 106 = 4.8 kms–1 05. g½jMÖ‡ni e¨vm 6000 km Ges Gi c„‡ô g Gi gvb 3.8 ms–2 n‡j H MÖ‡ni c„ô n‡Z GKwU e¯‘i gyw³‡eM KZ? [wm. †ev. 2017; h. †ev. 2016] A. 11.2 kms–1 B. 9.7 kms–1 C. 3.85 kms–1 D. 4.77 kms–1 S D info ve = 2gR= gD = 3.8 × 6000 × 103 = 4.77 kms–1 06. c„w_exi c„‡ô AwfKl©R Z¡iY hw` GKB _v‡K Ges c„w_exi e¨vmva© 1% e„w× †c‡j gyw³‡e‡Mi kZKiv cwieZ©bÑ [e. †ev. 2016] A. 1% evo‡e B. 1% Kg‡e C. 0.5% evo‡e D. 0.5% Kg‡e S C info Ve R; R2 = (100 + 1)% = 101% = 1.01 V2 = n × R = 1.01R = 1.005R V = (1.005 – 1) × 100% = 0.5% Concept 10 K…wÎg DcMÖ‡ni †eM I Zv‡`i ch©vqKvj msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. K…wÎg DcMÖ‡ni •iwLK †eM: v = R h GM Ges AveZ©b Kvj: T = v 2(R h) = 2 (R + h) R + h GM 02. f~-c„‡ôi Lye wbKU w`‡q †Kvb K…wÎg DcMÖn c„w_ex‡K cÖ`wÿb Ki‡j, v = GM R = gR Ges T = 2 g R 03. AveZ©b Kvj I D”PZvi g‡a¨ m¤úK©, h= GMT2 4 2 1 3 – R 04. a = 2R= 2 T 2 R = 4 2 T 2 R 05. MÖ‡ni e„ËvKvi MwZi Rb¨ †K›`ªgyLx ej, Fc = mv2 r = m r . GM R + h CONCEPTUAL MATH MEx 01 GKwU K…wÎg DcMÖn 7000km e¨vmva© wewkó e„ËvKvi Kÿc‡_ c„w_ex‡K cÖ`wÿY Ki‡Q, ch©vqKvj 2h n‡j, †K›`ªgyLx Z¡iY KZ = 2R= 2 T 2 R = 4 2 T 2 R = 4 (3.1416) 2 (23600) 2 7000 103 = 5.3ms–2 (Ans) MEx 02 GKwU K…wÎg DcMÖn c„w_exi Pvwiw`‡K f~-c„ô n‡Z 900 km Dci †_‡K e„ËvKvi c‡_ Nyi‡Q| c„w_exi e¨vmva© 6400 km Ges f~-c„‡ô ga¨Kl©Y RwbZ Z¡ib 9.81 m/sec2 n‡j DcMÖnwUi †eM KZ ? †ekx D”PZvq, v = GM R + h = gR2 R + h = R g R + h 7.3 9.81 / 6.4 6.4 0.9 9.81 6.4 v km s 7.42km/s
252 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. GKwU K…wÎg DcMÖn 7200km e¨vmva© wewkó e„ËvKvi Kÿc‡_ c„w_ex‡K cÖ`wÿY Ki‡Q, ch©vqKvj 2h n‡j, †K›`ªgyLx Z¡iY KZ? 02. 8000 km e¨vmv‡a©i Kÿc‡_ GKwU DcMÖ‡ni †eM KZ? 03. c„w_ex ‡_‡K 1600km D”PZvq GKwU K…wÎg DcMÖn c„w_ex‡K †K›`ª K‡i e„ËvKvi c‡_ `w¶Y Ki‡Q| Gi †eM †ei Ki04. 8000 km e¨vmv‡a©i Kÿc‡_ GKwU NyY©vqgvb DcMÖ‡ni c„w_ex n‡Z KZ D”PZvq Aew¯’Z| NOW PRACTICE SOLVE : 01. a = 2R= 2 T 2 R = 4 2 T 2 R = 4 (3.1416) 2 (23600) 2 7200 103 = 5.48ms–2 02. V = R h gR2 = 9.8 × (6.4 × 106 ) 2 8 × 106 = 7.08 103 ms–1 03. V = R h gR2 = 9.8 × (6.4 × 10 6 ) 2 (6.4 × 106 ) + (1.6 × 106 ) = 7.08 103 04. R + h = 8000 km , h = 8000 – R = 1600 km REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU K…wÎg DcMÖn 7000km e¨vmva©wewkó e„ËvKvi Kÿc‡_ c„w_ex‡K cÖ`wÿY Ki‡Q| DcMÖnwUi ch©vqKvj 2h n‡j †K›`ªgyLx Z¡iY KZ? [DU. 16-17] A. 1.331 m/s2 B. 2.663 m/s2 C. 5.325 m/s2 D. 10.650 m/s2 a = 2R = 2 T 2 R = 4 2 7000000 (2 3600) 2 = 5.3 m/s2 STEP 02 ANALYSIS OF JU QUESTION 01. GKwU K…wÎg DcMÖn c„w_exi Pviw`‡K me©v‡c¶v KZ Kg `ªæwZ‡Z cÖ`w¶Y Ki‡e? [JU. 13-14] A. 7.92 kms1 B. 6.4 kms1 C. 9.8 kms1 D. 320 ms1 DcMÖ‡ni ‡eM, v = gR = 9.864001000 = 7.92 kms–1 STEP 03 ANALYSIS OF RU QUESTION 01. GKwU K…wÎg DcMÖn c„w_exi c„ô †_‡K c„w_exi e¨vmv‡a©i wظY D”PZvq Ny‡i| H D”PZvq Gi MwZ‡eM KZ? [RU-C, Set-1. 19-20] A. gR B. 2 gR C. gR 3 D. gR 5 Avgiv Rvwb K…wÎg DcMÖ‡ni †eM, V = gR2 R + h = gR2 R + 2R = gR 3 02. GKwU K…wÎg DcMÖn c„w_exi c„ô †_‡K c„w_exi e¨vmv‡a©i A‡a©K D”PZvq Ny‡i| H D”PZvq Gi MwZ‡eM KZ? [RU. 16-17] A. 2gR B. 3gR C. 2 3gR D. 3 2gR DcMÖ‡ni †eM, v = R h gR2 = 3 2gR R gR R R gR 2 3 2 2 1 2 03. †Kvb GKwU D”PZv †hLv‡b AwfKl©R Z¡i‡Yi gvb gh = 8 ms–2.| ‡mLv‡b GKwU DcMÖ‡ni †eM 8 kms-1 | c„w_ex c„ô n‡Z KZ D”PZvq DcMÖnwU c„w_ex‡K cÖ`w¶b Ki‡Q? [c„w_exi e¨vmva© 6.4106 m] [RU. 16-17] A. 6400 km B. 3200 km C. 1600 km D. 1000 km DcMÖ‡ni †eM, v = g (R + h) 6 2 3 2 2 6.4 10 8 (8 10 ) ( ) R g v v g R h h h 8 10 6.4 10 1.6 10 1600km 6 6 6 STEP 04 ANALYSIS OF CU QUESTION 01. GKwU K…wÎg DcMÖn c„w_ex c„ô †_‡K 100km D”PZvq e„ËvKvi Kÿc‡_ Ae¯’vb Ki‡Q| c„w_exi fi 6 1024kg Ges e¨vmva© 6.4103 km| G = 6.67 10–11Nm2 kg2 n‡j DcMÖnwUi Kÿxq `ªæwZ KZ? [CU. 16-17] A. 8.91 kms1 B. 7.85 kms1 C. 11.19 kms1 D. 11.10 kms1 E. 9.81 kms1 V = R g R + h = 6.4 9.8 6.4 + 0.1 = 7.85 kms1 STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question BU 01. c„w_ex N~Y©‡Yi Rb¨ welye †iLvq Aew¯’Z †Kvb e¯‘i †K›`ªwfwËK Z¡iY KZ? (R = 6400km) [BU. 16-17] A. 9.8 m/s2 B. 4.9m/s2 C. 0.068m/s2 D. 0.034m/s2 a = 2R = 2 T 2 R = 2 86400 2 6.4 106 = 0.0338 = 0.034m/s2 PART B Analysis of Science & Technology Question SUST 01. c„w_exc„ô n‡Z 700.0km D‡×© GKwU K…wÎg DcMÖn c„w_ex‡K cÖ`w¶Y Ki‡Q| DcMÖnwUi Abyf~wgK ‡eM KZ ms–1 ? [SUST. 15-16] A. 4500 B. 4715 C. 5675 D. 7519 E. 8125 V = R R h g 6.4 0.7 9.8 6.4 = 7.5190km–1 =7519ms–1 PUST 01. c„w_exi †K›`ª †_‡K c„w_ex‡K wN‡i N~Y©vqgvb DcMÖ‡ni `~iZ¡ wظY Kiv n‡j Zvi MwZ‡eM- [PSTU. 14-15] A. 1.41 ¸Y evo‡e B. 1.41 ¸Y Kg‡e C. 2 ¸Y evo‡e D. 2 MY Kg‡e Ans B
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 253 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. c„w_exi e¨vmva© nVvr A‡a©K n‡q †M‡j w`‡bi mgqKvj n‡e- [PSTU. 14-15] A. 6hour B. 12hour C. 18hour D. †KvbwUB bq T = 24 – 2 2 24 = 24 6 = 18 N›Uv HSTU 01. f~-c„ó n‡Z 700km D”PZvi GKwU K…wÎg DcMÖn c„w_ex‡K cÖ`w¶Y Ki‡Q| K…wÎg DcMÖnwUi Abyf~wgK ‡eM KZ? (R = 6300km) [HSTU. 14-15] A. 6.44kms–1 B. 7.45kms–1 C. 8.45kms–1 D. 9.45kms–1 v = R R h g 6.3 0.70 9.8 6.3 7.0 9.8 6.3 = 7.454kms–1 STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION CUET 01. GKwU K…wÎg DcMÖn f~-c„ô †_‡K GKwU wbw`©ó D”PZvq 8km/Sec †e‡M Nyi‡Q, †hLv‡b AwfKl©R Z¡i‡Yi gvb gh = 8m/sec2 | f~c„ô †_‡K DcMÖnwUi D”PZv wbY©q Ki| [CUET. 15-16] A. 1600km B. 4000km C. 14400km D.8000km v = GM R + h = gh (R + h) R + h = 8000km h = 1600km BUTex 01. GKwU DcMÖn c„w_ex Z‡ji KvQ w`‡q Nyi‡Q| GwU‡K Amx‡g cvVv‡Z n‡j Gi MwZ evov‡Z n‡e- [BUTex. 16-17] A. 20% B. 30% C. 40% D.60% c„w_ex Z‡ji KvQ w`‡q Nyi‡j V = gR; Amx‡g cvVv‡Z n‡j, cÖ‡qvRbxq †eM, ve = 2gR MwZ evov‡Z n‡e = 2gR gR gR 100% =41.42% 40% 02. GKB Nb‡Z¡i `ywU MÖ‡ni e¨vmv‡a©i AbycvZ 2:1 n‡j G‡`i c„‡ô g Gi AbycvZ KZ n‡e? [BUTex. 16-17] A. 2:1 B. 1:2 C. 4:1 D.1:4 g = GM R 2 = 4 3 GR g1 g2 = R1 R2 = 2 1 03. hw` c„w_exi f‡ii wظY fi I 3 ¸Y e¨vmva© wewkó †Kvb MÖn _v‡K Zvn‡j D³ MÖ‡ni Z‡j 1kg f‡ii IRb KZ n‡e? [BUTex. 15-16] A. 1.1 N B. 2.2 N C. 4.4 N D. none of these g g = MR 2 R 2M = 2M(R 2 ) (3R) 2M = 2 9 g = 9.8 2 9 ms2 = 2.2ms2 W = mg = (1 2.2) = 2.2N STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. R I 4R e¨vmva©wewkó e„ËvKvi Kÿc‡_ cÖ`wÿYiZ `ywU K…wÎg DcMÖ‡ni ch©vqKv‡ji AbycvZ n‡eÑ [h. †ev. 2017] A. 8 : 1 B. 4 : 1 C. 1 : 4 D. 1 : 8 S D info T 2 R 3 T1 T2 = R1 R2 3 2 = R 4R 3 2 T1 T2 = 1 8 = T1 : T2 = 1 : 8 Dc‡ii DÏxcK jÿ Ki Ges 02 I 03 bs cÖ‡kœi DËi `vI: [Kz. †ev. 2015] fi M = 61024 kg, e¨vmva© R = 6.4106m 02. DcMÖnwUi Abyf~wgK †eM KZ? [Kz. †ev. 2015] A. 7509.43 ms1 B. 7510.43 ms1 C. 7508.43 ms1 D. 7507.43 ms1 S A info V = GM R + h = 6.673×10–11 × 6 × 1024 6.4 × 106 + 7 × 105 = 7509.43 ms–1 03. DcMÖnwUi ch©vqKvj KZ? [Kz. †ev. 2015] A. 1 hr. 39 min B. 1 hr. 40 min C. 1 hr. 41 min D. 1 hr. 42 min S A info T = 2 (R + h) R + h GM = 2 × 3.14 × (6.4 × 106 + 7×105 ) 6.4 × 106 + 7 × 105 6.673 × 10–11 × 6 × 1024 = 1 hr.39 min. 04. f~-c„ô n‡Z h D”PZvq c„w_ex‡K cÖ`wÿYiZ †Kv‡bv K…wÎg DcMÖ‡ni †eM- [wm.‡ev. 2017] A. v = GM R + h B. v = GM (R + h) 2 C. v = GM2 R + h D. v = GM R + h Ans D 05. †Kv‡bv e¯‘‡K gyw³‡e‡Mi KZ¸Y †e‡M wb‡ÿc Ki‡j K…wÎg DcMÖ‡n cwiYZ n‡e? [e. †ev. 2017] A. 1 2 ve B. 1 2 ve C. 2 ve D. 2 ve S A info mv2 R + h = GMm (R + h) 2 v = GM R + h v = g(R + h) = 2g (R + h) 2 = Ve 2 06. GKwU Kÿc‡_ AveZ©biZ `ywU DcMÖ‡ni GKwUi fi Ab¨wUi wظY n‡j fvix DcMÖ‡ni AveZ©bKvj AciwUi- [e. †ev. 2015] A. mgvb B. A‡a©K C. wظY D. Pvi¸Y Ans A 07. m~‡h©i Pviw`‡K c„w_exi Kÿc‡_i e¨vmva© 1.5 1011 m Ges AveZ©bKvj 3.14 107 †m., c„w_exi `ªæwZ KZ? [e. †ev. 2015] A. 2 10–7 ms–1 B. 4.7 103 ms–1 C. 15 102 ms–1 D. 30 103 ms–1 S D info V = 2 (R + h) T = 2×3.14 (6.4 × 106 + 1.5 × 1011) 3.14 × 107 = 30 × 103 ms–1
254 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 11 †Kcjv‡ii m~Î msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. 3 2 1 2 2 1 R R T T CONCEPTUAL MATH MEx 01 m~‡h©i Pvwiw`‡K ïµ I c„w_exi K¶ c‡_i e¨vmv‡a©i AbycvZ 54:75 c„w_ex‡Z 365 w`‡b GK eQi n‡j ﵇Z KZ w`‡b GK eQi n‡e? 2 3 e m e m R R T T ev, 2 3 75 54 365 Tm ev, 0.611 365 Tm T day m 223 [†R‡b ivLv fvj †h, ïµMÖ‡n 223 w`‡b 1 eQi] MEx 02 m~‡h©i gnvKl© e‡ji GMm r 2 Kvi‡Y MÖn¸‡jv m~h©‡K cÖ`w¶Y K‡i Mv2 r | kwbMÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i 3 ¸b n‡j m~h©‡K GKevi cÖ`w¶Y Ki‡Z kwbMÖ‡ni KZ eQi mgq jvM‡e? 3 2 1 2 2 1 R R T T 2 3 e s e s R R T T ev, Ts = RS RE 3 2 × Te = 3 1 3 2 × 1 = 5.2 eQi NOW START PRACTICE 01. g½j MÖn KZ eQ‡i GKevi m~h©‡K cÖ`w¶Y K‡i? (c„w_ex †_‡K m~‡h©i `~iZ¡ 150106 km. g½j †_‡K m~‡h©i `~iZ¡ 230106 km.) NOW PRACTICE SOLVE : 01. 2 3 e m e m R R T T ev, 3 2 6 6 150 10 230 10 1 Tm ev, Tm = 1.898 ev, Tm = 1.90 eQi REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF RU QUESTION 01. c„w_ex GK ïµ MÖ‡ni m~h©‡K cÖ`w¶Y Ki‡Z h_vµ‡g 365 Ges 224 w`b jvM‡j, m~h© n‡Z MÖn `ywUi `~i‡Z¡i AbycvZ n‡e- [RU. 12-13] A. 1.38 B. 1.63 C. 0.615 D. 2.653 1.384 224 365 3 2 3 2 2 1 2 1 T T R R 02. m~‡h©i gnvKl© e‡ji GMm r 2 Kvi‡Y MÖn¸‡jv m~h©‡K cÖ`w¶Y K‡i Mv2 r | kwbMÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i 3 ¸b n‡j m~h©‡K GKevi cÖ`w¶Y Ki‡Z kwbMÖ‡ni KZ eQi mgq jvM‡e? [RU. 11-12] A. 5 years B. 10 years C. 150.5 years D. 320.5 years E. 3000.5 years 3 2 1 2 2 1 R R T T 2 3 e s e s R R T T ev, e E S S T R R T . 2 3 = .1 1 3 2 3 = 5 eQi 03. m~‡h©i Pvwiw`‡K ïµ I c„w_exi K¶ c‡_i e¨vmv‡a©i AbycvZ 54:75 c„w_ex‡Z 365 w`‡b GK eQi n‡j ﵇Z KZ w`‡b GK eQi n‡e? [RU. 11-12, 13-14] A. 323 B. 223 C. 333 D. 233 2 3 e m e m R R T T ev, 2 3 75 54 365 Tm ev, 0.611 365 Tm T day m 223 STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question SUST 01. g½j MÖn KZ eQ‡i GKevi m~h©‡K cÖ`w¶b K‡i? (c„w_ex †_‡K m~‡h©i `~iZ¡ 150106 km. g½j †_‡K m~‡h©i `~iZ¡ 230106 km) [SUST. 08-09] A. 1.90 year B. 2.5 year C. 1.75 year D. 2.35 year 2 3 e m e m R R T T ev, 3 2 6 6 150 10 230 10 1 Tm ev, Tm = 1.898 ev, Tm = 1.90 eQi 02. †Kvb b¶‡Îi Pvwiw`‡K cÖ`w¶YiZ `ywU MÖ‡ni K¶ c‡_i e¨vmv‡a©i AbycvZ 2:3 n‡j Ges G‡`i AveZ©b Kvj T1 Ges T2 n‡j, wb‡Pi †KvbwU mwVK? [SUST. 07-08] A. 27 8 2 1 T T B. 3 2 2 2 1 T T C. 27 8 2 1 T T D. 27 8 1 2 T T 3 2 1 2 2 1 R R T T ev 3 2 1 2 1 R R T T 3 3 2 27 8 2 1 T T JUST 01. m~h© †_‡K c„w_exi Mo `~iZ¡ K‡g †M‡j eQ‡ii •`N©¨-[JUST-A, Set-Ka 19-20] A. K‡g hv‡e B. ‡e‡o hv‡e C. w¯’i _vK‡e D. Amxg n‡e †Kcjv‡ii 3q m~Î Abymv‡i T 2 R 3 R Kg‡j T Kg‡e
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 255 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 03 ANALYSIS OF HSC BOARD QUESTION 01. m f‡ii †Kv‡bv MÖn m~‡h©i Pviw`‡K r e¨vmv‡a©i e„ËvKvi c‡_ v mg`ªæwZ‡Z N~Y©vqgvb Ges MÖ‡ni AveZ©bKvj T n‡j, †Kcjv‡ii Z…Zxq m~Î n‡Z cvBÑ [Xv. †ev. 2016] A. T 2 = kr3 B. T 2 = k C. v = kT D. v = k T Ans A 02. MÖ‡ni ch©vqKvj T Ges m~h© n‡Z MÖ‡ni Mo `~iZ¡ r n‡j †Kcjv‡ii Z…Zxq m~Îvbymv‡i [iv. †ev. 2016] A. T r B. T r 2 C. T 2 r D. T 2 r 3 Ans D 03. MÖn¸‡jvi MwZc_ Dce„ËvKvi GB m~ÎwU †Kvb weÁvbxi? [iv. †ev. 2015; w`. †ev. 2015] A. U‡jgx B. †Kcjvi C. wc_v‡Mvivm D. M¨vwjwjI Ans B 04. wP‡Î †Kvb Ae¯’v‡b c„w_exi †eM me‡P‡q Kg? [P. †ev. 2015] B A D C m~h© A. A B. B C. C D. D Ans C 05. m~h© †_‡K c„w_exi `~iZ¡ hw` eZ©gvb `~i‡Z¡i `yB Z…Zxqvsk nq Z‡e GK eQ‡i w`‡bi msL¨v KZ?(c„w_ex‡Z 1 eQi = 365 w`b) [Kz.‡ev. 2019] A. 108.15 w`b B. 121.66 w`b C. 198.68 w`b D. 243.33 w`b Ans C Why T 2 R 3 T 2 1 R 3 1 = T 2 2 R 3 2 T 2 2 = R2 R1 3 × T2 1 = 2R 3 × 1 R 3 × T2 1 = 2 3 3 × (365)2 T2 = 198.68 days. 06. MÖn m¤úwK©Z †Kcjv‡ii Z…Zxq m~‡Îi MvwYwZK iƒc †KvbwU? [wm. †ev. 2016; Kz. †ev. 2017] A. T1 R1 2 = T2 R1 2 B. T1 2 R1 3 = T2 2 R2 3 C. r1 3 R1 2 = R2 3 R2 2 D. T1 2 R1 = T2 2 R2 Ans B 07. †Kcjv‡ii 3q m~‡Îi bvg †KvbwU? [e. †ev. 2017] A. Kÿc‡_i m~Î B. †ÿÎd‡ji m~Î C. ch©vqKv‡ji m~Î D. nvi‡gvwbK m~Î Ans C Concept 12 Acm~i I Abym~i msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. r1v1 = r2v2 [r = `~iZ¡, v = `ªæwZ] CONCEPTUAL MATH MEx 01 eya MÖ‡ni Abymyi I Acmyi `~iZ¡ h_vµ‡g 4.59 1010 m Ges 6.98 1010 m| Acmyi ¯’v‡b `ªæwZ 3.88 104 ms–1 n‡j Abymyi ¯’v‡b eya MÖ‡ni `ªæwZ KZ? 2 1 r1v1 = 2 1 r2v2,; V2 = 2 1 1 r r v = 10 10 4 4.59 10 6.98 10 3.88 10 = 5.90 104 ms–1 NOW START PRACTICE 01. c„w_exi Abymyi I Acmyi h_vµ‡g 1.471011 m Ges 1.521011 m| Abymyi ¯’v‡b c„w_exi MwZ Acm~‡ii KZ ¸Y? NOW PRACTICE SOLVE : 01. r1v1 = r2v2 2 1 v v = 1 2 r r = 11 11 1.47 10 1.52 10 = 1.034, Abym~i ¯’v‡b c„w_exi MwZ Acm~‡ii 1.034 ¸Y| SAQ Short Answer Questions WRITTEN PART BAQ Broad Answer Questions 01. GKwU †m‡KÛ †`vjK c„w_ex c„‡ô ï× mgq †`q| G‡K P›`ª c„‡ô †bqv n‡jv| P›`ª c„‡ô Gi ch©vqKvj wbY©q Ki| P›`ª A‡cÿv c„w_exi fi 81 ¸Y †ewk Ges e¨vmva© 4 ¸Y †ewk| Solve Avgiv Rvwb, ge = GMe Re 2 ; gm = GMm Rm 2 GLv‡b, Me = 81 Mm Re = 4 Rm gm ge = GMm Rm 2 Re 2 GMe = Mm Me Re Rm 2 gm ge = Mm 81Mm 4Rm Rm 2 = 16 81 Avgiv Rvwb, mij †`vj‡Ki †ÿ‡Î T2 T1 = g1 g2 Tm Te = ge gm = 81 16 = 9 4 Tm = 9 4 2 = 4.5 sec 02. c„w_ex c„‡ô GKwU †jv‡Ki IRb 90kg n‡j g½jc„‡ô Zvi IRb KZ n‡e? g½j Gi fi c„w_exi f‡ii 1 9 Ask Ges g½‡ji e¨vmva© c„w_exi e¨vmv‡a©i A‡a©K| Solve Avgiv Rvwb, g = GM R 2 GLv‡b, Me = 9Mm Re = 2Rm ge = GMe Re 2 ; gm = GMm Rm 2 gm ge = MmRe 2 MeRm 2 gm ge = 1 9 4 = 4 9 gm = 4 9 ge Wm = 4 9 We Wm = 4 9 90 = 40 kg 03. c„w_exi N~Y©b‡eM eZ©gvb N~Y©b‡e‡Mi KZ¸b n‡j wbiÿxq A‡j †Kvb e¯‘ fvinxb n‡e? [BUTex. 2020-21] Solve fvinxb = IRbnxb Aÿvs‡k IRb, mg = mg – m 2R cos2 wbiÿ‡iLvq IRb, mg = mg – m 2R awi, †KŠwbK †eM, n ¸Y n‡j fvinxb n‡e| mg = 0 mg – m(n) 2R = 0 n = g 2R GLv‡b, g = 9.8, = 2 243600 Ges R = 6.4 106m emv‡j n 17 Av‡m| c„w_exi N~Y©b‡eM eZ©gv‡bi 17 ¸Y n‡j wbiÿxq A‡j me e¯‘ fvinxb n‡e|
256 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 04. 80kg IR‡bi GKwU K…wÎg DcMÖn f~-c„ô †_‡K KZ D”PZvq ¯’vcb Ki‡j Zv cÖwZ 24 N›Uvq 2 evi GKB ¯’vb ch©‡eÿY Ki‡Z cvi‡e? [c„w_exi e¨vmva© 6400km I Zvi fi 6 1021Ton] Solve †`Iqv Av‡Q, c„w_exi e¨vmva© R = 6400km = 6.4 106m c„w_exi fi M = 6 1021 Ton = 6 1021 103 kg = 6 1024kg K…wÎg DcMÖ‡ni ch©vqKvj T = 24 2 hours = 12 hours = 12 3600s = 43200s awi, K…wÎg DcMÖ‡ni D”PZv = h ; Avgiv Rvwb, T 2 = 4 2 GM (R + h)3 T 2GM 4 2 = (R + h)3 R + h = T 2GM 4 2 1 3 h = T 2GM 4 2 1 3 – R = (43200) 2 6.67310–1161024 4 2 1 3 – 6.4 106m = h = 20.24976 106 m 05. GKwU w¯’i wjd‡Ui g‡a¨ ivLv GKwU mij †`vjbKvj T| hw` †`vjKwUi Dc‡ii w`‡K g/4 Z¡iY wb‡q D‡V, Zvn‡j †`vjKwUi †`vjbKvj KZ n‡e? Solve Avgiv Rvwb, T = 2 L g DwjøwLZ Ae¯’vq, T = 2 L g T T = g g T = g 5 4 g T = 2 5 T GLv‡b, g = ¯^vfvweK AwfKl©xq Z¡iY g = g + g 4 = 5 4 g awi, †`vj‡Ki Kvh©Kix •`N¨© = L cwiewZ©Z †`vjbKvj, T =? 06. GKwU mylg †Mvj‡Ki fi 1 104 kg Ges e¨vmva© 1m| †MvjK KZ…©K †Mvj‡Ki †K›`ª n‡Z 0.5m `~i‡Z¡ Aew¯’Z m1 f‡ii GKwU KYvi Dci gnvKl© e‡ji gvb KZ? [G = 6.673 10–11Nm2 /kg2 ] Solve †Mvj‡Ki NbZ¡ = 1104 4 3 0.5m m1 wP‡Î †`Lv‡bv KwíZ †MvjKwUi fi = 4 3 (0.5) 3 1104 4 3 = 104 8 kg KYvwUi Dci gnvKl© ej = 6.673 10–11104 m1 8(0.5) 2= 3.3365 10–7 m1N A_ev, KYvwUi Dci gnvKl© ej = GMm1 R 3 r = 6.673 10–11 104 0.5 1 3 m1 = 3.3365 10–7m1N 07. c„w_exi GKwU K…wÎg DcMÖn f~-c„ô †_‡K 850km E‡aŸ© c„w_ex‡K cÖ`wÿY Ki‡Q| DcMÖnwUi AveZ©b Kvj wbY©q Ki| [c„w_exi e¨vmva© R = 6400km Ges g = 9.8ms–1 ] Solve GLv‡b, h = 850 km = 850 103m; R = 6.4 106m Avgiv Rvwb, `ªæwZ, v = R g R+h = 6.4 106 9.8 6.4 106 + 850 103 = 7.440 103 ms–1 DcMÖ‡ni AveZ©b Kvj, T = 2 v (R + h) = 2(6.4106 +850103 ) 7.440103 = 6122.008s 08. g‡b Ki c„w_exi Kÿc_ e„ËvKvi hvi e¨vmva© 1.5 108 km| m~‡h©i fi wbY©q Ki| †`Iqv Av‡Q 1 eQi = 365 w`b Ges gnvKl© aªæeK G = 6.7 1011 N m2 kg2 | Solve GLv‡b, Kÿc‡_i e¨vmva©, r = 1.5 108 km = 1.5 1011 n ch©vqKvj, T = 365 w`b = 365 24 60 60 s gnvKl© aªæeK, G = 6.7 1011 N m2 kg2 m~‡h©i fi, M = ? Avgiv Rvwb, M = 4 2 r 3 GT2 = 4 2 (1.5 1011 m) 3 6.7 1011 N m2 kg2 (365 24 60 60 s) 2 = 2 1030 kg 09. 20 cm `~‡i _vKv Ae¯’vq `ywU e¯‘ ci¯úi‡K 1 108 N gnvKl©xq e‡j ci¯úi‡K AvKl©Y K‡i| hw` e¯‘؇qi mw¤§wjZ fi 5 kg nq, Z‡e cÖ‡Z¨K e¯‘i fi KZ wQj? Solve awi, GKwU e¯‘i fi = m1 ; myZivs Aci e¯‘wUi fi = 5 m1 F = G m1 m2 d 2 1 108 = 6.673 1011 m1 (5 m1) (6.20) 2 ev, 5m1 m1 2 = 5.99 6 ev, m1 2 5m1 + 6 = 0 ev, m1 = (3, 2) Ges m2 = (2, 3) kg 3 kg, 2 kg 10. cøy‡Uv MÖn n‡Z m~‡h©i Mo `~iZ¡ c„w_ex †_‡K m~‡h©i Mo `~i‡Z¡i 40 ¸Y| cøy‡Uvi AveZ©b Kvj KZ? Solve aiv hvK, cøy‡Uv I c„w_exi AveZ©bKvj h_vµ‡g T1 I T2 Ges m~h© n‡Z Zv‡`i Mo `~iZ¡ h_vµ‡g r1 I r2; †Kcjv‡ii Z…Zxq m~Î n‡Z cvB, T 2 1 T 2 2 = r1 r2 3 ev, T 2 1 = 40 r2 r2 3 T 2 2 ev, T 2 1 = 403 (1y)2 T1 = 253 y (cÖvq) GLv‡b, T2 = 1 y (eQi) r1 = 40 r2 T1 = ? 11. f‚c„‡ôi †h MfxiZvq AwfKl©R Z¡iY f‚c„‡ôi AwfKl©R Z¡i‡Yi 1 n Ask KZ n‡e? [c„w_exi e¨vmva© = R] Solve f‚c„‡ôi x MfxiZvq AwfKl©R Z¡iY gx n‡j, gx = g 1 x R gx g = 1 x R ev, 1 n = 1 x R gx g = 1 n ev, x R = 1 1 n ev, x = R 1 1 n = R n 1 n R n 1 n 12. †Kcjv‡ii cÖ_g I Z…Zxq m~Î `yÕwU wjL| [RU-C, Set-1. 19-20] Solve cÖ_g m~Î: Dce„‡Ëi m~Î: cÖwZwU MÖn m~h©‡K Dce„‡Ëi †dvKv‡m †i‡L GKwU Dce„ËvKvi c‡_ cÖ`wÿY Ki‡Q| 3q m~Î: mg‡qi m~Î: cÖwZwU MÖ‡ni ch©vqKv‡ji eM© m~h© n‡Z Zvi Mo `~i‡Z¡i Nbd‡ji mgvbycvwZK| T 2 R 3 13. wbDU‡bi gnvKl© m~Î †_‡K gnvKl©xq aªæe‡Ki gvÎv mgxKiY I GKK wbY©q Ki| [RU-C, Set-3. 19-20] Solve wbDU‡bi gnvKl© m~ÎwU n‡jv, F = G m1m2 r 2 G = Fr2 m1m2 [G] = MLT–2L 2 M 2 = [L3M –1T –2 ] Ges G Gi GKK = Nm2 kg2 =Nm2 kg–2
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 257 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 14. g‡b Ki c„w_exi †K›`ª w`‡q Gi e¨v‡mi GK cÖvšÍ †_‡K Aci cÖvšÍ ch©šÍ GKwU Uv‡bj Lbb Kiv n‡jv| †`LvI †h, GB Uv‡b‡j GKwU cv_i †dj‡j Gi MwZ mij †`vjb MwZ n‡e| cv_iwUi mij †`vjbMwZi ch©vqKvj wbY©q Ki| (c„w_exi NbZ¡ = 5.5 103 kg/m3 ) [RUET: 19-20] Solve f~-c„‡ô g = 4 3 RG †hLv‡b R c„w_exi e¨vmva© I c„w_exi Mo NbZ¡| f~c„ô †_‡K h MfxiZvq AwfKl©R Z¡iY a = 4 3 (R – h)G G‡ÿ‡Î (R – h) n‡jv c„w_exi †K›`ª †_‡K H we›`yi miY| G‡K x Øviv cÖwZ¯’vcb K‡i cvB, a = – 4 3 xG [.. . Z¡iY, mi‡Yi wecixZgyLx] cv_‡ii fi m n‡j wµqvkxj ej F = ma = – 4 3 Gmx = – kx [ .. . k = 4 3 mG = aªæeK] a = – k m x = – 2 x, hv mij †`vjb MwZi mgxKiY| cv_‡ii ch©vqKvj T = 2 = 2 m k = 2 m 4 3 mG = 3 G = 5068.64 s 15. f~-c„ô n‡Z 3.596 107 m D”PZvq GKwU K…wÎg DcMÖn ¯’vcb Kiv n‡jv hvi †eM 3.08 103 ms–1 cÖgvY Ki †h, DcMÖnwU f~-w¯’i DcMÖn? [NSTU-A, 19-20] Solve f~-w¯’i DcMÖ‡ni ch©vqKvj, T = 2 v (R + h) = 2 3.08 103 (6.4 106 + 35.96 106 ) = 2 106 3.08 103 (6.4 + 35.96) = 2 103 3.08 42.36 sec = 24 h ; †h‡nZz f~-w¯’i DcMÖ‡ni ch©vqKvj 24 NÈv ZvB D³ MÖnwU f~-w¯’i| 16. c„w_ex †_‡K 2600 wK‡jvwgUvi D”PZvq GKwU K…wÎg DcMÖn c„w_ex‡K †K›`ª K‡i e„ËvKvi c‡_ cÖ`wÿY Ki‡Q| Gi †eM NÈvq KZ n‡e? (c„w_exi e¨vmva© = 6400 wK‡jvwgUvi, fi = 6 1024 †KwR, Ges G = 6.67 1011 wbDUb-wg2 /†KwR2 )| [KUET; 19-20] Solve K…wÎg DcMÖ‡ni †eM, V = GM R + h = 6.67 10–11 6 1024 6400 103 + 2600 103 = 6668.33ms–1 = 6668.33 10–3 3600–1 kmh–1 = 24006 kmh–1 STEP 3 mKj cvV¨eB‡qi NCTB QUESTIONS ANALYSIS cÖkœ e¨vLvmn mgvavb cÖkœ wb‡Pi wPÎ I Z_¨ †_‡K 01 bs cÖ‡kœi DËi `vI : P R Q m m2 1 wP‡Î P I Q e¯‘؇qi fi h_vµ‡g m1 I m2 Ges G‡`i ga¨eZ©x `~iZ¡ R| G‡`i ga¨Kvi AvKl©Y ej F 01. ga¨eZ©x `~iZ¡ d-Gi gvb wظY Kiv n‡j AvKl©Y e‡ji gvb F (c~‡e©i gv‡bi) KZ ¸Y n‡e? [Bm&nvK, Avwgi] A. GK-Z…Zxqvsk B. GK-PZz_©vsk C. A‡a©K D. `yB-Z…Zxqvsk F2 = d1 d2 2 F1 = 1 2 1 F1 = 1 4 F1 02. †Kv‡bv ¯’v‡bi g-Gi gvb 9.832 n‡j wb‡Pi †Kvb Dw³wU mwVK? [Bm&nvK, Avwgi] A. ¯’vbwU †giæ A‡j Aew¯’Z B. ¯’vbwU Aÿvs‡k Aew¯’Z C. ¯’vbwU welyexq A‡j Aew¯’Z D. ¯’vbwU mgy`ªc„‡ô Aew¯’Z Ans A 03. gnvKl© e‡ji cÖK…wZ n‡jv- [Bm&nvK, Avwgi] i. gnvKl© ej e¯‘i ga¨Kvi cvi¯úwiK AvKl©Y ej ii. gnvKl© ej gva¨‡gi cÖK…wZi Ici wbf©i K‡i iii. gnvKl© ej e¯‘؇qi f‡ii ¸Yd‡ji mgvbycvwZK wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans C 04. †Kv‡bv we›`y‡Z gnvKl©xq †ÿÎ cÖvej¨- [Bm&nvK, Avwgi] (i) ‡f±i iwk (ii) dr dV E (iii) dr dV E wb‡Pi †KvbwU mwVK? A. i B. i I ii C. ii I iii D. i, ii I iii Ans B 05. †Kv‡bv e¯‘i Dr‡ÿcY †eM gyw³‡eM A‡cÿv †ewk n‡j e¯‘wU- [Bm&nvK, Avwgi] (i) Pvu‡`i g‡Zv DcMÖ‡n cwiYZ nq (ii) c„w_ex‡Z Avi wd‡i Av‡m bv (iii) cive„Ë c‡_ c„w_ex c„ô †Q‡o hvq wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans C 06. wb‡Pi †KvbwU mZ¨ bq? [Bm&nvK, Avwgi] A. c„w_exi evwl©K MwZi Rb¨ g-Gi gv‡bi cwieZ©b nq B. c„w_exi AvwýK MwZi Rb¨ g-Gi gv‡bi cwieZ©b nq C. Aÿvsk cwieZ©‡b g-G gv‡bi cwieZ©b nq D. D”PZvi Kvi‡Y g-Gi gv‡bi cwieZ©b nq Ans A 07. c„w_exi c„‡ô I Pvu‡`i c„‡ô AwfKl©xq Z¡i‡Yi AbycvZ KZ n‡e? [Bm&nvK, Avwgi] A. 81 : 4 B. 81 : 6 C. 81 : 10 D. 81 : 16 Pvu‡`, gm = 1.93 Ges c„w_ex‡Z ge = 9.81 16 81 1.93 9.81 m e g g 08. c„w_exi †K›`ª †_‡K f~-c„ô ch©šÍ g ebvg r †jLwPÎ †KvbwU? [Bm&nvK, Avwgi] A. g r = R B. g r = R C. g r = R D. g r = R Ans A DÏxc‡Ki Av‡jv‡K 09 I 10 bs cÖ‡kœi DËi `vI: K…wÎg DcMÖn c„w_ex h = 7 × 105 m (c„w_ex c„ô †_‡K K…wÎg DcMÖ‡ni D”PZv) 09. DcMÖnwUi Abyf~wgK †eM KZ? [Bm&nvK, Avwgi] A. 7509.43 ms–1 B. 7510.43 ms–1 C. 7508.43 ms–1 D. 7507.43 ms–1 S A info V = GM (R + h) = 6.673 × 10–11 × 6 × 1024 (6.4 × 106 + 7 × 105 ) = 7509.43 ms–1
258 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. DcMÖnwUi ch©vqKvj KZ? [Bm&nvK, Avwgi] A. 1 hr 39 min B. 1 hr 40 min C. 1 hr 41 min D. 1 hr 42 min S A info T = 2 (R + h) V = 2 × 3.1 (6.4 × 106 + 7 × 105 ) 7509.43 = 1 hr 39 min 11. wb‡Pi †KvbwU gnvKlx©q GKK wb‡`©k K‡i? [Bm&nvK, Avwgi] A. Nmkg-1 B. Jkg C. kgj-1 D. Nm-1 kg-1 Ans A 12. c„w_ex c„ô †_‡K gyw³‡eM 11.2 kms-1 | †h MÖ‡ni e¨vmva© c„w_exi wظY wKš‘ Mo NbZ¡ c„w_exi mgvb Zvi c„ô †_‡K gyw³‡eM n‡e- [Bm&nvK, Avwgi] A. 5.6 kms-1 B. 11.2 kms-1 C. 22.4 kms-1 D. Dc‡ii †Kv‡bvwUB bq Ans C 13. `ywU DcMÖn GKB e„ËvKvi c‡_ Nyi‡Q| Aek¨B Zv‡`iÑ [Bm&nvK, Avwgi] A. fi mgvb B. †KŠwYK fi‡eM mgvb C. MwZkw³ mgvb D. `ªæwZ mgvb Ans D 14. GKwU DcMÖn c„w_ex‡K †K›`ª K‡i e„ËvKvi c‡_ Nyi‡Q| nVvr K‡i AwfKl©xq ej hw` wejyß n‡q hvq Zvn‡j DcMÖnwU- [Zcb, AvwRR] A. GKB `ªæwZ‡Z GKB c‡_ Nyi‡Z _vK‡e B. GKB `ªæwZ‡Z Avw` Kÿc‡_i ¯úk©K eivei Pj‡Z _vK‡e C. ewa©Z `ªæwZ‡Z wb‡P c‡o hv‡e D. g~j Kÿc‡_ wKQzÿY P‡j †_‡g hv‡e Ans B 15. gnvKlx©q †ÿÎ cÖvej¨ n‡”Q- [Zcb, AvwRR] i. `~i‡Z¡i mv‡c‡ÿ gnvKlx©q wef‡ei n«vm nvi ii. – V iii. gnvKlx©q †ÿÎ cÖvej¨ n‡”Q GKK f‡ii Dci wµqvkxj ej wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans D 16. †Kvb we›`yi gnvKlx©q wefe- [Zcb, AvwRR] i. †¯‹vjvi ivwk ii. V = – GM r iii. V = w m A. ii B. i I ii C. ii I iii D. i, ii I iii V = – Gm R ; w m 17. Puv‡`i evqyk~b¨ ¯’v‡b w¯’ive¯’v †_‡K GKwU cvjK I GKwU mxmvi ej‡K †djv n‡jv| cvj‡Ki Z¡iY n‡e- [Zcb, AvwRR] A. mxmvi e‡ji †P‡q †ewk B. mxmvi e‡ji mgvb C. mxmvi e‡j †P‡q Kg D. 9.8 ms–2 Ans B 18. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظY| D³ MÖ‡ni AwfKl©R Z¡iY c„w_exi AwfKl©R Z¡i‡Yi AvU¸Y| D³ MÖ‡ni gyw³‡eM c„w_exi gyw³‡e‡Mi Zzjbvq KZ¸Y Zv wbY©q Ki| [Zcb, AvwRR] A. 2 ¸Y B. 4 ¸Y C. 8 ¸Y D. 16 ¸Y S B info VA = MA Me × Re RA × Ve = 8 × 1 2 Ve VA = 2Ve 19. Lvov Dc‡ii w`‡K g Gi gvb- [Zdv¾j, gwnDwÏb] A. k~b¨ B. 9.8ms-2 C. -9.8ms-2 D. †KvbwUB bq Ans C 20. `ywU e¯‘i g‡a¨ †h `~iZ¡ Av‡Q Zv A‡a©‡K †b‡g Avm‡j gnvKl© ejA. 2 kms-1 B. 9.8 kms-1 C. 11.18 kms-1 D. Amxg Ans B c„w_exi fi M, e¨vmva© R, gnvKl©xq aªæeK G, AwfKl©R Z¡iY g| DcwiD³ Z‡_¨i wfwˇZ 21 bs cÖ‡kœi DËi `vI: 21. G I g Gi g‡a¨ m¤úK© †KvbwU? [Zdv¾j, gwnDwÏb] A. 2 R gM G B. M gR G 2 C. GM R G 2 D. 2 gR M G g = 2 R GM M gR G 2 22. c„w_exi †K›`ª †_‡K †Kvb we›`yi `~iZ¡ r n‡j (r < R) AwfKl©R Z¡iY g Gi Rb¨ wb‡Pi †Kvb m¤úK©wU mwVK? (R c„w_exi e¨vmva©) [Mwb, mykvšÍ] A. g 1 r B. g 1 r 2 C. g r D. g r 2 Ans C 23. c„w_ex‡K NbZ¡ I r e¨vmv‡ai GKwU mylg †MvjK aiv †h‡Z cv‡i| Gi c„‡ô g Gi mv‡_ gnvKl© aªæeK G I D³ ivwk¸‡jvi m¤úK© †KvbwU? [Mwb, mykvšÍ] A. g = G r 2 B. 3G 4 r C. 4 r G 3 = g D. g = 4 2 G 3 Ans C 24. c„w_ex c„‡ôi †Kvb we›`y‡Z gnvKl© †¶Î cÖvej¨- [Mwb, mykvšÍ] i. H we›`y‡Z AwfKl©R Z¡i‡Yi mgvb ii. wef‡ei Xv‡ji (‡MÖwW‡q›U) FYvZ¥K gv‡bi mgvb iii. H we›`y‡Z Aew¯’Z †Kvb e¯‘i Dci wµqvkxj gnvKl© e‡ji mgvb wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans A 25. e¯‘i gyw³‡eM wbf©i K‡i wK‡mi Dci? [†gvK‡Q` m¨vi] i. e¯‘i fi ii. c„w_exi fi iii. c„w_exi e¨vmva© wb‡Pi †KvbwU mwVK? A. i B. i I ii C. i I iii D. ii I iii Ans D 26. weÁvbxMY c„w_exi Zzjbvq eo A_P NbZ¡ Kg Giƒc GKwU MÖ‡ni mÜvb †c‡q‡Qb| c„w_ex †_‡K GKRb e¨w³ H MÖn c„‡ô †M‡j Zvi IRb- [†gvK‡Q` m¨vi] i) evo‡ZI cv‡i ii) Kg‡ZI cv‡i iii) GKB _vK‡Z cv‡i wb‡Pi †KvbwU mwVK? A. i B. iii C. ii D. i I ii †h¯’v‡b NbZ¡ Kg, †m ¯’v‡b g Gi gvb Kg| Avevi g Gi gvb Kg‡j IRb K‡g| 27. welye AÂj A‡c¶v †giæ A‡ji e¨vmva© KZ wK‡jvwgUvi Kg? [†gvK‡Q` m¨vi] A. 43.47 m B. 43.47 km C. 22 m D. 22 km Ans D 28. c„w_ex c„ô †_‡K GKwU e¯‘‡K P›`ªc„‡ô †bIqv n‡”Q| e¯‘wUi IRb k~b¨ n‡e †h we›`y‡Z- [†gvK‡Q` m¨vi] A. c„w_exi AvKl©Y ej k~b¨ B. Puv‡`i AvKl©Y ej k~b¨ C. c„w_ex I Puv‡`i AvKl©Y ej mgvb I wecixZ D. c„w_ex I Puv` Df‡qi AvKl©Y ej k~b¨ Ans C 29. f~-c„ô n‡Z 400 km Af¨šÍ‡i Ges f~-c„‡ô AwfKl©xq Z¡i‡Yi AbycvZ KZ? [c„w_exi e¨vmva© 6400 km] [†gvK‡Q` m¨vi] A. 14:15 B. 15:14 C. 15:16 D. 16:15 6400 400 1 g g = 1– 1 16 = 15 16 30. c„w_exi fi GKB †_‡K hw` e¨vmva© 4% n«vm cvq Z‡e g- Gi gvb- [†gvK‡Q` m¨vi] A. e„w× cv‡e B. n«vm cv‡e C. w¯’i _vK‡e D. k~b¨ n‡e Ans A 31. hw` `ywU we›`y f‡ii e¯‘i ga¨eZ©x `~iZ¡ wظY Kiv nq, Z‡e gnvKl© ej n‡e- [igv m¨vi] A. GK-PZz_©vsk B. A‡a©K C. wظY D. Pvi ¸Y 1 1 2 1 2 2 1 2 F 4 1 F 2 1 F d d F 32. hw` Avgiv welyexq AÂj †_‡K †giæ A‡ji w`‡K hvB Z‡e AwfKl©xq Z¡iY- [igv m¨vi] A. GKB n‡e B. e„w× cv‡e C. K‡g hv‡e D. 45° A¶vsk ch©šÍ Kg‡e Ans B 33. c„w_ex Ges Ab¨ GKwU MÖ‡ni gyw³ †eM h_vµ‡g ve Ges vm MÖnwUi e¨vmva© c„w_exi e¨vmv‡a©i wظY Ges MÖnwUi Mo NbZ¡ c„w_exi Mo Nb‡Z¡i mgvb n‡j, wb‡Pi †KvbwU mwVK? [igv m¨vi] A. ve = 4 m v B. ve = 2 m v C. ve = vm B. ve = 2vm 2 v v 2 1 2R R R R v v m e e e m e m e 34. c„w_exi Pviw`‡K N~Y©biZ `ywU K…wÎg DcMÖ‡ni e¨vmv‡a©i AbycvZ 1:4 n‡j, G‡`i Avbylw½K ch©vqKv‡ji AbycvZ KZ n‡e? [igv m¨vi] A. 1:8 B. 1:4 C. 4:1 D. 8:1 : 1: 8 8 1 4 1 1 2 2 3 2 1 T T T T
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 259 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 35. E gnvKl©xq †ÿÎ cÖve‡j¨i †Kv‡bv we›`y‡Z m f‡ii e¯‘ ivL‡j Zvi Ici Kx cwigvY ej wµqv Ki‡e? [BKivg m¨vi] A. E m B. Em C. F m D. F m 2 gnvKl©q ‡¶Î cÖvej¨, E = F M F = Em 36. c„w_exi fi GKB †i‡L hw` e¨vmva© 40% Kgv‡bv nq Zvn‡j g Gi gvb- [BKivg m¨vi] i. e„w× cv‡e ii. n«vm cv‡e iii. AcwiewZ©Z _vK‡e wb‡Pi †KvbwU mwVK? A. i B. ii C. iii D. i I iii R2 = R – 0.4R = 0.6R g2 = g R R 2 2 1 = g 2 0.6 1 = 2.78g g Gi gvb e„w× cv‡e| 37. gnvKl©xq cÖvej¨, E = GM r 2 GB mgxKiY n‡Z eySv hvq †h- [BKivg m¨vi] i. M hZ †ewk n‡e, cÖvej¨ ZZ evo‡e ii. r hZ †ewk n‡e, cÖvej¨ ZZ Kg‡e iii. gnvKl©xq †ÿ‡Îi wewfbœ we›`y‡Z cÖvej¨ GKB n‡e wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii E = GM r 2 , wewfbœ we›`y‡Z cÖvej¨ wfbœ| M hZ ‡ewk n‡e cÖvej¨ ZZ ‡ewk, Avevi r hZ ‡ewk n‡e cÖvej¨ ZZ Kg n‡e| 38. ‘g’ Gi Dci c„w_exi AvwýK MwZi cÖfve me‡P‡q †ewk †Kv_vq? [kvgmyi m¨vi] A. DËi ‡giæ‡Z B. welyexq A‡j C. 45° A¶vs‡k D. `w¶Y †giæ‡Z c„w_exi wbR A‡¶ N~Y©‡bi Rb¨ g Gi gv‡bi cwieZ©b nq| AvwýK MwZi cÖfve me‡P‡q †ewk DËi †giæ‡Z| GKwU K…wÎg DcMÖn e„ËvKvi K¶c‡_ c„w_ex‡K AveZ©b Ki‡Q| Gi cÖ`w¶Y †eM, c„w_exi gyw³ †e‡Mi A‡a©K| c„w_exi Mo e¨vmva© R| Dc‡ii DÏxc‡Ki Av‡jv‡K 39 I 40 bs cÖ‡kœi DËi `vI| [kvgmyi m¨vi] 39. DcMÖnwUi D”PZvA. R 2 B. R C. 3R 2 D. 2R cÖ`w¶Y †eM, e v v 2 1 gR R H g R 2 2 1 gR R H g R 2 4 2 1 2 1 R H R 2R R H H 2R R R 40. DcMÖnwU‡K nVvr _vwg‡q G‡K c„w_exi w`‡K gy³ fv‡e co‡Z w`‡j †mwU f~-c„ô‡K KZ †e‡M AvNvZ Ki‡e? [kvgmyi m¨vi] A. gR B. vm C. gR 2 D. 2 gR GLv‡b, H = 0 , V = R g R+0 = R g R = gR STEP 4 MCQ CONCEPT TEST WRITTEN cÖkœ 01. GKRb †jvK f~c„‡ô 3m jvdv‡Z cv‡i| P‡›`ª KZ DuPz‡Z jvdv‡Z cvi‡e? A. 3m B. 6m C. 9m D. 15m 02. f~c„ô n‡Z KZ DuPz‡Z †M‡j AwfKl©R Z¡i‡Yi gvb c„w_exi Z¡i‡Yi gv‡bi 1 5 n‡e? A. 25 km B. 640 km C. 100 km D. 2560 km 03. c„w_exi c„‡ô †Kv‡bv e¯‘i IRb 1000 N n‡j c„w_ex c„ô n‡Z A‡a©K MfxiZvq IRb KZ n‡e? A. 1000 N B. 500 N C. 250 N D. 125 N 04. g½j MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i 0.532 ¸Y Ges fi c„w_exi f‡ii 0.11 ¸Y n‡j g½j MÖ‡ni c„‡ô AwfKl©R Z¡iY KZ? A. 4.8 ms–2 B. 3.8 ms–2 C. 5.8 ms–2 D. 3.44 ms–2 05. 100 kg f‡ii GKwU ¸iæfvi e¯‘i fvi †K›`ª †_‡K 10 m `~i‡Z¡ Aew¯’Z †Kv‡bv we›`yi gnvKlx©q wefe KZ? A. 6.67 × 10–10 J/kg B. – 6.67 × 10–10 J/kg C. 6.76 × 10–10 J/kg D. – 6.67 × 10–10 J/kg 06. `ywU e¯‘i ga¨eZx© `~iZ¡ cÖv_wgK `~i‡Z¡i 3 ¸Y Kiv nj| Zv‡`i ga¨Kvi PzovšÍ ej KZ? A. 1 2 F1 B. 1 4 F1 C. 1 9 F1 D. 1 3 F1 07. f~-c„ô n‡Z 800 km Af¨šÍ‡i I f~-c„‡ô AwfKlx©q Z¡i‡bi AbycvZ †ei Ki| A. 15 : 16 B. 7 : 8 C. 11 : 12 D. 12 : 13 08. gy³fv‡e cošÍ e¯‘i 3s I 4s †m‡K‡Û mi‡Yi AbycvZ KZ n‡e? A. 3 : 4 B. 4 : 9 C. 9 : 16 D. 3 : 4 09. GKwU e¯‘i IRb c„w_ex‡Z 95 N I P‡›`ª 16N| P›`ª A‡c¶v c„w_ex‡Z AwfKl©R Z¡iY KZ ¸Y? A. 4 ¸Y B. 5.94 ¸Y C. 5 ¸Y D. 1.96 ¸Y 10. c„w_exc„‡ô gnvKl©xq †¶Î cÖvej¨ E| KvíwbK GKwU MÖ‡ni NbZ¡ hw` c„w_exi Nb‡Z¡i mgvb nq Ges e¨vmva© hw` wظY nq Z‡e GB MÖ‡ni c„‡ô gnvKl©xq †¶Î cÖvej¨ KZ? A. 1 2 E B. 1 4 E C. 2E D. 1 2 E 11. mgvb fi wewkó `ywU MÖ‡ni e¨vmv‡a©i AbycvZ 1:3 n‡j MÖ‡ni Z¡i‡Yi AbycvZ KZ ? A. 1 : 3 B. 3 : 1 C. 2 : 1 D. 9 : 1 12. c„w_exi A‡a©K fi I A‡a©K e¨vmv‡a©i MÖ‡n g Gi gvb KZ? A. 19.6ms2 B. 9.8ms2 C. 4.9ms2 D. 1.96 ms2 13. f~-c„ô n‡Z KZ Mfx‡i AwfKlx©q Z¡i‡Yi gvb f~-c„‡ôi gv‡bi GK cÂgvsk n‡e? A. 5R 4 B. 4R 5 C. 2R 2 D. 5R 14. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظY| D³ MÖ‡ni c„‡ô AwfKl©R Z¡iY c„w_exi AwfKl©R Z¡i‡Yi AvU ¸Y| D³ MÖ‡n gyw³ †eM c„w_exi Zzjbvq KZ ¸Y? A. 2 e B. 2 e C. 4 e D. 3 e 15. GKwU K…wÎg DcMÖn c„w_exi Pvwiw`‡K me©v‡c¶v Kg KZ `ªæwZ‡Z cÖ`w¶Y Ki‡e? [R = 6400 km] A. 7.92 ms–1 B. 7.92 kms–1 C. 6.42 ms–1 D. 5.24 kms–1
260 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 16. R c„w_exi e¨vmva© n‡j f~-c„ô n‡Z KZ Mfx‡i g Gi gvb k~b¨ n‡e? A. R 2 B. R C. 2R D. 2R 3 17. f~-c„‡ô †Kvb †jv‡Ki IRb, 729N n‡j wZwb Pv‡` wM‡q KZUzKz IRb nviv‡eb| c„w_exi fi I e¨vmva© h_vµ‡g Pv‡`i fi I e¨vmva© 81 ¸Y I 4 ¸Y| A. 520N B. 585N C. 565N D. 625N 18. c„w_exi e¨vmva© 6400km n‡j KZ D”PZvq AwfKl©R Z¡ib f~-c„‡ô AwfKl©R Z¡i‡bi wZb-PZz_©vsk n‡e? A. 500km B. 600km C. 700km D. 800km 19. c„w_ex c„‡ô g Gi gvb 9.8 m/s2 c„w_exi †K‡›`ª g gvb k~b¨| c„w_exi c„ô †_‡K c„w_exi e¨vmv‡a©i A‡a©K MfxiZvq g Gi gvb KZ? A. 9.8 m/s2 B. 4.9 m/s2 C. 3.2 m/s2 D. 2.5 m/s2 20. c„w_exi Pviw`‡K N~Y©biZ `ywU K…wÎg DcMÖ‡ni e¨vmv‡a©i AbycvZ 1:4 n‡j, G‡`i Avbylw½K ch©vqKv‡ji AbycvZ KZ n‡e? A. 1:8 B. 1:4 C. 4:1 D. 8:1 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. c„w_exi fi I e¨vmva© Puv‡`i fi I e¨vmv‡a©i h_vµ‡g 81 I 4 ¸Y| f‚-c„‡ô 150 cvDÛ IR‡bi GKRb gvbyl Puv‡` wM‡q KZUzKz IRb nviv‡e? (Puv‡`i e¨vmva© 1000 gvBj) 22. evZv‡mi NbZ¡ 1.2kg/m3 n‡j 4.0 m 5.0 m †g‡S Ges 3.0 m Qv` ch©šÍ D”PZv wewkó GKwU N‡ii †fZ‡i _vKv evZv‡mi fi KZ? 23. 2 g 9.8ms n‡j 8000km e¨vmv‡a©i K¶c‡_ GKwU DcMÖ‡ni †eM KZ? 24. c„w_exi fi GKB †i‡L hw` e¨vmva© 2% n«vm Kiv nq Zvn‡j g-Gi gvb KZ cwieZ©b n‡e? 25. g½j MÖ‡ni fi c„w_exi f‡ii 0.108 ¸Y Ges e¨vmva© c„w_exi e¨vmv‡a©i 0.532 ¸Y n‡j g½jMÖ‡n GKwU e¯‘i gyw³ †eM KZ n‡e? [c„w_exi e¨vmva© 6400km] 26. GKwU MÖ‡ni e¨vmva© c„w_exi e¨vmv‡a©i wظb| D³ MÖ‡ni c„‡ô AwfKl©R Z¡ib c„w_exi AwfKl©R Z¡i‡bi 4 ¸b| D³ MÖ‡ni gyw³‡eM c„w_exi Zzjbvq27. m~‡h©i Pvwiw`‡K ïµ I c„w_exi K¶ c‡_i e¨vmv‡a©i AbycvZ 54:75 c„w_ex‡Z 365 w`‡b GK eQi n‡j ﵇Z KZ w`‡b GK eQi n‡e? 28. c„w_exi e¨vmva© A‡a©K n‡j AwfKl©R Z¡i‡Yi gvb n‡e29. evqy‡Z GK Kzj‡¤^i `ywU Avavb ci¯úi †_‡K 1 km e¨eav‡b Aew¯’Z n‡j, G‡`i ga¨Kvi ej KZ n‡e? 30. k~b¨ gva¨‡g `yBwU B‡jKUª‡bi ga¨Kvi Kzj¤^ ej FE Ges gnvKl© ej FG Gi AbycvZ n‡e? MCQ ANSWER ANALYSIS WRITTEN cÖkœ bs DËi e¨vL¨v 01 D 02 D gh = g 1 – 2h R g 5 = g 1 – 2h 6400 1 5 = 1 – 2h 6400 2h 6400 = 4 5 h = 6400 × 4 2 × 5 = 2560 km 03 B gd = g 1 – R 2 R = 9.8 1 – 1 2 = 4.9 ms–2 M = 1000 g × gd = 1000 9.8 × 4.9 = 500 N 04 B MM = 0.11 Me ; Rm = 0.532 Re gm = GMm Rm 2 Ges ge = GMe Re 2 gm = ge × Mm Me × Re Rm 2 = 9.8 × 0.11 × 1 0.532 2 = 3.8 ms–2 05 B V = – GM r = – 6.67 × 10–11 × 100 10 = – 6.67×10–10 J/kg 06 C 1 F 9 1 1 F 2 3 1 1 .F 2 2 d 1 d 2 F ¸b 07 B R d e g d g 1 ev, gd ge = 1 – 800 6400 1 – 1 8 = 7 8 7 : 8 08 C h1 h2 = t1 t2 2 = 3 4 2 = 9 : 16 09 B m e g g = m e W W = 16 95 = 5.94 10 C gnvKl©xq cÖvej¨, E = g ∵g = 4 3 RG; MÖn `ywUi NbZ¡ mgvb| E g R; E2 = R2 R1 E1 = 2E 11 D AwfKlx©q Z¡iY, g = 2 R GM MÖn `ywUi fi mgvb| g 1 R 2 g2 g1 = R1 R2 2 = 1 3 2 = 1 9
ASPECT PHYSICS cÖ_g cÎ gnvKl© I AwfKl© 261 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 12 B g = 2 R GM ; gα 2 R M g 2 9.8 19.6 1 2 2 1 g R R M M g 1 2 1 2 2 1 1 2 2 ms–2 13 B MfxiZv, h = e h g g 1 R = 5 1 1 R = 4R 5 14 C gyw³ †eM, v = 2gR vp = e p e p R R g g ve = 82ve = 4 ve 15 B DcMÖ‡ni †eM, v = gR = 6 9.86.410 = 7919.6 ms–1 = 7.92 kms1 16 B 17 B Wm= (e¨vmv‡a©i ¸b) 2 f‡ii ¸b We = (4) 2 81 729 = 144N; Puv‡` e¯‘i nvivb IRb = (729 – 144) = 585N 18 D R 2h 1 g g e h ev, 3 4 = 1 - 2h R ev, R 2h = 1 - 3 4 ev, 2h R = 1 4 ev, h = R 8 = 6400 8 = 800km 19 B gh = g R h 1 = 9.8 R 0.5R 1 = 4.9 ms–2 20 B d = 2 R n‡j, g = 2 g g 2R R 1 21 2 1 2 2 2 1 2 1 2 1 R R M M g g w w w 29.6 16 1 1 81 w 150 2 2 nviv‡bv IRb (150-29.6) = 120.4 lb-wt 22 = m v m = × v = 1.2 × (5 × 4 × 3) = 72 kg 23 V = GM R + h = gR2 R + h 6 2 6 8 10 9.8 6.4 10 V 3 7.0810 24 104 100 100 98 R R g g 2 2 1 2 2 2 1 = 0.96 g-Gi cwieZ©b (104-100) = 4% 25 GLv‡b, g½jMÖ‡ni gyw³‡eM vm vm = Mm Me Re Rm ve = 0.108 1 0.532 11.2 = 5.04kms–1 26 e m v v = e e m m 2g R 2g R = 1 2 1 4 = 2 2 ¸Y| 27 2 3 e m e m R R T T ev, 2 3 75 54 365 Tm ev, 0.611 365 Tm T day m 223 [†R‡b ivLv fvj †h, ïµMÖ‡n 223 w`‡b 1 eQi] 28 2 c e e c R R g g ev, 9.8 1 2 2 gc = 39.2ms-2 29 9000N 9KN (1000) 1 1 F 9 10 2 9 30 FE FG = K q1q2 d 2 G m1m2 d 2 = 9 109 (1.6 10–19) 2 6.673 10-11(9.1 10–31) 2 = 4.2 1042
262 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 AvšÍtAvYweK ej Ges wewfbœ cÖKvi ivmvqwbK eÜb 20% 25% 30% 35% 40% 10% 30% - CONCEPT-02 w¯ wZ¯ vcKZv m¤úwK©Z Z_¨vejx 35% 40% 45% 40% 30% 15% 45% - CONCEPT-03 cÖevnx c`v‡_©i aviYv 35% 30% 35% 35% 20% 10% 50% - CONCEPT-04 c„ôUvb m¤úwK©Z Z_¨vejx 50% 40% 35% 40% 35% 20% 55% - CONCEPT-05 mv›`ªZvi aviYv Ges cÖ‡qvRbxqZv 70% 65% 60% 65% 45% 05% 60% - CONCEPT-06 ¯úk© †KvY Ges •KwkKZv 75% 70% 65% 70% 70% 25% 80% - MvwYwZK cÖ‡qvM CONCEPT-01 cxob I weK…wZ msµvšÍ MvwYwZK cÖ‡qvM 80% 75% 75% 80% 80% 75% 90% CONCEPT-02 Bqs Gi w¯ wZ¯ vcK ¸YvsK, `„pZvi ¸YvsK I AvqZ‡bi ¸YvsK msµvšÍ MvwYwZK cÖ‡qvM 85% 80% 75% 85% 75% 80% 30% CONCEPT-03 mv›`ªZvsK, mv›`ªej I cÖvwšÍK †eM msµvšÍ MvwYwZK cÖ‡qvM 70% 65% 70% 75% 80% 55% 70% CONCEPT-04 cqm‡bi AbycvZ msµvšÍ MvwYwZK cÖ‡qvM 60% 70% 65% 70% 75% 65% 90% CONCEPT-05 c„ôUvb wbY©q msµvšÍ MvwYwZK cÖ‡qvM 70% 65% 60% 65% 60% 85% 60% CONCEPT-06 KvR wbY©q msµvšÍ MvwYwZK cÖ‡qvM 70% 75% 70% 70% 65% 80% 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 AvšÍtAvYweK ej Ges wewfbœ cÖKvi ivmvqwbK eÜb ✅ AvšÍtAvYweK ej: AvšÍtAvYweK `~i‡Z¡i cwigvY 10–9m †_‡K 10–10m| AvšÍtAvYweK ej AYy؇qi ga¨eZ©x `~i‡Z¡i Dci wbf©ikxj| ✅ AvšÍtAvYweK e‡ji cÖK…wZ: AvšÍtAvYweK ej AvKl©Yag©x n‡Z cv‡i, weKl©Yag©xI n‡Z cv‡i| AvšÍtAvYweK AvKl©Y ej‡KB w¯ wZ¯ vcK ej e‡j| mgw`K ag©x e¯‧ (Isotropic Body): avZe `Û, Zvi| Amg w`K ag©x e¯‧ (Anisotropic body): jeY, †KvqvU©R| w¯ wZ¯ vcK K¬vwšÍ: weÁvbx †Kjwfb G bvgKiY K‡ib| ✅ KwVb, Zij Ges evqexq c`v‡_©i AvšÍtAvYweK ejKwVb Zij evqexq 1. K¤úb MwZ _vK‡jI N~Y©b MwZ ‡bB 1. Zij c`v‡_©i KYv mg~n Kw¤úZ, AvewZ©Z Ges wbw`©ó mxgvi g‡a¨ ¯ vbvšÍwiZ n‡Z cv‡i| 1. AvYweK K¤úb, AveZ©b I ¯ vbvšÍi MwZ i‡q‡Q| 2. Pvc cÖ‡qv‡M AvqZb msKzwPZ nq bv| 2. Pvc cÖ‡qv‡M Zij c`v‡_©i D‡jøL‡hvM¨ ms‡KuvPb N‡U bv| 2. Pvc cÖ‡qv‡M M¨v‡mi AvqZb e¨vcKfv‡e n«vm cvq 3. AvšÍtAvYweK AvKl©Y ej MwZkw³i ‡P‡q †ekx 3. AvšÍtAvYweK AvKl©Y ej Ges MwZkw³i gvb KvQvKvwQ 3. MwZkw³i gvb AvšÍtAvYweK e‡ji †P‡q A‡bK ‡ekx| ✅ AvšÍtAvYweK ej m„wó nq `ywU kw³i `iæb- ◈ cvwicvwk¦©K AYy¸‡jvi wµqv cÖwZwµqvi d‡j m„ó wefe kw³ ◈ AYy¸‡jvi Zvcxq kw³ hv cÖK…Zc‡¶ AYy¸‡jvi MwZkw³| GB kw³ e¯‧i DòZvi Dci wbf©i K‡i| c`v‡_©i eÜb: cigvYymg~‡ni AvKl©b kw³ hv `ywU GKB ev wfbœ ‡g․‡ji cigvYyci®úi hy³ n‡q AYyMVb K‡i Zv‡K cvigvYweK eÜb e‡j| ◈ ivmvqwbK eÜb cÖavbZ 5 cÖKvi: (i) AvqwbK eÜb (ii) mg‡hvRx eÜb (iii) nvB‡Wªv‡Rb eÜb (iv) avZe eÜb (v) f¨vbWviIqvjm ej RwbZ eÜb ◈ AvqwbK eÜb: avZe I AavZe ‡g․‡ji ivmvqwbK wewµqv Kv‡j avZzi cigvYyi ewnt¯Íi ‡_‡K AavZz cigvYyi ewnt¯Í‡i GK ev GKvwaK B‡jKUªb ¯ vbvšÍwiZ nIqvi gva¨‡g m„ó abvZ¥K Avqb I FYvZ¥K Avq‡bi g‡a¨ w¯ i •e`y¨wZK AvKl©Y Øviv †h eÜb MwVZ nq, Zv‡K ivmvqwbK eÜb e‡j| AvqwbK eÜb Øviv m„ó ‡h․M‡K AvqwbK †h․M e‡j| D`v: NaCl ◈ mg‡hvRx eÜb: AYyMV‡bi mgq hw` cigvYywbR wbR ewnt¯Í‡i wbw®…q M¨v‡mi w¯ wZkxj B‡jKUªb KvVv‡gv AR©‡bi D‡Ï‡k¨ mgvb msL¨K AhyMj B‡jKUªb mieivn K‡i GK ev GKvwaK B‡jKUªb †Rvo m„wó K‡i Ges Dfq cigvYyZv mgvb fv‡e †kqvi K‡i, Z‡e cigvYy؇qi g‡a¨ †h eÜb MwVZ nq Zv‡K mg‡hvRx eÜb e‡j| D`v: H2 c`v‡_©i MvVwbK ag© STRUCTURAL PROPERTIES OF MATTER cÖ_g cÎ 07 Aa¨vq c„ôv 262
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 263 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ mg‡hvRx e܇bi kZ©: `ywU AavZe cigvYyi g‡a¨ mg‡hvRx eÜb MwVZ nq| Dfq AavZe cigvYymgmsL¨K B‡jKUªb †hvMvb w`‡q GK ev GKvwaK B‡jKUªb hyMj m„wó K‡i Zv Dfq cigvYymgfv‡e †kqvi K‡i _v‡K| ◈ avZe eÜb: †Kvb avZzi g‡a¨ †h AvKl©Y ej cigvYy¸‡jv‡K ci®ú‡ii mv‡_ AvU‡K iv‡L Zv‡K avZe eÜb e‡j| ◈ avZe e܇bi •ewkó¨: e܇bi cÖK…wZ mg‡hvRx ai‡bi wKš‧ Am¤ú„³ Ges AwaK msL¨K cigvYyi mv‡_ Ave× _vKvi my‡hvM i‡q‡Q| cwji eR©b bxwZ Abymv‡i hZwU my‡hvM i‡q‡Q cigvYy¸‡jvi gv‡S B‡jKUªb NbZ¡ Zvi †P‡q Kg| ✅ f¨vbWvi Iqvjm eÜb: KvQvKvwQ Aew¯ Z cigvYymg~‡ni g‡a¨ GKwU mve©Rbxb `ye©j AvKl©Y ej wµqv K‡i| ‡h cvi¯úwiK wµqvi d‡j G ej m„wó nq Zv‡K f¨vbWviIqvjm cvi®úwiK wµqv e‡j Ges G ej‡K f¨vbWviIqvjm eÜb e‡j| f¨vbWviIqvjm cvi¯úwiK wµqvi d‡j m„ó G eÜb‡K f¨vbWviIqvjm eÜb e‡j| G ai‡bi e܇b eÜb kw³ LyeB `~e©j| D`v: wbw¯…q M¨vm e܇bi aib kw³ cÖK…wZ D`vniY AvqwbK mej B‡jKUªb ¯ vbvšÍi (Electron transfer) Lvevi jeY (sodium salt) mg‡hvRx mej B‡jKUªb Ask MÖnY (Electron sharing) nxiK avZe `ye©j B‡jKUªb Ask MÖnY avZeLÛ f¨vÛvi Iqvjm `ye©j wØ-‡giæ wg_w¯…qv (Bipolar interaction) KwVb wbqb (solid neon) ✅ f¨vÛvi Iqvjm eÜb Gi cÖKvi‡f` : 3 cÖKvi:- WvB‡cvj-WvB‡cvj eÜb WvB‡cvj- Avweó WvB‡cvj eÜb Avweó WvB‡cvj - Avweó WvB‡cvj eÜb (f¨vÛvi Iqvjm e‡ji cÖavb Ask) ✅ cøvRgv Ae¯’v: AZ¨vwaK ZvcgvÎvq evqexq c`v_© AvqwbZ n‡q mg msL¨K ab I FY Avqb m„wó nq| c`v‡_©i GB Ae¯ v‡KB cøvRgv Ae¯ v e‡j| GwU c`v‡_©i PZz_© Ae¯ v G Ae¯ vq c`v‡_©i cigvYyi wbDwK¬qvm‡K wN‡i _vKv B‡jKUª‡bi GK ev GKvwaK B‡jKUªb gy³ n‡q hvq| REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF JU QUESTION 01. c`v‡_©i ivmvqwbK e܇bi gv‡S me‡P‡q `ye©j eÜb- [JU. 16-17] A. AvqwbK eÜb B. mg‡hvRx eÜb C. f¨vbWviIqvjm eÜb D. avZe eÜb me‡P‡q †ekx kw³kvjx eÜb AvqwbK, me‡P‡q `ye©j eÜb f¨vbWviIqvjm| 02. Amsi¶Ykxj ej- [JU. 12-13] A.mv›`ª ej B.gva¨Kl©Y ej C.Zwor ej D.†KvbwUB bq Ans A 03. hw` p Ges b h_vµ‡g cxob Ges weK…wZi gvb nq, Z‡e G‡`i g‡a¨ mwVK m¤úK© nj- [JU. 10-11] A. p = b B. pb C. p b 1 D. p 2 1 b Ans B STEP 02 ANALYSIS OF RU QUESTION 01. ÔcøvRgvÕ c`v‡_©i †Kvb ai‡bi Ae¯’v? [RU. 18-19] A. KwVb B. AvqwbZ C. Zij D. M¨vmxq Ans B 02. AvqwbK I mg‡hvRx e܇bi Zzjbvq avZe eÜb- [RU. 17-18] A. kw³kvjx B. mgvb C. `ye©j D. †KvbwUB bq Ans C 03. †Kvb eÜbwU †ekx kw³kvjx nq? [RU. 16-17] A. mg‡hvRx B. AvqwbK C. avZe D. f¨vbWviIqvjm me‡P‡q †ekx kw³kvjx eÜb AvqwbK, me‡P‡q `ye©j eÜb f¨vbWviIqvjm| STEP 03 ANALYSIS OF CU QUESTION 01. e¯‘i †Kvb a‡g©i Rb¨ Kuv‡Pi Mv‡q cvwbi †dvUv †j‡M _v‡K- [CU. 07-08] A. mv›`ªZv B. c„ôUvb C. AvmÄb D. msbg¨Zv E. AvšÍtAvYweK ej Ans B STEP 04 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. PK I †ev‡W©i AYyi g‡a¨ AvKl©b ej? [KU. 09-10] A. mskw³ ej B. AvmÄb ej C. AwfKl© ej D.mv›`ª ej mskw³ ej: GKB c`v‡_©i wewfbœ Abyi g‡a¨ AvKl©Y| AvmÄb ej: wewfbœ c`v‡_©i Abyi wfZi cvi¯úwiK AvKl©Y ej| IU 01. wfbœag©x Avq‡bi g‡a¨ Kzj¤^ AvKl©‡bi d‡j eÜb m„wó nq- [IU. 15-16] A. avZe eÜb B. AvqwbK eÜb C. f¨vÛvi Iqvjm eÜb D. mg‡hvRx eÜb Ans B PART B Analysis of Science & Technology Question BSMRMU 01. cvwbi AYyi ga¨Kvi eÜb‡K Kx e‡j? [BSMRMU. 2020-21; MBSTU. 14-15] A. Ionic Bond B. Covalent Bond C. Dipole-Dipole Bond D. Metalic Bond S C info cvwbi AYy‡Z w؇cvj (Dipole) we`¨gvb| d‡j AYy-AYy‡Z we`¨gvb eÜb n‡jv Dipole-Dipole Bond. STEP 05 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. AvšÍtAvYweK e‡ji gvÎv ev AYy‡Z e܇bi cÖK…wZi Dci wbf©i K‡i †h․M mg~n‡K fvM Kiv hvq [MAT. 02-03] A. fvM Kiv hvq bv B. `yB fv‡M C. wZb fv‡M D. Pvi fv‡M Ans B STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. [Xv. †ev. 2016] i. mskw³ ej > AvmÄb ej ii. AvmÄb ej > mskw³ ej iii. 60 nj ¯úk© †KvY wb‡Pi †KvbwU mwVK? A. i B. i I iii C. i I ii D. ii I iii Ans A 02. †Kv‡bv c`v‡_©i AYy¸wji g‡a¨ wbU ej k~b¨ nq hLb-[Xv.†ev.2015; P.‡ev.2015] A. r = r0 B. r < r0 C. r > r0 D. r >> r0 Ans A 03. wewfbœ c`v‡_©i Aby¸‡jvi g‡a¨ cvi¯úwiK AvKl©b ej‡K ejv nqÑ [iv.‡ev. 2019] A. c„ôUvb B. AvmÄb C. mskw³ ej D. mv›`ª ej Ans B
264 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 04. †Kvb Ae¯’vq AYymg~‡ni g‡a¨ AvšÍtAvYweK AvKl©Y ej me©wb¤œ nq?[h. †ev. 2016] A. Zij B. cøvRgv C. KwVb D. evqexq Ans D 05. AvšÍtAvYweK AvKl©Y I weKl©Y ej mgvb nq hLb [P. †ev. 2015] A. r > r0 B. r < r0 C. r = 0 D. r = r0 Ans D 06. †KvbwUi †ÿ‡Î f¨vbWvi Iqvjm ej we`¨gvb? [iv. †ev. 2015] A. Aw·‡Rb AYyi eÜb B. †mvwWqvg I †K¬vwib cigvYyi eÜb C. wmwjKb cigvYyi eÜb D. Zvgvi cigvYy eÜb Ans A Concept 2 w¯’wZ¯’vcKZv m¤úwK©Z Z_¨vejx ✅ w¯’wZ¯’vcKZv m¤úwK©Z ivwkgvjv: K. w¯’wZ¯’vcKZv: †h a‡g©i d‡j evB‡i †_‡K cÖhy³ ej AcmvwiZ n‡j weK…Z e¯‧ Zvi c~‡e©i Ae¯ vq wd‡i Av‡m, Zv‡K w¯ wZ¯ vcKZv e‡j| ◈ †jvnv ev B¯úvZ ivev‡ii †P‡q †ekx w¯ wZ¯ vcK | L. bgbxq e¯‘/cøvw÷K e¯‘/Aw¯’wZ¯’vcK e¯‘: weK…ZKvix ej Acmvi‡Yi ci hw` e¯‧i Ae¯ vb cybtcÖvwß bv N‡U Z‡e Zv‡K bgbxq e¯‧ e‡j| ev¯Í‡e c~Y© bgbxq e¯‧ cvIqv hvq bv| Z‡e AvUvi `jv, gvwUi `jv/ Kv`v, PzBsMvg, †gvg cªf„wZ‡K bgbxq e¯‧ wn‡m‡e aiv nq| M. c~Y© w¯’wZ¯’vcK e¯‘: †Kvb e¯‧i Dci ej cÖ‡qvM Kivi ci H ej AcmviY Kiv n‡j e¯‧wU hw` cy‡ivcywi c~‡e©i Ae¯ v wd‡i cvq Z‡e Zv‡K c~Y© w¯ wZ¯ vcK e¯‧ e‡j| mvaviYZ ivevi‡K c~Y© w¯ wZ¯ vcK e¯‧ aiv nq| N. c~Y© `„p e¯‘: †Kvb e¯‧i Dci †h †Kvb ej cÖ‡qv‡M K‡i hw` Zvi weK…Z ev KvwqK cwieZ©b NUv‡bv bv hvq, Z‡e H e¯‧‡K c~Y© `„p e¯‧ e‡j| c~Y©`„p e¯‧ ev¯Í‡e cvIqv hvq bv| Z‡e e¨envwiK Kv‡R mvaviYZ; KuvP, B¯úvZ‡K c~Y© `„p e¯‧ wn‡m‡e aiv nq| O. w¯’wZ¯’vcK mxgv: B¯úvZ I nxivi w¯ wZ¯ vcK mxgv Lye †ekx Avevi `¯Ívi w¯ wZ¯ vcK mxgv Lye Kg| P. Amn fvi Ges Amn cxob: Amn cxob = Amn fvi ‡¶Îdj = F A Q. weK…wZ: †Kvb GKwU e¯‧i Dci ej cÖ‡qv‡M Gi GKK gvÎvq †h cwieZ©b N‡U Zv‡K weK…wZ e‡j| weK…wZ †¯‥jvi ivwk| gvÎv I GKK †bB| S. cxob: †Kvb GKwU e¯‧i GKK †¶Îd‡ji Dci wµqvg~jK ev cÖwZwµqvg~jK e‡ji gvb‡K cxob e‡j| cxob = F A ✅ cxo‡bi cÖKvi‡f`: (i) •`N©¨ cxob (ii) AvKvi ev K…šÍb cxob ev †gvPo cxob/ e¨eZ©b cxob (iii) AvqZb cxob ◈ cxo‡bi GKK: Nm-2 ev Pa ◈ cxo‡bi gvÎv: [ML-1T -2 ] ✅ w¯’wZ¯’vcK ¸YvsK wZb cÖKvi h_v: (i) Bqs Gi w¯ wZ¯ vcK ¸YvsK (ii) `„pZvi w¯ wZ¯ vcK ¸YvsK (iii) AvqZ‡bi w¯ wZ¯ vcK ¸YvsK ✅ wewfbœ c`v‡_©i Bqs (Y), KvwVb¨(n) I AvqZb (K) ¸YvsK: (KwVb c`v‡_©i wZbwUB _vK‡e, Zij I M¨v‡mi ïay AvqZb ¸YvsK _vK‡e ) c`v_© Y(Nm-2 )1010 n(Nm-2 )1010 K(Nm-2 )1010 Zvgv 12.6 4 14 †jvnv (XvjvB) 11 4.4 9 B¯úvZ 20 8.4 18 A¨vjywgwbqvg 7 2.6 7.5 cvwb - - 0.2 cvi` - - 2.6 ✅ msbg¨Zv: msÁv ‡Kv‡bv e¯‧i Pviw`K †_‡K mgvb Pvc cÖ‡qvM Ki‡j e¯‧wUi AvqZb K‡g hvq| e¯‧i G ag©‡K msbg¨Zv e‡j| MvwYwZK msÁv AvqZb ¸Yv‡¼i wecixZ ivwk‡K msbg¨Zv e‡j| msbg¨Zv = 1 k = 1 AvqZb ¸Yv¼ = AvqZb weK…wZ AvqZb cxob we‡kl Z_¨ KwVb I Zij c`v‡_©i Zzjbvq M¨v‡mi msbg¨Zv A‡bK †ewk| KLbI KLbI AvqZb ¸YvsK‡K Amsbg¨Zv e‡j | ✅ weK…wZ: †¯‥jvi iwk ( l L , v V , ), GKK: GKK bvB| ✅ cxob: †¯‥jvi ivwk (F/A), GKK: Nm-2 ev c¨vm‡Kj, gvÎv: ML-1T -2 ✅ û‡Ki m~Î, cxob weK…wZ = aªæeK ✅ cqm‡bi AbycvZ = cvk¦© weK…wZ ‣`N©¨ weK…wZ = 1 L d D ✅ cqm‡bi Abycv‡Zi GKK bvB| Gi gvb – 1 0.5 ✅ avZe c`v‡_©i †ÿ‡Î: 0 0.5 ✅ AwaKvsk avZe c`v‡_©i †¶‡Î cqm‡bi Abycv‡Zi gvb 0.3 ✅ B¯úvZ ivevi A‡c¶v †ekx w¯ wZ¯ vcK
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 265 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ w¯ wZ¯ vcKZv wb‡gœv³ wel‡qi Dci wbf©i K‡i- AvNvZ (Hammering): AvNv‡Zi d‡j e¯‧ fv½‡Z PvB‡j w¯ wZ¯ vcKZv e„w× cvq| Lv` (Impurity): Lv‡`i Dcw¯ wZ‡Z e¯‧i w¯ wZ¯ vcKZv e„w× cvq| ZvcgvÎv (Temperature): ZvcgvÎv e„w× †c‡j w¯ wZ¯ vcKZv n«vm cvq| e¨wZµg: B¯úvZ, †KvqvUR© 500C ZvcgvÎvq cvwbi AvqZ‡bi w¯ wZ¯ vcK ¸YvsK me©vwaK| ✅ Amn fvi/IRb = cxob †¶Îdj = PA ✅ KwVb c`v_© 2 cÖKvi: ùwUK: hvi wbw`©ó MjbvsK Av‡Q| AwbqZvKvi: wbw`©ó MjbvsK bvB| ✅ Amn cxob: Amn e‡ji Rb¨ †h cxob nq ZvB Amn cxob| ✅ w¯ wZ¯ vcK K¬vwšÍ Gi Avwe®‥viK K¨vjwfb| Amn cxob: †Kvb GKwU e¯‧i GKK †¶Îd‡ji Dci cÖhy³ Amn fvi‡K Amn cxob e‡j| cxob w¯ wZ¯ vcK ¸YvsK A‡c¶v Kg n‡j weK…wZ ¯ vqx n‡e| Amsbg¨Zv: AvqZb ¸YvsK‡K A‡bK mgq Amsbg¨Zv e‡j| c`v‡_©i Amsbg¨Zvi µg Ñ KwVb > Zij > M¨vm ✅ Bqs Gi ¸YvsK Y Gi gvÎv [ML-1T -2 ] REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. w¯’wZ¯’vcK ¸Yvs‡Ki gvÎv Kx? [DU. 20-21] A. MLT1 B. ML1T 2 C. MLT2 D. ML2T 2 S B info Y = cxob weK…wZ = F A weK…wZ = Nm–2 weK…wZ = kg ms–2m –2 = ML1T 2 weK…wZi †Kv‡bv GKK ev gvÎv mgxKiY nq bv| 02. GKwU Zv‡ii Dci Uvb F n‡j •`N©¨e„w× nq x| ZviwU hw` û‡Ki m~Î †g‡b P‡j Ges Zv‡ii Dcv`v‡bi Bqs ¸YvsK Y nq Z‡e Zv‡i mwÂZ wefe kw³ KZ? [DU. 12-13] A. 1 2 Yx B. Yx C. 1 2 Fx D. Fx Uvb/ej = F, ‣`N©¨ e„w× l = x Ges Bqs Gi ¸YvsK = Y mwÂZ wefe kw³ Ep = ? W = Ep = 1 2 Fx STEP 02 ANALYSIS OF JU QUESTION 01. Pv‡ci gvÎv †KvbwU? [JU-A, Set-E. 20-21; H, Set-1. 19-20] A. [MLT–1 ] B. [ML–1T –2 ] C. [ML2T –2 ] D. [ML2T –1 ] S B info Avgiv Rvwb, P = F A MLT –2 L 2 = ML–1 T –2 Ges GKK Nm–2 ev Pa| 02. GKwU Av`k© `„p e¯‘i Rb¨ Bqs Gi ¸Yv¼- [JU-A, Set-A. 19-20] A. 0 B. C. 1 D. –1 Bqs Gi ¸Yv¼, Y = FL Al `„pe¯‧i †ÿ‡Î, l = 0 myZivs Y = 03. 50 kg f‡ii GKRb evjK Lvwj cv‡q `uvwo‡q Av‡Q| Zvi cv‡qi cvZvi †ÿÎdj 200 cm2 n‡j evjK KZ…©K f~wgi Dci Pvc KZ? [JU-H, Set-1. 19-20] A. 2.45 Pa B. 245 Pa C. 24.5 kPa D. 24.5 Pa P = F A = mg A = 50 9.8 200 104 . Pa = 24500 Pa = 24.5 kPa 04. Pv‡ci SI GKK n‡”Q- [JU. 16-17] A. N.m-2 B. N.m2 C. N/m D. N2m 2 Ans A 05. Bqs Gi ¸YvsK n‡e [JU. 13-14] A. A F Y B. L 1 Y C. Al FL Y D. A FL Y Ans C 06. ZvcgvÎv e„w× †c‡j B¯úv‡Zi w¯’wZ¯’vcK ¸YvsK [JU. 12-13] A. Increases B. Decreases C. Remain Unchanged D.None Ans A STEP 03 ANALYSIS OF RU QUESTION 01. K„šÍb weK…wZ n‡j, e¯‘i- [RU. 17-18] A. AvqZb cwiewZ©Z nq B. ‣`N©¨ cwiewZ©Z nq C. †¶Îdj cwiewZ©Z nq D. ‡KvbwUB bq K…šÍb weK…wZ = Av‡cw¶K miY e¨eavb `yiZ¡ = A Fh 02. w¯’wZ¯’vcK mxgv me‡P‡q †ewk- [RU. 17-18] A. nxivi B. B¯úv‡Zi C. `¯Ívi D. ivev‡ii Ans B 03. AvqZb ¸Yv‡¼i wecixZ ivwk‡K e‡j- [RU. 17-18 ; CoU. 09-10; Xv.†ev. 2017; P.†ev. 2016] A. AvqZb weK…wZ B. cxob C. AvqZb ¸Yv¼ D. msbg¨Zv Ans D 04. `yBwU wfbœ cÖ¯’‡”Q‡`i Zv‡ii Bqs-Gi w¯’wZ¯’vcK ¸YvsK GKB| Zvi `yBwU- [RU. 17-18] A. wfbœ •`‡N©¨i B. wfbœ Dcv`v‡bi C. GKB Dcv`v‡bi D. ‡h †KvbwU n‡Z cv‡i Bqs Gi ¸YvsK nq Zv‡ii Dcv`v‡bi| GKB Dcv`v‡bi Zv‡ii Bqs Gi ¸YvsK cwiewZ©Z nq bv 05. †Kvb w¯’wZ¯’vcK ¸Yv¼wU Zi‡ji Rb¨ cÖ‡hvR¨? [RU. 15-16] A. Bqs ¸Yv¼ B. `„pZvi ¸Yv¼ C. AvqZb ¸Yv¼ D. †KvbwUB bv Ans C 06. wb‡gœi †Kvb ivwkwUi †Kvb gvÎv ev GKK bvB? [RU. 14-15] A. Bqs ¸Yv¼ B. AvqZb ¸Yv¼ C. `„pZv ¸Yv¼ D. cqm‡bi AbycvZ Ans D 07. AvqZb weK…wZi GKK- [RU. 08-09] A. m3 B. bvB C. Nm–2 D. m Ans B 08. e¯‘i cÖwZ GKK gvÎvq †h cwieZ©b nq Zv‡K e‡j- [RU. 06-07] A. cxob B. weK…wZ C. w¯ wZ¯ vcKZv D. w¯ wZ¯ vcK mxgv Ans B STEP 04 ANALYSIS OF CU QUESTION 01. cxob (Stress) Gi gvÎv (Dimension) †KvbwU? [CU. 14-15, JnU. 14-15, IU. 01-02, 12-13; PSTU. 14-15, JnU. 13-14, CoU. 13-14; MBSTU-B. 15-16; JU. 12-13; 13-14] A. [ML-1T -2 ] B. [ML-2T -2 ] C. [ML-1T -1 ] D. [ML1T 2 ] S A info Avgiv Rvwb, cxob = ej †ÿÎdj = fi Z¡iY (•`N©¨) 2 = MLT –2 L 2 = [ML–1T –2 ] 02. GKwU wbLyuZ Amsbg¨ e¯‘i cqmb AbycvZ- [CU. 14-15] A. -1 B. -0.5 C. 0 D. 0.25 Ans C 03. wb‡Pi †KvbwU gvÎvMZfv‡e w¯’wZ¯’vcK ¸Yvs‡Ki mgZzj? [CU. 13-14] A. cxob B. weK…wZ C. c„ôUvb D. mv›`ªZv E. Z¡iY Ans A 04. L ‣`N©¨ Ges A cÖ¯’‡”Q` wewkó GKwU Zv‡ii •`N©¨ eivei F ej cÖ‡qvM Kivq •`N©¨ l cwigvY e„w× cvq| ZviwU‡Z K…Z Kv‡Ri cwigvY KZ? [CU. 13-14] A. FL B. F L 2 C. EL D. Fl 2 E. F 1 2A . 2 1 2 2 2 Fl Ans L YAl l L YAl w
266 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. GKwU Zvi ej cÖ‡qv‡M m¤cÖmvwiZ Ki‡j, ZviwUi GKK AvqZb mwÂZ kw³i ivwk †KvbwU? [CU. 10-11] A. 1 2 ej weK…wZ B. 1 2 ej weK…wZ C. 1 2 cxob ej D. cxob weK…wZ E. 1 2 •`N¨© cxob •`N¨© weK…wZ Ans E 06. mvaviYZ ZvcgvÎv e„wׇZ †Kvb GKwU e¯‘i w¯’wZ¯’vcKZv- [CU. 06-07] A. e„w× cvq B. n«vm cvq C. AcwiewZ©Z _v‡K D. gvcv hvq bv E. gvcv A_©nxb Ans B 07. ivev‡ii †P‡q B¯úv‡Zi Bqs Gi ¸YvsK- [CU. 05-06] A. †QvU B. eo C. wظY D. wZb¸Y E. Pvi¸Y Ans B STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. cqm‡bi AbycvZ (Possion's ratio) Gi gvÎv (dimension) †KvbwU? [JnU. 17-18] A. [ML–1T –2 ] B. [MLT–2 ] C. [ML–1T –1 ] D. †KvbwUB bq †h‡nZz cqm‡bi AbycvZ GKB cÖKvi `ywU ivwki AbycvZ ZvB Gi †Kvb gvÎv I GKK †bB| 02. 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 KU 01. †h c`v‡_©i evav`vbKvix ej †ekx †m c`v‡_©i- [KU. 13-14] A. w¯ wZ¯ vcKZv ‡ekx B. w¯ wZ¯ vcKZv Kg C. AvšÍtKYv AvKl©Y Kg D. NbZ¡ Kg Ans A 02. wb¤œwjwLZ †KvbwU mv›`ªZv Gi D`vniY- [KU. 11-12] A. f¨vbvwWqvg 1890C ZvcgvÎvq M‡j B. gay axi MwZ m¤úbœ Zij C. mvjdvi Gi MÜ cPv wW‡gi gZ D. `ya A¯^”Q Ans B 03. mv›`ªZv mnM () Gi gvÎv- [KU. 11-12] A. ML-1T -1 B. MLT-1 C. ML-2T -2 D. ML-2T -1 Ans A 04. gvbe †`‡n wkiv Dcwkiv w`‡q i‡³i PjvPj †Kvb a‡g©i Dci n‡q _v‡KA. cøveZv B. mv›`ªZv C. •KwkKZv D. c„ôUvb Ans B 05. •`N©¨ eivei mgvb ej cÖ‡qv‡M GKB c`v‡_©i wb‡¤§v³ 4wU Zv‡ii g‡a¨ †KvbwUi •`N©¨ e„w× me‡P‡q †ekx n‡e? [KU. 09-10] A. •`N©¨ 4m Ges e¨vmva© 4mm B. •`N©¨ 3m Ges e¨vmva© 3mm C. •`N©¨ 2m Ges e¨vmva© 2mm D. •`N©¨ 1m Ges e¨vmva© 1mm Ans D CoU 01. w¯’wZ¯’vcK mxgvi g‡a¨ cxob weK…wZi mgvbycvwZK GwU- [CoU. 15-16] A. wbDU‡bi m~Î B. û‡Ki m~Î C. cqm‡bi AbycvZ D. Bqs Gi ¸YvsK Ans B IU 01. gvÎvwenxb ivwk- [IU-D, 19-20] A. weK…wZ B. cxob C. Bqs ¸Yv¼ D. `„pZvi ¸Yv¼ weK…wZi gvÎv I GKK †bB| 02. GKwU m¤ú~Y© `„pe¯‘i AvqZb ¸Yv¼- [IU. 16-17; BRUR. 15-16] A. 0 B. 1 C. D. –1 AvqZb ¸YvsK, K = A v FV ; GKwU m¤ú~Y© `„pe¯‧i, Δv = 0; K= A0 FV 03. GKwU Zvi cÖmvi‡Yi d‡j GKK AvqZ‡b K…Z KvR- [IU. 15-16] A. cxob×weK…wZ B. 1 2 ×cxob×weK…wZ C. cxob/ weK…wZ D. weK…wZ=cxob Ans B 04. Amn cxob- [IU. 14-15] A. Amnej †¶Îdj B. †¶Îdj Amnej C. Amnej †¶Îdj D. weK…wZ Amnej Ans A 05. K…šÍb cxob I K…šÍb weK…wZi AbycvZ‡K ejv nq- [IU. 13-14] A. KvwVb¨ ¸bv¼ B. K„šÍb ¸bv¼ C. Bqs ¸bv¼ D. msbgZv Ans B 06. †jvnvi w¯’wZ¯’vcKZv ivev‡ii †P‡q- [IU. 12-13] A. ‡ekx B. Kg C. mgvb D. †KvbwU bq Ans A 07. w¯’wZ¯’vcK K¬vwšÍi Avwe®‥viK n‡jb weÁvbx- [IU. 10-11] A. ûK B. Bqs C. wbDUb D. ‡Kjwfb Ans D 08. w¯’wZ¯’vcKZvq A F †K wK e‡j? [IU. 05-06] A. Bqs ¸YvsK B. `„pZvi ¸YvsK C. AvqZb ¸YvsK D. ZvcgvÎvi ¸YvsK Ans B BRUR 01. †KvbwU ûK-Gi m~Î- [BRUR. 13-14; mKj †evW© 2018] A. Stress/Strain = Constant B. Strain/stress = constant C. Stress=Strain D. Stressconstant = Strain Ans A BU 01. GKK AvqZ‡b w¯’wZkw³i mgxKiY- [BU. 14-15] A. E = cxob weK…wZ 2 B. E = cxob weK…wZ C. E = 1 2 cxob / weK…wZ D. E = 1 2 cxob weK…wZ Ans D 02. wb‡Pi †Kvb&wU mwVK? [BU. 12-13] A. U = 1 2 YAl2 L /Al B. U = 1 2 YAl2 L C. U = 1 2 l L Yl D. U = W A Ans B JKKNIU 01. A¨vjywgwbqvg Bqs ¸Yv¼ KZ? [JKKNIU-B, Set-3, 19-20] A. 7 1010 B. 10 1010 C. 1.6 1010 D. None Ans A PART B Analysis of Science & Technology Question SUST 01. Bqs Gi ¸YvsK †Kvb c`v‡_©i me‡P‡q †ekx? [SUST. 07-08; w`. †ev. 15] A. ivevi B.Zvgv C. ¯^Y© D. B¯úvZ Ans D JUST 01. wb‡Pi †Kvb c`v‡_©i w¯’wZ¯’vcKZv me‡P‡q †ewk? [JUST. 14-15] A. cvwb B. wcP C. ivevi D. †jW E. B¯úvZ Ans E MBSTU 01. Bqs Gi ¸bv‡¼i gvÎv- [MBSTU. 14-15; Kz.†ev. 2019] A. ML–1T –2 B. MLT–2 C. ML–2T –1 D. MLT–1 Ans A HSTU 01. †KvbwUi AvqZb ¸YvsK me‡P‡q †ewk? [HSTU. 15-16] A. B_vbj B. wMømvwib C. †c‡Uªvwjqvg D. cvwb Ans B
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 267 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION RUET 01. bx‡Pi †KvbwU gvÎvMZfv‡e w¯’wZ¯’vcK ¸Yvs‡Ki mgZzj¨ [RUET. 11-12] A. Stress B. Strain C. Surface tension D. Acceleration E. None Ans A STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. mev©‡cÿv w¯’wZ¯’vcK e¯‘ †KvbwU? [MAT. 17-18, JUST-A, Set-Ka 19-20, KU. 18-19] A.Zvgv B. †jvnv C. †KvqvU©R D. KvV Ans C 02. SI c×wZ‡Z cxo‡bi GKK †KvbwU? [MAT. 15-16] A. Nm–1 B. Nm C. Nm–2 D. m N Ans C 03. GKwU w÷‡ji Zv‡ii ZvcgvÎv evov‡j Bqs Gi ¸Yv¼- [MAT. 14-15] A. e„w× cv‡e B. n«vm cv‡e C. GKB _vK‡e D. cÖ_‡g e„w× ‡c‡q c‡i Kg‡e Ans A 04. w¯’wZ¯’vcKZv m¤ú‡K© bx‡Pi †Kvb Dw³wU mZ¨ bq? [MAT. 12-13] A. w¯ wZ¯ vcK ¸YvsK = cxob weK…wZ B. w¯ wZ¯ vcK ¸YvsK = weK…wZ cxob C. Amn cxob = Amnfvi ‡¶Îdj D. w¯ wZ¯ vcK ¸Yvs‡Ki GKK Nm-2 Ans B 05. †h me e¯‘i w¯’wZ¯’vcK ag© wewfbœ w`‡K wewfbœ Zv‡K e‡j? [MAT. 12-13] A. c~Y© `„p e¯‧ B. Amgw`K ag©x e¯‧ C. mgw`K ag©x e¯‧ D. c~Y© w¯ wZ¯ vcK e¯‧ Ans B 06. w¯’wZ¯’vcK ¸Yvs‡Ki †¶‡Î wb‡gœi †KvbwU mwVK bq? [MAT. 11-12] A. cxob GKK weK…wZ GKK = Nm-2 B. B¯úvZ ivev‡ii †P‡q †ekx w¯ wZ¯ vcK C. e¨eZ©b cxob = F/A D. Pvc e„wׇZ me mgq e¯‧ ms‡KvwPZ nq bv Ans A 07. wb‡gœi †KvbwU w¯’wZ¯’vcKZvi Rb¨ mwVK mgxKiY? [MAT. 10-11; Zcb m¨vi] A. Y = PV v B. W = dl DL C. = F A D. = MgL r 2 l Ans C 08. cqm‡bi AbycvZ n‡j wb‡gœi †KvbwU mwVK? [MAT. 02-03; CU. 09-10; 15-16] A. = cvk¦© weK…wZ ‣`N©¨ weK…wZ B. = •`N©¨ weK…wZ cvk¦© weK…wZ C. = cvk¦© weK…wZ •`N©¨ weK…wZ D. = cvk¦© weK…wZ Ans A 09. †h me e¯‘i w¯’wZ¯’vcK ¸Y me w`‡K mgvb bq Zv‡`i e‡j- [MAT. 01-02] A. Perfectly rigid body B. perfectly elasitc body C. Isotropic body D. Anisotropic body Ans D DAT 01. w¯’wZ¯’vcK mxgvi g‡a¨ e¯‘i e¨eZ©b cxob I e¨eZ©b weK…wZi AbycvZ n‡”QÑ [DAT. 2020-21] A. `„pZvi ¸bv¼ B. AvqZb ¸bv¼ C. Bqs-Gi ¸bv¼ D. cqm‡bi AbycvZ S A info w¯ wZ¯ vcK mxgvi g‡a¨ e¨eZ©b cxob I e¨eZ©b weK…wZi AbycvZ n‡jv `„pZvi ¸Yv¼| `„pZvi ¸YvsK = e¨eZ©b cxob e¨eZ©b weK…wZ cxob weK…wZ = aªæeK STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. w¯’wZ¯’vcK mxgvi g‡a¨ •`N©¨ cÖmviY ebvg fvi-Gi mwVK †jLwPÎ †KvbwU? [Xv. †ev. 2019] A. m¤úªmviY fvi B. m¤úªmviY fvi C. m¤úªmviY fvi D. m¤úªmviY fvi Ans A 01. cÖfve †Mvj‡Ki e¨vmva© †KvbwU? [iv. †ev. 2015] A. 10–15 m B. 10–10 m C. 10–9 m D. 10–8 m Ans B Concept 3 cÖevnx c`v‡_©i aviYv ✅ cÖevnx c`v_©: †h mKj c`v_© cÖevwnZ nq Zv‡`i cÖevnx c`v_© e‡j| ✅ cÖevnxi cÖevn wewfbœ cÖKvi n‡Z cv‡i: mg-cÖevn: hw` me©¶Y cÖevnxi †eM aªæe _v‡K, Z‡e Zv‡K mgcÖevn e‡j| Amg cÖevn: hw` me©¶Y cÖevnxi †eM GKB bv _v‡K, Z‡e Zv‡K AmgcÖevn e‡j| w¯’i cÖevn/ †mªvZ cÖevn/ aviv‡iL cÖevn: hw` me©Î cÖevnxi †eM mgvb _v‡K, Z‡e Zv‡K w¯ i cÖevn e‡j| Aw¯’i cÖevn: hw` me©Î cÖevnxi †eM mgvb bv _v‡K, Z‡e Zv‡K Aw¯ i cÖevn e‡j| mg‡iL cÖevn: hw` cÖevnxi ¯Íi ci¯ú‡ii mgvšÍiv‡j P‡j Z‡e Zv‡K mg‡iL cÖevn e‡j| wew¶ß cÖevn: hw` cÖevnxi ¯Íi ci¯ú‡ii mgvšÍiv‡j bv P‡j eis MwZ‡Z AveZ© ev N~Y©vq‡bi m„wó nq Zv‡K wew¶ß cÖevn e‡j| ✅ †mªvZ‡iLv ev aviv†iL cÖevn: †h cÖevnxi †e‡Mi wewfbœ we›`y‡Z cÖevnxi KwYKv¸‡jvi MwZ‡eM mg‡qi mv‡_ AcwiewZ©Z _v‡K Zv‡K aviv‡iL cÖevn e‡j| ✅ Avgv‡`i kix‡i agbx‡Z i³ mÂvjb cÖevn GKwU aviv‡iL cÖevn| ✅ msKU †eM: me©vwaK †h ‡eM ch©šÍ †Kvb Zij aviv‡iL cÖevn eRvq iv‡L †m †eM‡K msKU †eM e‡j| ✅ aviv‡iL cÖev‡ni •ewkó¨: cÖevnxi ¯Íi¸‡jv ci¯úi mgvšÍivj nq bv| wew¶ß cÖev‡n Pjgvb cÖevnxi KYv¸wji †eM wewfbœ| †eM †ekx n‡j MwZ †iLv¸wj N~wY©cv‡Ki gZ nq| wew¶ß cÖevn me©`v AkvšÍ cÖevn| ✅ †÷vKm& Gi cÖvwšÍK †e‡Mi mgxKiY: v = 2r2 ( – )g 9 ✅ cÖvwšÍK †e‡Mi wbf©ikxjZv: cÖvwšÍK †eM †Kvb wbw`©ó ¯ v‡b Zi‡ji mv›`ªZvs‡Ki e¨¯ÍvbycvwZK| v 1 e¯‧ I Zi‡ji Nb‡Z¡i mgvbycvwZK| v , cošÍ †Mvj‡Ki e¨vmv‡a©i e‡M©i mgvbycvwZK| v r 2
268 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 †K․wkK bj cvwb‡Z AvswkK †Wvev‡bv| Gi A‡a©K e¨vmv‡a©i Avi GKwU †K․wkK bj cvwb‡Z AvswkK †Wvev‡bv n‡j Zvi g‡a¨Kvi cvwbi ¯Í‡¤¢i D”PZv cÖ_gwUi g‡a¨Kvi cvwbi ¯Í‡¤¢i D”PZvi Zzjbvq KZ n‡e? [DU. 07-08, KU. 11-12] A. GK PZz_©vsk (one quarter) B. A‡a©K (Half) C. wظY (Double) D. Pvi¸Y (quadruple) h1r1 = h2 r2 ev h2 h1 = r1 r2 ev h2 h1 = 1 1 2 = 2 (wظY) 02. cvwb fwZ© GKwU weKv‡ii g‡a¨ GK UzKiv eid fvmgvb i‡q‡Q| eid Mjvi ci cvwbi †j‡fj- [DU. 05-06] A. rises B. falls C. remains same D. first increases and then decreases Ans C STEP 02 ANALYSIS OF RU QUESTION 01. cÖevnxi mv›`ªZvi gvb Abymv‡i †KvbwU mwVK? [RU. 17-18] A. AvjKvZiv > `ya > †Zj B. ‡Zj > AvjKvZiv > `ya C. AvjKvZiv > †Zj > `ya D. `ya > †Zj > AvjKvZiv Ans C STEP 03 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. bx‡Pi †KvbwU cÖevnx c`v‡_©i Rb¨ mwVK bq? [MAT. 09-10] A. weÁvbx j¨vcøvm me© cÖ_g AvYweK Z‡Ë¡i mvnvh¨ c„ô Uv‡bi e¨vL¨v †`b B. AvšÍ:AvYweK ej `yÕiKg msmw³ ej I AvmÄb ej C. DòZv ev ZvcgvÎvi Dci c„ô Uvb wbf©i K‡i| DòZv e„w× †c‡j c„ôUvb n«vm cvq D. ZvcgvÎv Kgvi mv‡_ mv‡_ M¨v‡mi mv›`ªZv e„w× cvq Ans D Concept 4 c„ôUvb m¤úwK©Z Z_¨vejx ✅ c„ôUvb: Zi‡ji DcwiZj GKwU Uvb Uvb w¯ wZ¯ vcK wSwjøi b¨vq AvPiY K‡i Ges Zij c„‡ô ¯úk©K eivei me w`‡K GKwU ej wµqv K‡i| Zij c„‡ô ¯úk©K eivei wµqvkxj G ej‡K Zi‡ji c„ô Uvb e‡j| ✅ c„ôUv‡bi e¨envi: cvwbi Z‡j ‡cvKvgvK‡oi PjvPj Mv‡Q cvwbi cwienb (msmw³ I AvmÄb e‡ji mw¤§wjZ wµqvq Mv‡Qi cvZvq cvwb ‡cu․‡Q hvq) mu~‡Pi cvwb‡Z fvmv (MÖxR gvLv‡bv mu~P) ✅ c„ô Uv‡bi e¨envwiK cÖ‡qvM: c„ô Uv‡bi Rb¨B cvwbi †duvUv, cvi‡`i †duvUv BZ¨vw` †MvjvKvi avib K‡i c„ô Uv‡bi Rb¨B Zi‡ji DcwiZj AeZj †`Lvq c„ô Uv‡bi Rb¨B cvi‡`i DcwiZj DËj †`Lvq ✅ GKwU KuvP c„‡ôi Dci cvwb Xvj‡j Zv hZUv Qovq `ya ZZUv Qovq bv mv›`ªZv I c„ôUvb Gi Rb¨| ✅ mv›`ªZv I ZvcgvÎvi g‡a¨ m¤úK© nj- T (M¨v‡mi †¶‡Î) ✅ c„ôUv‡bi AvYweK ZË¡: weÁvbx j¨vcjvm me© cÖ_g AvYweK Z‡Ë¡i mvnv‡h¨ Zi‡ji c„ôUvb e¨vL¨v K‡ib| ✅ AvšÍtAvYweK ej 2 cÖKvi : msmw³ ej (Cohesive Force) AvmÄb ej (Adhesive Force) ✅ msmw³ ej (Cohesive Force): GKB c`v‡_©i wewfbœ AYyi g‡a¨ cvi¯úwiK AvKl©Y ej‡K mskw³ ej e‡j| ✅ AvmÄb ej (Adhesive Force): wewfbœ c`v‡_©i AYyi †fZi cvi¯úwiK AvKl©Y ej‡K AvmÄb ej e‡j| ✅ Kj‡gi wb‡ei Kvwji cÖevn c„ôUv‡bi D`vniY| dvD‡›Ub †cb Kvwji mv›`ªZv a‡g©i Dci wfwË K‡i cÖ¯‘Z Kiv nq| m¤úK© ¯úk©‡KvY c„ôZj †fRv‡bv Qov‡bv •KwkK b‡j Zi‡ji msmw³ ej > 2 × AvmÄb ej ¯ ~j‡KvY DËj †fRv‡e bv Qwo‡q co‡e bv Ae‡ivnb NU‡e msmw³ ej < 2 × AvmÄb ej m~ÿ‡KvY AeZj †fRv‡e Qwo‡q co‡e Av‡ivnb NU‡e msmw³ ej = 2 × AvmÄb ej mg‡KvY mgZj ✅ msKU ZvcgvÎvq Zi‡ji c„ô Uvb k~b¨ ✅ mv›`ªZv Zij I evqexq c`v‡_©i mvaviY ag© ✅ ‡hme Zi‡ji NbZ¡ KwV‡bi NbZ¡ A‡c¶v Kg †mme Zij KwVb‡K wfRvq| Avi †hme Zi‡ji NbZ¡ KwV‡bi NbZ¡ A‡c¶v †ewk †mme Zij KwVb‡K wfRvq bv| ‡hgb: cvi`, KvuP ✅ cÖevnx: (K) ms‡KuvPbxq cÖevnx: Zij c`v_©; (L) Ams‡KuvPbxq cÖevnx: M¨vm ✅ we‡kl ag©: ïay gvÎ wbw`©ó c`v‡_© †`Lv hvq| KwVb c`v‡_©i we‡kl ag©: ZviZv, cviZv, f½yiZv, `„pZv Zij c`v‡_©i we‡kl ag©: c„ô Uvb, m‡gv”PkxjZv mv›`ªZv Zij I M¨vmxq c`v‡_©i mvaviY ag©| ✅ c„ôUvb I c„ô kw³: c„ôUvb: Zi‡ji g‡a¨ †h e‡ji cÖfv‡e Zij c„ô me©`v ms‡KvwPZ n‡q me©wbgœ †¶Îd‡j Avm‡Z Pvq †mB ej‡K c„ôUvb e‡j| c„ôUvb, T = F L c„ôUvb : GKK Nm-1 , gvÎv [MT-2 ] ✅ c„ôUv‡bi •ewkó¨: c„ôUvb Zi‡j Zj‡K msKzwPZ Kivi †Póv K‡i| Zij Z‡ji †¶Îdj evovevi †Póv Ki‡j c„ô Uvb Zv cÖwZ‡iva Kivi †Póv K‡i| Pvcmv›`ªZvsK ZvcgvÎvmv›`ªZvsK Zij c`v_© Pv‡ci †Kvb cÖfve †bB ZvcgvÎvmv›`ªZvsK M¨vmxq c`v_©
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 269 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ c„ô kw³ ev Zj kw³: Zij Z‡ji †¶Îdj GK GKK e„w× Ki‡Z †h cwigvY KvR mvwaZ nq Zv‡K c„ô kw³ (E) e‡j| E=T (cÖK…Z c‡¶ c„ô Uvb I c„ô kw³ GKB| ✅ GKK Nm-1 ev, Jm-2 ✅ gvÎv mgxKiY [MT-2 ] ✅ Zi‡ji c„ô Uvb wbY©q: ◈ Zi‡ji c„ô Uvb wbY©‡qi D‡jøL‡hvM¨ c×wZ: •KwkK bj c×wZ KzBs‡Ki c×wZ R¨vMv‡ii c×wZ ey`ey` c×wZ K e R KzBs‡Ki c×wZ, •KwkK bj c×wZ ey`ey` c×wZ R¨vMv‡ii c×vwZ ✅ c„ô Uv‡bi Dci cÖfvewe¯ÍviKvix welq: `~wlZ KiY: Zij hw` Pwe©, †Zj Øviv `~wlZ nq Z‡e c„ô Uvb n«vm cvq| Zi‡ji gy³ Zj: Zi‡ji gy³ Z‡ji mv‡_ Ab¨ †Kvb e¯‧ mshy³ _vK‡j c„ô Uvb n«vm cvq| ZvcgvÎv: ZvcgvÎv e„w× †c‡j Zi‡ji c„ô Uvb n«vm cvq| ZvcgvÎv n«vm †c‡j c„ôUvb e„w× cvq| Zwor AvwnZ: Zij Zwor AvwnZ n‡j c„ô Uvb n«vm cvq| A‣Re c`v_© `ªexf~Z Ki‡Y c„ô Uvb e„w× cvq| ✅ c„ô Uvb m¤úwK©Z NUbv: m~P cvwb‡Z fvmv †Zj †X‡j mgy`ª kvšÍ Kiv Kc©y‡ii cvwb‡Z bvPvbvwP Kiv cvwbi Dci †Zj Qwo‡q cov Kj‡gi wb‡e Kvwj cÖevn QvZvi Kvc‡o e„wói cvwb wfZ‡i bv Avmv †Mvmj Kivi ci †Zvqv‡j Øviv Mv †gvQv| gvwUi cv‡Î cvwb VvÛv _v‡K miæ KuvP b‡ji cÖv‡šÍ Zvc w`‡j cÖvšÍwU †MvjvKvi n‡q hvq evwZi mj‡Z †e‡q †Zj DVv gvwU mvaviYZ wfRv Ges ‡e‡j gvwU Kv`v gvwUi †P‡q ï®‥ cvi` e¨v‡ivwgUv‡ii cvV me mgqB cÖK…Z cv‡Vi ‡P‡q Kg nq †Kvb cwi®‥vi KuvP c„‡ô cvwb Qwo‡q c‡o wKš‧ cvi` †dvUvi AvKvi avib K‡i| ◈ c„ôUvb m¤úwK©Z NUbv g‡b ivLvi †K․kj: QvZvq Kvwj jvM‡j Kc~©i †Zj Avi cvwb w`‡q my›`i K‡i ay‡Z n‡e| QvZvq QvZvi Kvc‡o e„wói cvwb wfZ‡i bv Avmv Kvwj jvM‡j Kj‡gi Kvwj cÖevn Kc~©i Kc~©‡ii cvwb‡Z bvPvbvwP ‡Zj Avi †Zj †X‡j mgy`ª kvšÍ Kiv| cvwb w`‡q cvwbi Ici †Zj Qwo‡q cov my›`i K‡i ay‡Z n‡e| m~u‡Pi cvwb‡Z fvmv REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU KvuP c„‡ôi Dci cvwb Xvj‡j Zv hZUv Qovq `ya ZZUv Qovq bv| Gi KviY- [DU. 09-10] A. mv›`ªZv B. c„ôUvb C. Dfq D. †KvbwUB bq Ans B STEP 02 ANALYSIS OF JU QUESTION 01. GKB c`v‡_©i wewfbœ AYyi g‡a¨ cvi¯úwiK AvKl©Y ej‡K wK e‡j? [JU-A, Set-C. 20-21] A. c„ôkw³ B. mskw³ ej C. AvmÄb ej D. c„ôUvb S B info mskw³ ej: GKB c`v‡_©i wewfbœ AYyi ga¨eZ©x AvKl©b ej| AvmÄb ej: wewfbœ c`v‡_©i AYyi ga¨eZ©x AvKl©b ej| 02. c„ôkw³i GKK †KvbwU? [JU-A, Set-B. 20-21, JUST-B, 19-20, BSFMSTU-A, 19-20] A. Nm–1 B. N–1m C. Nm D. Nm–2 S A info Avgiv Rvwb, T = F L c„ôUvb/c„ôkw³i GKK = N m = Nm–1 03. c„ôUvb n‡jv- [JU. 12-13] A. ej/†¶Îdj B. ej/•`N©¨ C. ej ‣`N©¨ D. cxob †ÿÎdj Ans B STEP 03 ANALYSIS OF RU QUESTION 01. msKU ZvcgvÎvq Zi‡ji c„ôUvb- [RU. 17-18] A. k~b¨ B. Amxg C. mmxg D. †KvbwUB bq Surface tension decreases with the increase oftemperature, reaching a value of zero at the critical temperature. 02. c„ôUv‡bi Gm. AvB GKK n‡jv- [RU. 16-17, SUST. 06-07; Xv. †ev. 2016;e. †ev. 2016] A. Nm-2 B. Nm C. Nm2 D. Nm-1 Ans D 03. Bernoulli Gi m~Î n‡”Q [RU. 09-10] A. kw³i msi¶Y bxwZ B. †iv‡ai m~Î C. c„ô Uv‡bi m~Î D. None of them Ans C 04. ej c‡q›U †cb †Kvb bxwZ‡Z KvR K‡i? [RU. 06-07; KU. 09-10] A. c„ôUvb B. mv›`ªZv C. ga¨vKl©Y D. †KvbwUB bq Ans A
270 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 04 ANALYSIS OF CU QUESTION 01. mvev‡bi ey`ey‡`i Pvc P, c„óUvb T Ges e¨vmva© r| G‡`i g‡a¨ m¤úK© Kx? [CU-A, Set-3. 20-21, BU. 10-11] A. P = 4T/r B. P = 2/r C. P = 4T/3r D. P = T/r S A info cvwbi Rb¨ Pvc = 2T/r| Ab¨w`‡K, mvev‡bi ey`ey‡`i Rb¨ Pvc = 4T/r | KviY, mvevb ey`ey‡`i `yBwU c„ô| 02. †Kvb Zi‡ji ¯úk©‡KvY 90 †_‡K Kg n‡j H Zi‡ji Zj Kxiƒc n‡e? [CU-A, Set-1. 20-21] A. DËj B. AeZj C. mgveZj D. m‡gvËj S B info Zi‡ji ¯úk© †KvY 90 A‡cÿv Kg n‡j ZijwU m~²‡Kv‡Y •KwkK b‡ji mv‡_ AeZjc„ô n‡q †j‡M _v‡K| 03. wb‡Pi †KvbwU c„ôUvb (T) Ges c„ôkw³ (E) Gi g‡a¨ m¤úK© eySvq? [CU-A, Set-1. 20-21] A. E = 2T B. E = T C. E = T/2 D. E = T/4 S B info msL¨vMZ w`K †_‡K ev mvswL¨K gv‡bi w`K †_‡K †Kv‡bv wbw`©ó Zi‡ji c„ôUvb Zvi wb‡Ri c„ôkw³i mgvb| 04. Zi‡ji c„‡ô †Kvb †Zj ev Pwe© RvZxq c`v_© fvmgvb _vK‡j Zi‡ji c„ôUvb wK nq? [CU. 15-16, 12-13, JUST. 15-16, IU. 16-17] A. †e‡o hvq B. wظY ev‡o C. K‡g hvq D. mgvb _v‡K E. k~b¨ _v‡K Zi‡j A‣Re c`v_© fvmgvb _vK‡j c„ôUvb e„w× cvq| Zi‡j ‣Re c`v_© fvmgvb _vK‡j c„ôUvb n«vm cvq| 05. mvaviY ZvcgvÎvq cvwbi c„ô kw³ KZ? [CU. 13-14, CU. 20-21] A. 6210–3 J/m2 B. 8210–4 J/m2 C. 7210–3 J/m2 D. 7210–4 J/m2 E. 6210–4 J/m2 Ans C 06. †h ZvcgvÎvq †Kvb GKwU Zi‡ji c„ôUvb k~b¨ nq Zv‡K e‡j- [CU. 12-13] A. k~b¨ ZvcgvÎv B. msKU ZvcgvÎv C. wngPK ZvcgvÎv D. cigk~b¨ ZvgcvÎv Ans B 07. c„ôUv‡bi gvÎv mgxKiY †Kvb&wU? [CU. 11-12, CU. 14-15] A. [ML2 ] B. M T 2 C. M L 2 D. [MT2 ] E. M L Ans B 08. c„ô Uv‡bi GKK †KvbwU? [CU. 09-10] A. Nm-1 B. Nm-2 C. NT-2 D. NT-1 E. Nm Ans A STEP 05 ANALYSIS OF DU-7 Clg QUESTION 01. ZjUv‡bi gvÎv †KvbwU? [DU. 7Clg-A: 20-21; CU-A, Set-4: 20-21, 13-14; DU. 00-01] A. MT–2 B. MT2L –1 C. MLT–1 D. ML–1T –2 S A info c„ôUvb/ZjUv‡bi GKK Nm–1 =1kgms–2 .m–1 =1 kg s2 [M T2 ] ZjUv‡bi gvÎv = [MT–2 ] STEP 06 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. ZvcgvÎv e„w× ‡c‡j Zj Uvb- [JnU. 15-16; DU-Tech.19-20] A. e„w× cvq B. w¯ i _v‡K C. n«vm cvq D. k~b¨ nq mvaviYfv‡e ZvcgvÎv e„w× †c‡j Zi‡ji c„ôUvb n«vm Ges ZvcgvÎv n«vm †c‡j Zi‡ji c„ôUvb e„w× cvq| e¨wZµg: MwjZ Zvgv I K¨vWwgqvg KU 01. Zi‡j c„ôUvb †Kvb wel‡qi Dci wbf©i K‡i bv? [KU. 10-11] A. ZvcgvÎv B. Zi‡ji Dcwiw¯ Z gva¨g C. `~lY D. AwfKl©xq ej Ans D IU 01. msKU ZvcgvÎvq Zi‡ji c„ôUv‡bi cwigvY- [IU. 13-14] A. me©vwaK B. me©wbgœ C. 5 GKK D. 0 Ans D 02. ZvcgvÎv evov‡j Zi‡ji c„ôUvb- [IU. 04-05] A. n«vm cvq B. e„w× cvq C. AcwiewZ©Z _v‡K D. †KvbwUB bq Ans A PART B Analysis of Science & Technology Question SUST 01. cvwbi Dci B¯úv‡Zi GKwU ‡eøW †f‡m _vKvi KviY- [SUST. 08-09] A. cvwbi c„ôUvb B. cvwbi DaŸ©Pvc C. cvwbi NYZ¡ B¯úv‡Zi Nb‡Z¡i †P‡q Kg nIqvi Rb¨ D. cvwbi viscosity Gi Kvi‡Y Ans A JUST 01. NaCl cvwb‡Z wgwkÖZ Ki‡j c„ôUvb- [JUST-C, 19-20] A. ev‡o B. K‡g C. k~b¨ D. AcwiewZ©Z _v‡K Zi‡j A‣Re c`v_© `ªexf~Z Ki‡j Ges ZvcgvÎv n«vm Ki‡j c„ôUvb ev‡o| Avi mKj †ÿ‡Î K‡g| NaCl GKwU A‣Re c`v_©| ZvB c„ôUvb evo‡e| 02. †Kvb GKwU Zij Z‡ji †¶Îdj GK GKK e„w× Ki‡Z †h cwigvY KvR mvwaZ nq, Zv‡K wK ejv nq? [JUST. 14-15] A. c„ôUvb B. mv›`ªZv C. c„ôkw³ D. K…šÍb cxob E. ej Ans C MBSTU 01. †Kvb a‡g©i Kvi‡Y cvwbi †duvUv ‡MvjvK…wZ nq? [MBSTU. 15-16, DU-Tech.19-20, CU. 12-13, 15-16; NSTU-A. 19-20] A. mv›`ªZv B. w¯ wZ¯ vcKZv C. c„ôUvb D. bgbxqZv c„ôUv‡bi Rb¨ cvwbi †dvUv †MvjvK…wZ nq Ges mv›`ªZvi Rb¨ cošÍ e„wói †dvUvi †eM Aev‡a e„w× cvqbv| HSTU 01. hLb cvwb‡Z wKQz wWUvi‡R›U †gkv‡bv nq ZLb Gi c„ôUvbÑ [HSTU. 15-16, NSTU. 14-15] A. AcwiewZ©Z _v‡K B. n«vm cvq C. e„w× cvq D. n«vm ev e„w× †c‡Z cv‡i Zi‡j A‣Re c`v_© `ªexf~Z _vK‡j c„ôUvb e„w× cvq, •Re c`v_© `ªexf~Z _vK‡j c„ôUvb n«vm cvq| 02. ‡Kvb a‡g©i Kvi‡b Zij Zj msKzwPZ nIqvi ‡Póv K‡i? [HSTU. 14-15] A. c„óUvb B. mv›`ªZv C. ‡K․wkKZv D. me¸wj Ans A STEP 07 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. 2:1 Abycv‡Zi e¨vm wewkó `ywU mvevb cvwbi ey`ey‡`i wfZi Kvi AwZwi³ Pv‡ci AbycvZ KZ n‡e? [BUET. 14-15] A. 1:2 B. 1:4 C. 2:1 D. 4:1 Ans A 02. ‡Kvb a‡g©i Kvi‡Y cvwbi †duvUv †MvjvK…wZ nq- [BUET :09-10] A. viscosity B. elasticity C. surface tenison D. capillarity Ans C STEP 08 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. Zi‡ji c„ôUv‡bi Dci wb‡Pi †KvbwUi cÖfve bvB? [MAT. 18-19] A. ZvcgvÎv B. Pvc C. `~wlZKiY D. `ªexf~Z e¯‧i Dcw¯ wZ Ans B 02. wb‡Pi †KvbwU Zi‡ji c„ô Uv‡bi Dci cÖfve we¯Ívi K‡i? [MAT. 10-11] A. gy³ Z‡ji ms¯ú‡k© †h gva¨g _v‡K Zvi Dci c„ô Uv‡bi gvb wbf©i K‡i bv B. Zi‡ji c„‡ô †Zj RvZxq c`v_© fvmgvb _vK‡j Zi‡ji c„ôUvb K‡g hvq C. mvevb `ªexf~Z Ki‡j cvwbi c„ô Uvb 7210-3Nm-1 nq D. DòZvi Dci c„ôUvb wbf©i K‡i bv Ans B
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 271 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 03. Zi‡ji c„ô Uv‡bi †¶‡Î †KvbwU mwVK bq? [MAT. 06-07] A. Zi‡ji gy³ Z‡j ev c„‡ô †Kv‡bv wKQz fvmgvb _vK‡j Zi‡ji c„ôUvb K‡g hvq B. Zij ZwoZvwnZ n‡j Gi c„ô Uvb n«vm cvq C. DòZv e„w× †c‡j c„ôUvb e„w× cvq D. gy³Z‡ji ms¯ú‡k© †h gva¨‡g _v‡K Zvi Dci c„ô Uvb wbf©i K‡i ZvcgvÎv e„w× †c‡j c„ô Uvb n«vm cvq| 04. wb‡gœi †KvbwU c„ô Uv‡bi Dci cÖfve we¯Ívi K‡i bv| [MAT. 07-05] A. `~wlZ KiY B. †P․¤^KZ¡ C. ZwoZvwnZKiY D. ZvcgvÎv Ans B 05. wb‡gœi †KvbwU c„ôUvb m¤úwK©Z NUbv bq? [MAT. 04-05] A. ej‡c‡b †jLv nIqv B. m~P cvwb‡Z fvmv C. Kc©~‡ii cvwb‡Z bvPv D. QvZvi Kvco Ans A DAT 01. ÒKc~©‡ii cvwb‡Z bvPvÓ- c`v‡_©i †Kvb a‡g©i R‡b¨ N‡U? [DAT. 19-20; KU. 07-08] A. ZjUvb B. w¯ wZ¯ vcKZv C. cwievwnZv D. mv›`ªZv c„ôUvb ev ZjUvb m¤úwK©Z K‡qKwU NUbv: Zi‡ji c„‡ô myB †f‡m _vKv| cvwbi Dci w`‡q †cvKvgvK‡oi nuvUv| mvev‡bi †dbv| AkvšÍ mgy`ª‡K kvšÍ Kiv| Mv‡Q cvwbi cwienY| †Kvb cwi®‥vi KvPc„‡ô cvwb Qwo‡q c‡o, Kc~©‡ii cvwb‡Z bvPv| wKš‧ cvi` †duvUvi AvKvi aviY K‡i| STEP 09 ANALYSIS OF HSC BOARD QUESTION 01. †QvU †QvU Zij we›`y †MvjvKvi aviY Kivi KviY n‡jvÑ [Xv. †ev. 2019] A. gnvKl©xq ej B. AvmÄb ej C. mKj w`K †_‡K mgvb Pvc D. c„ôUvb ej Ans D 02. e„wói GKwU eo †duvUv †f‡½ A‡bK¸‡jv †QvU †duvUvq cwiYZ n‡j †duvUv¸wji me©‡gvU- [Xv. †ev. 2016] A. †ÿÎdj n«vmcvq B. †ÿÎdj e„w× cvq C. AvqZb n«vm cvq D. †ÿÎdj AcwiewZ©Z _v‡K Ans B 03. †KvbwU c`v‡_©i mvaviY ag©? [h. †ev. 2017] A. c„ôkw³ B. mv›`ªZv C. w¯ wZ¯ vcKZv D. c„ôUvb Ans D 04. Zi‡ji c„ô Uvb wbf©i K‡i [e. †ev. 2017] i. ‣KwkK b‡ji e¨vmva© ii. mskw³ ej iii. Zi‡ji NbZ¡ wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans D DÏxc‡Ki Av‡jv‡K cÖ‡kœi DËi `vI: [gv. †ev. 2018] wP‡Î r e¨vmv‡a©i GKwU Kuv‡Pi •KwkK bj weï× cvwb‡Z Wzev‡j Zv cvwb‡Z h D”PZvq D‡V| c‡i 2r e¨vmv‡a©i Dci GKwU •KwkK bj cvwb‡Z Wzev‡bv n‡jv| h T sin T cos 05. 1g †ÿ‡Î c„ôUvb T1 I wØZxq †ÿ‡Î c„ôUvb T2 n‡j †Kvb m¤úK©wU mwVK? A. T2 > T1 B. T2 < T1 C. T1 = T2 D. 2H1 = T2 S C info Df‡qi c„ôUvb mgvb n‡e, T1 = T2 Concept 5 mv›`ªZvi aviYv Ges cÖ‡qvRbxqZv ✅ mv›`ªZv: ‡h a‡g©i `iæb ‡Kvb cÖevnxi wewfbœ ¯Í‡ii Av‡cw¶K MwZ‡Z evavi m„wó nq Zv‡K H cÖevnxi mv›`ªZv e‡j| †Zj, `ya I AvjKvZivi g‡a¨ AvjKvZivi mv›`ªZv me‡P‡q ‡ekx cvwbi Zzjbvq gayi mv›`ªZv ‡ekx ✅ mv›`ªZv µg: AvjKvZiv > †Zj > `ya Ges gay > cvwb| ✅ mv›`ªZv: Bnv c`v‡_©i we‡kl ag©| GwU Zij I evqexq c`v‡_©i mvaviY ag©| ✅ mv›`ªZv ¸Yvs‡Ki GKK: 1. Nsm-2 , 2. kgms-1 , (1Nsm-2 = 10poise) ✅ mv›`ªZv ¸Yvs‡Ki Dci ZvcgvÎv I Pv‡ci cÖfve| ✅ ZvcgvÎv: Zij c`v‡_©i mv›`ªZvsK ZvcgvÎv e„w×i mv‡_ `ªæZ n«vm cvq| M¨v‡mi mv›`ªZvsK ZvcgvÎv e„w×i mv‡_ e„w× cvq ✅ Pv‡ci cÖfve: Pvc e„w× †c‡j Zij c`v‡_©i mv›`ªZvsK e„w× cvq| M¨v‡mi mv›`ªZvsK Pv‡ci Dci wbf©ikxj bq| ✅ msKU ZvcgvÎv : †h ZvcgvÎvq †Kvb GKwU Zi‡ji c„ô Uvb k~b¨ nq, Zv‡K msKU ZvcgvÎv e‡j| ✅ mv›`ªZv mnM: GKK †eM Aeµ‡g GKwU cÖevnxi GKK †¶Îd‡ji Dci †h cwigvY mv›`ªZv ej wµqv K‡i, Zv‡K H cwievnxi mv›`ªZv ¸YvsK e‡j| ✅ GB ej cÖevnxi ¯úk©K eivei wµqv K‡i| ✅ Gi gvb cªevnxi cªK…wZ I ZvcgvÎvi Dci wbf©i K‡i| ✅ gvÎv mgxKib: [ML-1T -1 ] ✅ mv›`ªZv msµvšÍ NUbvejx: kxZj cvwbi †P‡q Mig cvwbi MwZ `ªæZZi nq| Aevafv‡e cZbkxj e„wói †duvUv AšÍ‡eM/ cÖvwšÍK †eM (aªæe †eM) cÖvwßi Kvi‡b D”P †eM cÖvß nq bv| ✅ mv›`ªZvi cÖ‡qvRbxqZv: MwZkxj ‡b․Kv, ÷xgvi, jÂ, Rvnv‡Ri Dci cvwbi Ges MwZkxj gUi Mvwo I wegv‡bi Dci evqyi mv›`ªZvRwbZ evav j¶¨ K‡iB G mg¯Í h‡š¿i b·v •Zix Kiv nq| kxZj cvwbi †P‡q Mig cvwbi MwZ `ªZZi nq| dvDb‡Ub ‡cb Kvwji mv›`ªZv a‡g©i Dci wfwË K‡iB cÖ¯‧Z Kiv nq| AšÍ‡eM cÖvwßi Kvi‡Y Aevafv‡e cZbkxj e„wói †duvUv D”P †eM cÖvß nq bv aªæe †eM wb‡q co‡Z _v‡K| wkiv Dcwkiv w`‡q i‡³i PjvPj GB a‡g©i Dci n‡q _v‡K| AvKv‡k Nywo Dov|
272 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. Zi‡ji †¶‡Î mv›`ªZvi mn‡Mi m‡½ ZvcgvÎvi m¤úK© n‡”Q- [DU. 08-09; BU. 11-12] A. T B. T C. T 2 D. None of this Zij c`v‡_©i mv›`ªZv mnM ZvcgvÎv e„w×i mv‡_ `ªæZ n«vm cvq| A_©vr mv›`ªZvi mnM cig ZvcgvÎvi e¨¯ÍvbycvwZK| STEP 02 ANALYSIS OF JU QUESTION 01. mv›`ªZv ¸Yvs‡Ki GKKÑ [JU-A, Set-G. 20-21; Xv.†ev. 2016] A. Nm–1 S B. Nsm–1 C. N–1m –1 S D. Nsm–2 S D info mv›`ªZv ¸Yvs‡Ki GKK: Nsm–2 Ges gvÎv : ML–1T –1 02. †h a‡g©i d‡j Zij Zvi wewfbœ ¯Í‡ii Av‡cwÿK MwZi we‡ivwaZv K‡i Zv‡K Zi‡ji Kx e‡j? [JU-A, Set-E. 20-21] A. mv›`ªZv B. mv›`ªZv ej C. mv›`ªZv¼ D. c„ôUvb S A info mv›`ªZv: †h a‡g©i d‡j Zi‡ji wewfbœ ¯Í‡ii Av‡cwÿK MwZ evavcÖvß nq Zv‡K Zi‡ji mv›`ªZv e‡j| 03. M¨v‡mi mv›`ªZv ¸Yv¼ ZvcgvÎv e„w×i mv‡_- [JU-A, Set-D. 19-20] A. e„w× cvq B. K‡g C. AcwiewZ©Z _v‡K D. †KvbwUB bq M¨v‡mi †ÿ‡Î mv›`ªZv¸bv¼ ZvcgvÎv e„w×i mv‡_ e„w× cvq| T ; Zi‡ji †ÿ‡Î mv›`ªZv ¸bYv¼ ZvcgvÎv e„w×i mv‡_ n«vm cvq| 04. mv›`ªZvq cigvYyi- [JU. 13-14] A. AvšÍtAvYweK ej AwfKl© e‡ji wecix‡Z KvR K‡i B. AvšÍtAvYweK ej AwfKl© e‡ji w`‡K KvR K‡i C. AvšÍtAvYweK ej AwfKl© e‡ji mgvšÍivj KvR K‡i D. †KvbwUB bq Ans A 05. mv›`ªZvi ¸YvsK n‡e- [JU. 12-13] A. = F A dv dy B. = F A dl dL C. = a D. = F A l L Ans A STEP 03 ANALYSIS OF RU QUESTION 01. mv›`ªZv wK‡mi mv‡_ Zzjbxq? [RU. 17-18] A. ej B. KvR C. Z¡iY D. Nl©Y Ans D STEP 04 ANALYSIS OF CU QUESTION 01. KuvPc„‡ô cvwb Qwo‡q c‡o wKš‘ cvi` †duvUvq cwiYZ nq †Kb? [CU. 07-08] A. mv›`ªZvi Rb¨ B. mg‡qi Rb¨ C. Aw¯ i cÖev‡ni Rb¨ D. Zi‡ji c„ôUv‡bi Rb¨ E. mv›`ªZv Ges c„ôUv‡bi Rb¨ Ans E STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. mv›`ªZv ¸Yvs‡Ki gvÎv mgxKiY †KvbwU? [JnU. 09-10, 13-14, RUET. 06-07, RU. 14-15, 10-11, CU-A, 19-20, JU. 13-14; iv.†ev. 20017] A. [ML–1T -3 ] B. [M-1L -1T -1 ] C. [ML-1T 2 ] D. [ML-1T -1 ] S D info mv›`ªZv¼, = Fdx Adv KU 01. me©v‡cÿv w¯’wZ¯’vcK †KvbwU? [KU. 18-19] A. †jvnv B. Zvgv C. †KvqvU©R D. KvV Ans C IU 01. c‡qR (Poise) wK‡mi GKK? [IU. 11-12; w`. †ev. 2017] A. c„ôUvb B. c„ôkw³ C. ZjUvb D. mv›`ªZv Ans D BRUR 01. Zi‡ji †ÿ‡Î mv›`ªZv mn‡Mi mv‡_ ZvcgvÎvi m¤úK©- [BRUR-F, Set-2, 19-20] A. T B. T C. 1 T D. T 2 Zi‡ji †ÿ‡Î 1 T Ges M¨v‡mi †ÿ‡Î T A_©vr Zi‡ji †ÿ‡Î ZvcgvÎv evo‡j mv›`ªZv n«vm cvq Ges M¨v‡mi †ÿ‡Î e„w× cvq| 02. cqm‡bi AbycvZ cÖKvk Kiv nq? [BRU. 12-13] A. = r L L B. = Lr rt C. = Lr rl D. = rr Ll Ans C 03. 1.0 c‡qR (Poise)=? [BRU. 15-16] A. 1.0 wbDUb/†m‡KÛ B. 1.0 wbDUb. †m‡KÛ C. 0.1 wbDUb/†m‡KÛ D. 0.1 wbDUb. †m‡KÛ Ans D BU 01. M¨v‡mi †¶‡Î mv›`ªZv ¸YvsK ZvcgvÎv T Gi mv‡_ wbgœiƒ‡c ev‡o- [BU. 11-12] A. T B. T C. T 1 D. T 1 Ans A PART B Analysis of Science & Technology Question SUST 01. GKwU Zij I GKwU KwVb c`v‡_©i ga¨Kvi ¯úk© †KvY †KvbwU n‡j Zij c`v_©wU KwVb c`v_©wU‡K †fRv‡e bv? [SUST. 18-19] A. 0 B. 30 C. 60 D. 45 E. 110 Ans E HSTU 01. Aevafv‡e cZbkxj e„wói ‡dvUv D”P‡eM cÖvß nq bv †Kb? [HSTU. 15-16] A. ‘g’ aªæe‡Ki gv‡bi Rb¨ B. Zij c`v‡_©i Rb¨ C. c„ôUv‡bi Rb¨ D. AšÍ‡eM cÖvwßi Rb¨ e„wói †duvUv evqy gÛ‡ji wfZi w`‡q cZ‡bi mgq AwfK‡l©i Kvi‡Y Gi †eM e„w× †c‡Z _v‡K Ges mv›`ªZvi Kvi‡Y Gi Dci evqy gÛ‡ji evav`vbKvix ejI e„w× cvq| GK mgq †dvUvi wbU Z¡ib k~b¨ nq| †dvUvwU ZLb aªæe †e‡M co‡Z _v‡K| GB †eMB AšÍ‡eM| AšÍ‡eM cÖvwßi Kvi‡Y Aev‡a cZbkxj e„wói †dvUv D”P †eM cÖvß nq bv| NSTU 01. †Kvb e¯‘ hw` AwfK‡l©i cÖfv‡e †Kvb Zi‡ji ga¨w`‡q cwZZ nq, Zvn‡j ZvÑ [NSTU-A, 19-20] A. †÷vK‡mi m~Î B. Pvj©‡mi m~Î C. mv›`ªZvi m~Î D. Pv‡ci m~Î F = 6rv STEP 06 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. mv›`ªZvs‡Ki Dci ZvcgvÎv I Pv‡ci cÖfv‡ei †¶‡Î †Kvb&wU mwVK bq? [MAT. 13-14] A. Zij c`v‡_©i mv›`ªZv ZvcgvÎv e„w×i mv‡_ `ªæZ n«vm cvq B. M¨v‡mi mv›`ªZv M¨vm AYymg~‡ni Mo ‡e‡Mi mgvbycvwZK C. M¨v‡mi mv›`ªZvsK Pv‡ci Dci wbf©ikxj D. Pvc e„w× †c‡j Zij c`v‡_©i mv›`ªZvsK e„w× cvq M¨v‡mi mv›`ªZvi Dci Pv‡ci ‡Kvb cÖfve †bB| GwU Pv‡ci we¯Í…Z cvjøvi ‡¶‡Î cÖ‡hvR¨| Z‡e wbgœ Pv‡ci †¶‡Î e¨vwZµg †`Lv hvq|
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 273 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. Zi‡j cZbkxj e¯‘i Rb¨ †Kvb †jLwPÎ mwVK? (†eM v, MfxiZv = h) [Xv. †ev. 2017] A. B. C. D. Ans A 02. M¨v‡mi mv›`ªZv ¸Yv¼ ZvcgvÎvi- [P. †ev. 2016] A. mgvbycvwZK B. e¨¯ÍvbycvwZK C. eM©g~‡ji mgvbycvwZK D. eM©g~‡ji e¨¯ÍvbycvwZK S C info Zij: Log = A + B T T 03. Vc= msKU †eM, = Zi‡ji mv›`ªZv¼, = Zi‡ji NbZ¡, r b‡ji e¨vmva© n‡j †Kvb †jLwPÎwU mwVK? [Kz.‡ev. 2019] A. B. C. D. Ans D 04. †Kvb c`v‡_©i mv›`ªZv me‡P‡q †ewk? [wm. †ev. 2015] A. †Zj B. `ya C. gay D. cvwb Ans C Concept 6 ¯úk© †KvY Ges •KwkKZv ✅ ¯úk© †KvY: mskw³ ej I AvmÄb e‡ji mw¤§wjZ wµqvq m„ó †KvY‡K ¯úk© †KvY e‡j (†KvY m„wó nq)| ✅ ¯úk© †Kv‡Yi cÖKvi‡f`: m~² ¯úk© †KvY (0 0 < < 900 ) Zi‡ji gy³ Zj AeZj nq ¯’~j ¯úk© †KvY (90 < < 1800 ) Zi‡ji gy³ Zj DËj nq| ✅ ¯úk© †Kv‡Yi gvb: ¯úk© †Kv‡Yi me©wbgœ gvb 0 m‡e©v”P gvb 180 mvaviY cvwb I Kuv‡Pi weï× cvwb I cwi®‥vi Kuv‡Pi ¯úk© †Kvb cÖvq 0 0 †fZiKvi ¯úk© †KvY 8 AvmÄb ej hZ †ekx n‡e ¯úk© †KvY ZZ m~² †KvY n‡e cvi` I Kuv‡Pi †fZiKvi ¯úk© †KvY 140 mskw³ ej hZ †ekx n‡e ¯úk© †KvY ZZ ‡ekx n‡e| iƒcv I cvwbi †fZiKvi ¯úk© †KvY 90 ✅ ¯úk©‡KvY wb‡gœv³ welq¸‡jvi Dci wbf©i K‡i: KwVb I Zi‡ji cªK…wZ Zi‡ji Dcwiw¯ Z gva¨g KwVb I Zi‡ji weï×Zv ✅ ¯úk© †KvY m~² †KvY n‡j ( < 90) Zij I KwV‡bi ga¨eZ©x AvmÄb ej > Zi‡ji mskw³ ej| ‣KwkK b‡j Zi‡ji Av‡ivnY nq ✅ ¯úk© †KvY ¯’~j nq ( > 90) ‣KwkK b‡j Zi‡ji Aebgb N‡U ✅ ‡K․wkKZv ev †K․wkKZ¡: †K․wkK b‡j Zi‡ji DVv bvgv‡K ‡K․wkKZv e‡j| Gi KviY Zi‡ji c„ôUvb| ✅ wbf©iZv: †K․wkKZv wbf©i K‡i †K․wkK b‡ji e¨vmva© hZ Kg nq Zij c`v‡_©i DVv bvgvi cwigvb ZZ AwaK; Zij c`v‡_©i cÖK…wZi Dci| ✅ Zij KwVb‡K wfRv‡j (cvwb I KuvP) ¯úk©‡KvY m~²‡KvY nq A_©vr 0 < < 90 Ges Zij KwVb‡K bv wfRv‡j (cvi` I KuvP) ¯úk© †KvY ¯ ~j †KvY nq A_©vr 90 < < 180| ✅ ¯úk© †KvY †h †h wel‡qi Dci wbf©i K‡i: 1. KwVb Zi‡ji cÖK…wZ 4. KwVb Zi‡ji Av‡cw¶K MwZ 2. Zi‡ji Dcwiw¯ Z gva¨g 5. KwVb Zi‡ji ga¨Kvi Nl©Y 3. KwVb Zi‡ji weï×Zv ◈ ¯úk©‡KvY †h †h wel‡qi Dci wbf©i K‡i Zv g‡b ivLvi †K․kj: MNP M N P Medium (Zi‡ji Dcwiw¯ Z gva¨‡g) Nature (KwVb I Zi‡ji cÖK…wZ) Purity (KwVb I Zi‡ji weï×Zv)
274 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. `yBwU †K․wkK b‡ji g‡a¨ GKwU AciwU A‡c¶v †ekx miæ| `yBwU‡KB Lvovfv‡e cvwbi g‡a¨ AvswkK †Wvev‡bv n‡j †ekx miæwUi wfZi cvwbi ¯Í‡¤¢i D”PZv †ekx nq| Gi KviY [DU. 08-09] A. miæ b‡j evqy Pvc K‡g hvq B. miæ b‡j cvwbi NbZ¡ K‡g hvq C. GLv‡b cvwbi c„ôUvb b‡ji e¨v‡mi e‡M©i Dci wbf©ikxj wKš‧ cvwbi ¯Í‡¤¢i IRb e¨v‡mi Dci wbf©ikxj D. GLv‡b cvwbi c„ôUvb b‡ji e¨v‡mi Dci wbf©ikxj wKš‧ cvwbi ¯Í‡¤¢i IRb e¨v‡mi e‡M©i Dci wbf©ikxj bq| Ans D STEP 02 ANALYSIS OF JU QUESTION 01. K‡ji cvwb I ‡cwi‡¯‥v‡ci ga¨eZ©x ¯úk© †KvY KZ? [JU. 12-13] A. 00 B. 900 C. 1400 D. 80 Ans D STEP 03 ANALYSIS OF RU QUESTION 01. ¯úk©‡KvY 120 n‡j •KwkK b‡j Zij- [RU. Sinovac, Set-1. 20-21; e.†ev. 2015] A. Dc‡i DV‡e B. wb‡P bvg‡e C. DfqB n‡Z cv‡i D. AcwiewZ©Z _vK‡e S B info •KwkK b‡ji †`qv‡ji mv‡_ ¯úk© †KvY ¯ ‚j‡KvY n‡j Zij wb‡P bvg‡Z ïiæ K‡i| KviY T = rhg 2cos h 2Tcos rg | mgxKi‡Y ¯ ~j‡KvY n‡j cos Gi gvb FYvZ¥K nq, d‡j h Gi gvbI FYvZ¥K nq| A_©vr, Zij wb‡P bvg‡Z ïiæ K‡i| 02. cvi` I Kuv‡Pi ga¨Kvi ¯úk© †KvY KZ n‡e? [RU. 16-17; CoU. 09-10] A. 80 B. 90 C. 140 D. 160 weï× cvwb I cwi®‥vi Kuv‡Pi wfZiKvi ¯úk©‡KvY: cÖvq k~b¨| cvwb I Kuv‡Pi wfZiKvi ¯úk© †KvY = 8˚ iƒcv I cvwbi wfZiKvi ¯úk© †KvY = 90˚,cvi` I Kuv‡Pi ga¨Kvi ¯úk© †KvY = 140˚ 03. hw` ¯úk© †KvY 90Gi Kg nq, Z‡e •KwkK b‡j Zi‡ji Ae¯’v †Kgb n‡e? [RU. 16-17] A. Dc‡i DV‡e B. bx‡P bvg‡e C. Dc‡i DV‡e ev bx‡P bvg‡e D. AcwiewZ©Z _vK‡e ¯úk©‡KvY my¶¥‡KvY A_©vr 0˚ < < 90˚ n‡j ev, Zij KwVb‡K wfRv‡j Zij c`v_© Dc‡ii w`‡K DV‡e| 04. cvi` c~Y© cv‡Î Kuv‡Pi •KwkK bj Wyev‡j b‡ji Af¨šÍ‡i cvi`- [RU. 10-11] A. Dc‡i DV‡e B. w¯ i _vK‡e C. wb‡P bvg‡e D. DVvbvgv Ki‡e Ans C STEP 04 ANALYSIS OF CU QUESTION 01. hw` †Kvb Zi‡ji Zj mgZj nq, Avav‡ii cÖvPx‡ii mv‡_ Zi‡ji ¯úk©K ‡KvY- [CU. 15-16] A. 0° B. 90° C. m~² ‡KvY D. ¯ ~j ‡KvY E. ‡Kv‡bvwUB bq Ans B STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. †h me Zij KuvP‡K †fRvq bv Zv‡`i ¯úk© †KvY KZ wWMÖx? [KU-A, Set-Ka. 19-20; w`.†ev. 2015] A. 0 B. < 90 C. 90 D. > 90 Zij KuvP‡K wfRv‡j ¯úk©‡KvY m~²‡KvY ev, 0°< < 90° Zij KuvP‡K bv wfRv‡j ¯úk©‡KvY ¯ zj‡KvY ev, 90°< <180° 02. Mv‡Qi †Mvovq evwj R‡g _vK‡j MvQ g‡i hvq KviY- [KU. 12-13] A. evwj AwaK cvwb a‡i iv‡L B. evwj •KwkK b‡j KvR K‡i bv, d‡j cvwb a‡i ivL‡Z cv‡i bv, ï®‥ _v‡K C. evwj AwaK DËß nq D. evwj evZv‡mi ej mn¨ Ki‡Z cv‡i bv Ans A IU 01. ¯úk©‡KvY KZ n‡j †Kvb †K․wkK bj w`‡q Zij c`v_© Dc‡ii w`‡K DV‡e? [IU. 16-17] A. 120 B. 180 C. >90 D. < 90 Zij KwVb‡K wfRv‡j (cvwb I KuvP) Zij c`v_© Dc‡ii w`‡K DV‡e| G‡ÿ‡Î ¯úk©‡KvY m~¶¥‡KvY nq A_©vr 0 < < 90| 02. •Zjv³ c`v_©hy³ KvuP I weï× cvwbi ga¨Kvi ¯úk© †Kvb- [IU. 12-13] A. cÖvq ïb¨ B. FYvZ¡K C. Aek¨B ï‡b¨i †P‡q Kg D. ‡KvY •Zix nq bv Ans A 03. cvi` I Kuv‡Pi ga¨Kvi ¯úk© †Kvb nq- [IU. 10-11] A. ¯ ~j †Kvb B. m~¶è‡Kvb C. k~b¨ D. mg‡Kvb Ans A BRUR 01. cwi®‥vi KuvP I weï× cvwbi ga¨Kvi ¯úk© †KvY (cÖvq) KZ? [BRU. 15-16] A. 0 B. 8 C. 90 D. 139 cvwb I Kvu‡Pi wfZiKvi ¯úk© †Kvb 8° weï× cvwb I cwi®‥vi KuvP Gi ¯úk© †Kvb 0° PART B Analysis of Science & Technology Question SUST 01. GKwU †K․wkK b‡ji GK cÖvšÍ Lvovfv‡e cvwb‡Z wbgw¾Z Ki‡j cvwb b‡ji wfZi Av‡ivnb K‡i| b‡ji e¨vmva© r Ges Av‡ivwnZ cvwbi D”PZv h (hLb r << h) G `ywUi g‡a¨ m¤úK© n‡”Q- [SUST-B, 19-20] A. h r –1 B. h r 2 C. h r D. h r 3 E. h r –2 Avgiv Rvwb, T = hrg 2cos ; h 1 r ; h r 1 MBSTU 01. hw` ¯úk© †KvY 90 Gi †ewk nq, Z‡e Zi‡ji c„ô n‡eÑ [MBSTU-A, Set-2 19-20, ] A. AeZj B. DËj C. mgZjveZj D. mgZj DËj ¯úk©‡KvY 0˚ < < 90˚ n‡j Zi‡ji gy³ Zj AeZj Ges ¯úk©‡KvY 90˚ < < 180˚ n‡j Zi‡ji gy³ Zj DËj| 02. KuvP I weï× cvi‡`i †ejvq ¯úk© †Kv‡Yi gvbÑ [MBSTU-A, Set-2 19-20] A. 0 B. cÖvq 139 C. cÖvq 90 D. cÖvq 8 ● cvi` I Kuv‡Pi g‡a¨ ¯úk© †KvY 140˚ ● mvaviY cvwb I Kuv‡Pi g‡a¨ ¯úk© †KvY 8˚ ● weï× cvwb I Kuv‡Pi g‡a¨ ¯úk© †KvY 0˚ ● iæcv I cvwbi g‡a¨ ¯úk© †KvY 90˚ [Ref. Avwgi m¨vi] STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. hLb †Kvb •KwkK b‡j Zi‡ji gy³ Zj Avbyf~wgK _v‡K, ZLb ¯úk© ‡KvY n‡j- [KUET. 13-14] A. DËj B. AeZj C. mgAeZj D. †KvbwUB bq Ans A
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 275 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. hw` ¯úk© †KvY 90 Gi Kg nq, Z‡e Zi‡ji c„ô †Kgb n‡e? [MAT. 16-17] A. AeZj B. DËj C. mgveZj D. †KvbwU bq Ans A 02. ¯úk© †Kvb hw` m~² †Kvb nq Z‡e wb‡gœi †Kvb •ewkó¨ mwVK bq? [MAT. 14-15] A. †Kv‡nwmf ej, GW‡nwmf e‡ji †P‡q eo nq B. ZijwU b‡ji Mv‡q ¯úk© K‡i bv C. †K․wkK b‡j Zi‡ji Ae‡¶c nq D. †K․wkK b‡j Zi‡ji c„ô‡`k AeZj nq Ans B DAT 01. eøvwUs †ccvi †Kvb a‡g©i Rb¨ cvwb ï‡l †bq? [DAT. 18-19] A. mv›`ªZvi wµqvq B. c„ôUv‡bi wµqvq C. •KwkK wµqvq D. c„ôkw³i wµqvq Ans C 02. hw` ¯úk© †KvY 90 Gi Kg nq Z‡e Zi‡ji c„ô wK n‡e? [DAT. 17-18, RU-C, Set-1. 19-20, MBSTU. 19-20, BRUR. 12-13] A. DËj B. AeZj C. mgZjveZj D. mg‡ZvËj ● cvi` I Kuv‡Pi g‡a¨ ¯úk© †KvY 140˚ ● mvaviY cvwb I Kuv‡Pi g‡a¨ ¯úk© †KvY 8˚ ● weï× cvwb I Kuv‡Pi g‡a¨ ¯úk© †KvY 0˚ ● iæcv I cvwbi g‡a¨ ¯úk© †KvY 90˚ STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. †h mg¯Í Zij Øviv KuvP wf‡R bv Zv‡`i ¯úk© †KvY n‡eÑ [w`. †ev. 2017] A. m~²‡KvY B. ¯ ~j‡KvY C. k~b¨ D. mg‡KvY Ans B 02. wb‡Pi †Kvb Nb‡Z¡i Zi‡ji g‡a¨ KvPbj Wzev‡bv n‡j ¯’~j ¯úk© †KvY n‡e? [h. †ev. 2016] A. 0.8 103 kgm–3 B. 0.87 103 kgm–3 C. 1 103 kgm–3 D. 13.6 103 kgm–3 Ans D 03. iƒcv I weï× cvwbi ga¨Kvi ¯úk© †KvY (cÖvq) KZ? [h. †ev. 2015] A. 0˚ B. 8˚ C. 90˚ D. 140˚ Ans A Z‡_¨i Av‡jv‡K 04bs cÖ‡kœi DËi `vI: [wm.‡ev. 2019] GKwU •KwkK b‡ji e¨vm 0.4mm| G‡K 72×10-3N/m c„ôUvb Ges 103 kg m-3 Nb‡Z¡i cvwb‡Z Wzev‡bv n‡jv| 04. b‡ji KZ D”PZvq cvwb DV‡e? A. 7.3469 m B. 0.73469 m C. 0.073469 m D. 0.00734 m S C info h = 2T rg = 2 × 72 × 10–3 2 × 10–4 × 103 × 9.8 = 0.073469 m 05. ¯úk© †KvY wbf©i K‡i- [wm. †ev. 2016] A. KwVb I Zi‡ji cÖK…wZi Ici B. Zi‡ji D”PZvi Ici C. KwVb I Zi‡ji weï×Zvi Ici D. A I C DfqB Ans D 06. GKwU •KwkK bj‡K wMømvwi‡b Wzev‡j [wm. †ev. 2015] i. KvP I wMømvwi‡bi ¯úk© †KvY m~² †KvY nq ii. Zij c„ô AeZj AvKvi aviY K‡i iii. KvP I wMømvwi‡bi ¯úk© †KvY ¯ ~j †KvY nq wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans A STEP 2 mg„× †ewmK MvwYwZK cÖ Ö‡qvM (MATH) kU©KvU †UKwbK Concept 1 cxob I weK…wZ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. cxob = ej †ÿÎdj = F A = mg A [m f‡ii e¯‧‡K Szwj‡q w`‡j] 02. weK…wZ = •`N©¨ e„w× Avw` e„w× = l L [‣`N©¨ weK…wZ] = †ÿÎdj e„w× Avw` †ÿÎdj = a A [†ÿÎdj ev K…šÍb weK…wZ] = AvqZb e„w× Avw` AvqZb = v V [AvqZb weK…wZ] 03. Amn cxob = Amn ej (Amn IRb) †ÿÎdj 04. Amn fvi = Amn IRb = Amn fi AwfKl©R Z¡iY 05. NbZ¡ = fi AvqZb CONCEPTUAL MATH MEx 01 1m j¤^v Ges 1mm e¨vmva©wewkó GKLÛ Zvi 20kg ej Øviv Uvbv †`qv n‡j ZviwUi •`N©¨ 0.6mm cwigvY e„w× cvq| ZviwUi (i) cxob (ii) weK…wZ wbY©q Ki| i. cxob F A = mg r 2 = 20 9.8 (1 10–3 ) 2 = 62.42 106 Nm2 ii. weK…wZ, l L = 0.6 10–3 1 = 0.6 10–3 MEx 02 GKwU Zv‡ii •`N©¨ 4m cÖ¯’‡”Q‡`i †ÿÎdj 0.003m2 , Amn cxob 3.267 105Nm2 | ZviwUi (i)Amn fi (ii) Amn fvi KZ? Amn IRb = Amncxob cÖ¯ ‡”Q‡`i †ÿÎdj = 3.267 105 0.003 = 9.8 102N i. Amn fi = Amn IRb AwfKl©R Z¡iY = 9.8 102 9.8 = 102 kg ii. Amn fvi = Amn IRb = 9.8 102N| MEx 03 10m j¤^v GKwU `Û‡K nvZzwo Øviv AvNvZ Kivq `ÛwU 10.1m n‡jv, `ÛwUi weK…wZ KZ? weK…wZ = cwieZ©b Avw` = 0.1 10 = 0.01 NOW START PRACTICE 01. GKwU Zv‡ii Dcv`v‡bi Bqs ¸YvsK 2 1011Nm–2 | ZviwUi •`N©¨ 25% evov‡Z cÖhy³ cxob wbY©q Ki| 02. GKwU Zv‡ii •`N©¨ 3m, e¨vm 0.002m Amn cxob 6 107Nm2 | ZviwUi Amn IRb KZ?
276 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 PRACTICE SOLVE : 01. cÖhy³ cxob F A = Y l L = 2 1011 25 100 = 5 1010Nm2 02. Amn IRb = Amncxob cÖ¯ ‡”Q‡`i †ÿÎdj = Amncxob R 2 = 6 107 (0.001)2 = 188.5N REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF JU QUESTION 01. GKwU Zv‡ii •`N©¨ 3m. e¨vm 0.002m. Amn cxoY 6107 Nm–2 | ZviwUi Amn IRb KZ? [JU. 17-18] A. 190.4 N B. 170.4N C. 188.4 N D. 200.4 N e¨vmva©, r =(0.002 /2)= 0.001 m Amn IRb = Amncxob cÖ¯ ‡”Q‡`i †¶Îdj = Amncxob r 2 =6107 (0.001)2 = 188.4 N 02. GKwU Zv‡ii Dcv`v‡bi Bqs Gi ¸Yvs‡K 21011N/m2 | ZviwUi •`N©¨ 15% e„w× Ki‡Z cÖhy³ cxob KZ n‡e? [JU. 16-17, 15-16] A. 31010Nm-2 B. 31010Nm-1 C. 31011Nm2 D. 3109Nm-2 cxob = Y L = 21011 100 15 = 31010Nm-2 03. GKwU MÖvbvBU cv_‡ii cÖ¯’‡”Q‡`i †¶Îdj 0.03m2 , Amncxob 3.267 105 Nm2 | Amnfvi KZ? [JU. 13-14, RU. 11-12] A. 9801 N B. 8801 N C. 7801 N D. 6801 N Amn fvi ev IRb = Amncxob cÖ¯ ‡”Q‡`i †¶Îdj = 3.267 105 0.03= 9801 N STEP 02 ANALYSIS OF RU QUESTION 01. GKB c`v‡_©i •Zwi `ywU Zv‡ii e¨vmv‡a©i AbycvZ 3:1| hw` Zvi `y‡Uv‡K mgvb ej Øviv Uvbv nq Z‡e Zv‡`i cxo‡bi AbycvZ KZ n‡e? [RU. 16-17] A. 9:1 B. 1:9 C. 3:1 D. 1:3 cxob = 2 r F A F F Constant n‡j, cxob 2 2 3 1 r 1 02. ‡KvbwU mwVK? [RU. 15-16; JU. 13-14] A. cxob = FA B. cxob = F A C. cxob = A F D. cxob = F+A Ans B 03. 1 sq mm cÖ¯’‡”Q‡`i †¶Îdj wewkó Zv‡ii GK cÖv‡šÍ 1 N ej cÖ‡qvM Kiv n‡j cxob n‡e- [RU. 13-14; CU. 13-14] A. 106 Nm–2 B. 104 Nm–2 C. 100 Nm–2 D. 50 Nm–2 cxob = A F = 6 1 10 1 =106 Nm–2 STEP 03 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. 0.8 Av‡cw¶K ¸iæ‡Z¡i GKwU e¯‘i Zi‡j GKwU e¯‘i nvivb IRb 20g n‡j e¯‘wUi evwn¨K AvqZb- [KU. 11-12] A. 20cc B. 25cc C. 16cc D. 36cc e¯‧i Avt¸iæZ¡ = NbZ¡ = 0.4 e¯‧i nvivb IRb = e¯‧i KZ…©K AcmvwiZ mg AvqZb Zi‡ji IRb = 20g AvqZb = fi NbZ¡ = 20 0.8 = 25cc CoU 01. GKwU e¯‘i evZv‡m IRb 100 gm I cvwb‡Z IRb 80 gm| e¯‘i NbZ¡- [CoU. 13-14] A.5 gm/cc B. 0.7gm/cc C. 4.5 gm/cc D. 7.5 gm/cc wa = 100gm, ww = 80gm nvivb IRb=AvqZb=(100–80)=20gm ev, 20cc NbZ¡ = 100 20 = 5gm/cc IU 01. GKwU Zv‡ii •`N¨ 4m; cÖ¯’‡”Q‡`i †¶Îdj 0.003m2 , Amncxob 3.267×105 Nm–2 n‡j ZviwUi Amn fi- [IU. 13-14] A. 9.82 kg B. 983 kg C. 102 kg D. 103 kg Amn fi = Amn cxob‡¶Îdj g = 3.267 105 0.003 9.8 = 102 kg PART B Analysis of Science & Technology Question SUST 01. mgy‡`ªi cvwb‡Z (NbZ¡ 1.025gm/cL| GKwU U¨v¼v‡ii 10% cvwbi Ic‡i _v‡K| U¨v¼viwU‡K †WW mx ‡Z (NbZ¡ 1.3gm/cL) fvmv‡bv n‡j KZ kZvsk cvwbi Ic‡i _vK‡e| [SUST. 08-09] A. 0.7% B. 0.5% C. 12.7% D. 12.5% h2 = 1.3 1.025 0.1 = 0.127 = 12.7% 02. GKwU e¯‘i Av‡cw¶K ¸iæZ¡ 6 Ges evZv‡m IRb 36 gm cvwb‡Z Gi IRb KZ n‡e| [SUST. 07-08] A. 5 gm B. 30 gm C. 6 gm D. 36 gm e¯‧i Dcv`v‡bi NbZ¡ = 6 gm/cc = e¯‧i AvqZb v = fi NbZ¡ = 36 6 6 cc cvwb‡Z e¯‧i AvcvZZ nvivb IRb = 6 gm cvwb‡Z e¯‧i IRb = (36-6) gm = 30 gm MBSTU 01. 1m `xN© GKwU Zv‡ii •`N©¨ 0.01m e„w× †c‡j ZviwUi Aby‣`N©¨ weK…wZ n‡e- [MBSTU. 15-16] A. 1 B. 1m C. 0.01 D. 0.01m weK…wZ = l L = 0.01 1 = 0.01 STEP 04 ANALYSIS OF HSC BOARD QUESTION 01. ‡Kv‡bv c`v‡_©i Amncxob 4.9108 Nm2 | H c`v‡_©i •Zwi GKwU Zv‡ii cÖ¯’‡”Q‡`i †ÿÎdj 1mm2 n‡j ZviwU‡Z me©wb¤œ KZ fi Szjv‡j ZviwU wQ‡o hv‡e? [P. †ev. 2016] A. 0.5 kg B. 5 kg C. 10 kg D. 50 kg S D info P = F A F = PA F = 4.9 × 108 × 10–6 mg = 490 m = 50 kg 02. ‡Kv‡bv Zvi‡K †K‡U mgvb `yB UzKiv Kiv n‡jv| GZ Zv‡ii Amn fvi n‡eÑ [h.‡ev. 2019] A. c~‡e©i A‡a©K B. c~‡e©i mgvb C. c~‡e©i mgvb D. c~‡e©i GK PZz_©vsk Ans B
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 277 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES wb‡Pi DÏxcKwU c‡ov Ges 03 I 04 bs cÖ‡kœi DËi `vI : [h. †ev. 2015] 2m •`N©¨ Ges 1 mm2 cÖ¯ ‡”Q‡`i †ÿÎdjwewkó Zv‡i 20 kg fi Ryjv‡j ZviwU 1mm cÖmvwiZ nq| 03. ZviwUi cxob KZ? A. 1.96 108 Nm2 B. 2.0 107 Nm2 C. 1.96 105 Nm2 D. 1.96 102 Nm2 S A info cxob = F A = 20 × 9.8 1 × 10–3 = 1.96 × 108 Nm–2 04. D³ ZviwUi i. •`N©¨ weK…wZ 0.5 102 ii. Bqs Gi ¸YvsK 3.92 1011 Nm2 iii. K…ZKv‡Ri cwigvY 0.098 J wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C.ii I iii D. i, ii Iiii S C info •`N©¨ weK…wZ = ℓ L = 1 × 10–3 2 = 0.5 × 10–3 05. †Kv‡bv Zv‡ii Amn cxob wbf©i K‡i Zv‡ii- [Kz. †ev. 2019] A. e¨vmv‡a©i Ici B. ‣`‡N©¨i Ici C. cÖ¯ ‡”Q‡`i AvK…wZi Ici D. Dcv`v‡bi Ici Ans C wb‡Pi DÏxcKwU c‡ov Ges 06bs cÖ‡kœi DËi `vI : [e. †ev. 2015] mgvb •`‡N©¨i wZbwU Zvi A, B Ges C G GKB gv‡bi cxob 51012 Nm2 cÖ‡qv‡Mi d‡j •`N©¨ e„w× h_vµ‡g 5%, 2% Ges 1% nj| 06. B Zv‡ii weK…wZ [e. †ev. 2015] A. 2 B. 0.2 C. 0.02 D. 0.002 S C info weK…wZ = 0.02 1 = 0.02 07. GKwU Zv‡ii cÖ¯’‡”Q‡`i †ÿÎdj 1 mm2 Ges Amn fi 40 kg| Zv‡ii Amn cxob- [mKj †evW© 2018] A. 4 × 10–6 Nm–2 B. 3.92 × 10–4 Nm–2 C. 4 × 107 Nm–2 D. 3.92 × 108 Nm–2 S D info Amn cxob = F A = 40 × 9.8 1 × 10–6 = 3.92 × 108 Nm–2 Concept 2 Bqs Gi w¯’wZ¯’vcK ¸YvsK, `„pZvi ¸YvsK I AvqZ‡bi ¸YvsK msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. Bqs Gi w¯ wZ¯ vcK ¸YvsK, Y = •`N©¨ cxob ‣`N©¨ weK…wZ = FL Al = mgL r 2 l [m f‡ii e¯‧‡K Szwj‡q w`‡j] 02. `„pZvi ¸YvsK, n = Fh A = F A 03. AvqZb ¸YvsK, K = FV Av = PV v [P = Pvc] 04. msbg¨Zv, B = 1 AvqZ‡bi ¸YvsK = 1 K 05. K…ZKvR ev wefe kw³ = 1 2 Fl = 1 2 YAl L l = YAl 2 2L 06. GKK AvqZ‡b mwÂZ kw³ = 1 2 × cxob weK…wZ = (cxob) 2 2Y = 1 2 Y (weK…wZ)2 CONCEPTUAL MATH MEx 01 1mm2 cÖ¯’‡”Q‡`wewkó GKwU B¯úvZ Zv‡ii •`N©¨ 5% e„w× Ki‡j KZ ej cÖ‡qvM Ki‡Z n‡e? (Y = 2 1011Nm–2 ) F = Y 1 L A = 2 1011 5 100 1 10–6 = 1 104N MEx 02 cvi‡`i AvqZ‡bi w¯’wZ¯’vcK ¸Yv¼ 2.6 1010 Pa n‡j msbg¨Zv wbY©q Ki| msbg¨Zv, B = 1 K = 1 2.6 1010 = 3.846 10–11Pa NOW START PRACTICE 01. GKB Dcv`v‡bi •Zwi wØZxq Zv‡ii •`N©¨ cÖ_g Zv‡ii •`‡N©¨i wظY wKš‘ e¨vmva© cÖ_g Zv‡ii A‡a©K n‡j Ges mgvb fvi cÖ‡qvM Ki‡j wØZxq Zvi I cÖ_g Zv‡ii •`N©¨ cÖmvi‡Yi AbycvZ KZ? 02. 1 eM©wgwjwgUvi cÖ¯’‡”Q` wewkó GKwU B¯úv‡Zi Zv‡ii cÖv‡šÍ 2 104N ej cÖ‡qvM Ki‡j Gi •`N©¨ kZKiv KZ e„w× cv‡e? [Y = 2 1011 Nm2 ] 03. evqyi AvqZb ¸Yv¼ 1.015 104Nm–2 | †Kvb wbw`©ó AvqZ‡bi evqy‡K k~b¨ AvqZ‡b wb‡q Avm‡Z Gi Dci wK cwigvY Pvc cÖ‡qvM Ki‡Z n‡e? NOW PRACTICE SOLVE : 01. l2 l1 = L2 L1 r1 2 r2 2 = 2 4 = 8 (Ans) 02. Y = FL Al l L = F YA l L = 2 104 (2 1011) 10–6 l L = 10% 03. k = PV v P = Kv V = (1.015 104 ) (v 0) V = 1.015 104Nm2
278 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 Zv‡ii Bqs Gi ¸YvsK 41011 N/m2 | ZviwUi •`N©¨ 7.5% evov‡Z Kx cwigvY cxob cÖ‡qvRb n‡e? [DU. 17-18] A. 7.51011 N/m2 B. 31010 N/m2 C. 5.331010 N/m2 D. 41010 N/m2 cÖhy³ cxob = A F =Y L =41011 100 7.5 = 31010 Nm–2 02. 1m `xN© I 10-2 cm2 cÖ¯’‡”Q` wewkó GKwU Zvi‡K 2kg IRb Øviv Uvbv nj| ZviwUi •`N©¨ e„w× wbY©q Ki| (Bqs Gi ¸YvsK Y=21011Nm-2 ) [DU. 11-12] A. 9.810-5m B. 9.810-2m C. 210-5m D. 210-2m Al mgL Y ev, AY mgL l = 2 4 11 10 10 2 10 2 9.8 1 = 9.810-5m 03. 5m `xN© Ges 1mm2 cª¯’‡”Q` wewkó GKwU Zv‡i 20 kg fi Szwj‡q †`qv nj| G‡Z Zv‡ii •`N¨© 2mm e„w× †c‡j ZviwUi Bqs Gi ¸YvsKi gvb KZ? [DU. 09-10] A. 51011 dyne/cm2 B. 201012dyne/cm2 C. 5.6 1012 dyne/cm2 D. 4.91012dyne/cm2 Y = mgL Al = 20103 9805102 110-2 210-1 = 4.91012 dyne/cm2 04. 1.0 m `xN© I 1.0 mm e¨vmv‡a©i †ejbvKvi Zv‡ii Dci 100 N ej cÖ‡qvM Ki‡j ‣`N©¨ e„w× ‡c‡q 1.001 m nq| Zv‡ii Bqs Gi ¸YvsK KZ? DU. 06-07] A. 1 1011 Nm2 B. 1011 Nm-2 C. x 1011 Nm-2 D. 2 x 1011 Nm-2 r l FL Y 2 ev, 0.001 2 ) 3 (10 100 1 Y 11 2 10 1 Nm 05. 5m ‣`N©¨ Ges 0.5 mm e¨vmwewkó GKwU Zvi‡K 98 N ej Øviv Uvbv n‡jv| ZviwU KZUzKz e„w× cv‡e? (Y =2x1011Nm-2 )| [DU. 03-04; JU. 11-12] A. 2x10-2 m B. 1.25 x 10-2 m C. 4 x 10-2 m D. 1.5 x 10-2 m Y = Fl r 2 l ev, l = FL Yr 2 = 98 5 2 1011 3.14 (0.25 10–3 ) 2 = 1.25 102m 06. GKB Dcv`v‡bi •Zwi wØZxq Zv‡ii •`N¨© cÖ_g Zv‡ii •`‡N¨©i wظY wKš‘ e¨vmva© cÖ_g Zv‡ii A‡a©K n‡j Ges mgvb fvi cÖ‡qvM Ki‡j wØZxq Zvi I cÖ_g Zv‡ii •`N¨© cÖmvi‡Yi AbycvZ KZ? [DU. 01-02] A. Same B. 2 C. 1/2 D. 8 Bqs-Gi ¸YvsK, Y = A FL •`‡N©¨i cwieZ©b, ℓ = 2 Y r FL ; ℓ 2 r L 1 2 = 1 2 L L 2 2 1 r r = 24 = 8 07. 4m •`N¨© Ges 0.05 cm e¨v‡mi GKwU Zvi‡K 5 kg IR‡bi ej Øviv Uvbv n‡j •`N¨© cÖmviY n‡Z n‡e- [DU. 00-01] (†`Iqv Av‡Q Bqs‡qi ¸Yv¼ = 2.041010 kgNm–2 ) A. 410-2 m B. 410-4 m C. 510-3 m D. 510-2 m Bqs-Gi ¸YvsK, Y = A FL •`‡N©¨i cwieZ©b, ℓ = 2 Y r mgL = 10 4 2 2.04 10 (2.5 10 ) 5 9.8 4 = 510–2 m STEP 02 ANALYSIS OF JU QUESTION 01. GKwU Zv‡ii Dcv`v‡bi Bqs Gi ¸Yv¼ 21011 Nm–2 Ges cÖ¯’‡”Q‡`i †¶Îdj 110–4 m 2 | Zv‡ii •`N©¨ 10% e„w× Ki‡Z cÖhy³ ej n‡e- [JU. 15-16] A. 410–6 N B. 2106 N C. 21010 N D. †KvbwUB bq Bqs-Gi ¸YvsK, Y = A FL cÖhy³ ej, F = Y L A = 21011 100 10 110–4 = 2106 N 02. 1m •`N©¨ Ges 510–4 m e¨vmwewkó B¯úv‡Zi Zv‡i 19.6N ej cÖ‡qvM Ki‡j GwU e„w× †c‡q 1.02m nq| Zv‡ii Bqs ¸Yv¼ KZ n‡e? [JU. 15-16] A. 4.99109 Nm–2 B. 5.99109 Nm–2 C. 4.99107 Nm–2 D. 5.99107 Nm–2 2.5 10 0.02 19.6 1 2 4 2 r l FL Y 9 4.9910 03. 1m •`N©¨ Ges 5104 m e¨vm wewkó GKwU B¯úv‡Zi Zv‡i 19.6 N ej cÖ‡qvM Ki‡j GwU e„w× †c‡q 1.02 m nq| Zv‡ii Bqs Gi ¸YvsK KZ? [JU. 13-14; RU. 13-14, CU. 13-14] A. 4.99109 Nm2 B. 5104Nm2 C. 1.02106 Nm2 D. 4.991011 Nm2 Bqs-Gi ¸YvsK, Y = A FL = 2 r FL = (2.5 10 ) 0.02 19.6 1 4 2 = 4.99109 Nm2 04. M¨v‡mi AvqZb ¸YvsK 6103N/m2 M¨v‡mi AvqZb 10% Kgv‡Z n‡j wK cwigvY AwZwi³ Pvc cÖ‡qvM Ki‡Z n‡e? [JU. 12-13] A. 300N/m2 B. 400N/m2 C. 1000N/m2 D. 600N/m2 K = PV v p = Kv V = 6103 10 100 = 600N/m2 05. 1mm2 cÖ¯’‡”Q` wewkó GKwU B¯úvZ Zv‡ii •`N©¨ 5% e„w× Ki‡Z e‡ji cÖ‡qvRb [Y=21011 Nm2 ] [JU. 11-12, 10 -11; HSTU. 15-16] A. 104 N B. 108 N C. 105 N D. 107 N Bqs-Gi ¸YvsK, Y = A FL cÖhy³ ej, F = Y L A = 21011 100 5 110–6 = 1104 N 06. 10m j¤^v Ges 1mm e¨vm wewkó GKwU Zvi‡K 100N ej Øviv Uvbv n‡jv| ZviwUi •`N©¨ e„w× cv‡e? (Y = 2.2 1011 Nm2 ) [JU. 11-12] A. 6410-3m B. 4.610-3m C. 6.410-3m D. 5.78710-3m r l FL Y 2 ev, 2 11 3 2 2.2 10 (0.5 10 ) 100 10 Y FL l 3 5.787 10 m STEP 03 ANALYSIS OF RU QUESTION 01. Y = 2.0×1011 Nm–2 ¸Yv¼ wewkó c`v‡_©i GKwU Zv‡ii •`N©¨ w¯’wZ¯’vcK mxgvi g‡a¨ 5% evov‡Z m„ó cxobÑ [RU. Moderna, Set-2. 20-21] A. 1.0×1010 Nm–2 B. 1.0×1011 Nm–2 C. 2.0×1010 Nm–2 D. 5.0×1010 Nm–2 S A info Bqs Gi ¸YvsK, Y = FL Al = FL A(0.05L) F A = 0.05Y = 0.05 2 1011 = 1 1010Nm–2 [l = L Gi 5% = 0.05L]
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 279 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. w¯’wZ¯’vcK mxgvi g‡a¨ 1sq mm cÖ‡”Q‡`i †ÿÎdj wewkó Zv‡ii GK cÖv‡šÍ 1N ej cÖ‡qvM Kiv n‡j cxob n‡eÑ [RU. Astrazeneca, Set-1. 20-21] A. 106 Nm2 B. 104 Nm2 C. 100 Nm2 D. 50 Nm2 S A info cxob, P = F A = 1Nm–2 1 10–6 = 106 Nm2 [A = 1mm2 = 1 10–6 m 2 ] 03. B¯úv‡Zi Rb¨ Bqs Gi ¸Yv¼ y = 2.0 1011 N m–2 n‡j, w¯’wZ¯’cK mxgvi g‡a¨ 1 mm2 cÖ¯’‡”Q` wewkó GKwU B¯úv‡Zi Zv‡ii •`N©¨ 10% e„w× Ki‡Z cÖ‡qvRbxq ejÑ [RU. Sinovac, Set-1. 20-21, RU. 16-17; w`. †ev. 2016] A. 2 104 N B. 104 N C. 1.2 10 4 N D. 0 N S A info Avgiv Rvwb, Bqs Gi ¸Yv¼, Y = FL Al F = Y Al L = 2.0 1011 1 10–6 0.1L L = 2 104N [ZviwUi •`N©¨ e„w×, l = L Gi 10%] 04. mgvb •`‡N©¨i wZbwU Zvi A, B Ges C-†Z cxo‡bi gvb mgvb Ges •`N©¨ e„w× lA > lB > lC n‡j wb‡Pi †KvbwU mwVK? [†hLv‡b Y Bqs Gi ¸Yv¼] [RU. 18-19] A. YA > YB > YC B. YC > YB > YA C. YA = YB = YC D. B I C DfqB Ans B 05. GKwU `‡Ûi w¯’wZ¯’vcK ¸YvsK 20×1010 N/m2 Ges NbZ¡ 8000 kg/m3 | Ab¨ GKwU `‡Ûi w¯’wZ¯’vcK ¸YvsK 15×1010 N/m2 Ges NbZ¡ 9000 kg/m3 | †Kvb `ÛwU‡Z k‡ãi †eM †ewk n‡e? [RU. 17-18] A. cÖ_gwU‡Z B. wØZxqwU‡Z C. mgvb D. †KvbwUB bq Ans A 06. wcZ‡ji w¯’wZ¯’vcK ¸Yv¼- [RU. 16-17] A. 91011Nm–2 B. 910–11Nm–2 C. 910–16Nm–2 D. 910–10Nm–2 Ans A 07. 3 m `xN© Ges 1 cm2 cÖ¯’‡”Q` wewkó GKwU Zv‡ii Bqs Gi ¸YvsK 51010dyne/cm2 | GKB ZviwUi •`N©¨ A‡a©K Ki‡j Bqs Gi ¸YvsK KZ dyne/cm2 n‡e? [RU. 16-17] A. 2.51010 B. 51010 C. 1011 D. †KvbwUB bq Bqs Gi ¸YvsK nq Zv‡ii Dcv`v‡bi| GKB Dcv`v‡bi Zv‡ii Bqs Gi ¸YvsK cwiewZ©Z nq bv| 08. 1 eM© wgwjwgUvi cÖ¯’‡”Q` wewkó GKwU B¯úv‡Zi Zv‡i KZ ej cÖ‡qvM Ki‡j Gi •`N©¨ wظY n‡e? (Y=2 1012 WvBb/eM© †m.wg.) [RU. 15-16] A. 2 1010 WvBb B. 3 10 10 WvBb C. 4 1010 WvBb D. 5 1010 WvBb Bqs-Gi ¸YvsK, Y = A FL cÖhy³ ej, F = Y L A = 21012 1 1 10-2 = 2 1010 WvBb 09. 1 cm2 cÖ¯’‡”Q` wewkó GKwU Zv‡i KZ ej cÖ‡qvM Ki‡j Gi •`N©¨ wظY n‡e? [Y=21012 dyne cm-2 ] [RU. 13-14, 08-09] A. 41012 dyne B. 21012 dyne C. 3.51012 dyne D. 151012 dyne Bqs-Gi ¸YvsK, Y = A FL cÖhy³ ej, F = Y L A = 21012 L L 1 = 21012 dyne 10. cvwbi AvqZb ¸YvsK 0.22 1010 Nm2 | 1L cvwbi AvqZb 0.1% cwieZ©b Ki‡Z KZ Pv‡ci cÖ‡qvRb? [RU. 12-13] A. 0.221010Nm-2 B. 0.221013Nm-2 C. 2.2106Nm-2 D. 2.2Nm-2 K = PV V P = KV V = 0.22 1010 0.1 100 = 2.2 106Nm 2 11. GKwU B¯úv‡Zi Zv‡ii •`N¨© 4m, e¨vm 0.5mm Ges Bqs Gi w¯’wZ¯’vcK ¸YvsK 2 × 1011 N/m2 . GLb GKB B¯úv‡Zi Avi GKwU Zv‡ii •`N©¨ 5m. Ges e¨vm 4mm n‡j Bqs Gi w¯’wZ¯’vcK ¸YvsK KZ n‡e? [RU. 11-12] A. 1 1011 N/m2 B. 2 1011N/m2 C. 5 1010N/m2 D. 2 105N/m2 GKB Dcv`vb nq Gi w¯ wZ¯ vcK ¸Yv¼ GKB _vK‡e| A_©vr, Bqs Gi w¯’wZ¯’vcK ¸YvsK 2 × 1011 N/m2 | 12. cvwbi AvqZb ¸YvsK = 0.2×1010Nm-2 ej‡Z eySvq †h cvwbi Avw` AvqZb mgvb AvqZb n«vm Ki‡Z cvwbi cÖwZ eM©wgUvi †¶Îd‡ji Dci j¤^fv‡e Pvwiw`‡K †_‡K †h ej cÖ‡qvM Ki‡Z n‡e Zvi cwigvY n‡e- [RU. 11-12] A. 0.5×1010N B. 0.2×1010N C. 0.3×1010N D. 0.4×1010N Ans B 13. 1 m `xN© Ges 106 m 2 cÖ¯’‡”Q` ‡¶Îdj wewkó GKwU Zvi‡K •`N©¨ eivei 19.6 N e‡j Uvbv nj| Zv‡ii •`N©¨ e„w×i cwigvb KZ? (Y = 2 1011 N/m2 ) [RU. 09-10] A. 8.8105m B. 9.8106m C. 9.8105m D. 8.8104m Bqs-Gi ¸YvsK, Y = A FL •`‡N©¨i cwieZ©b, ℓ = YA FL = 11 6 2 10 10 19.6 1 = 9.810–5 m STEP 04 ANALYSIS OF CU QUESTION 01. hw` P cxob Ges Y †Kvb Zv‡ii Dcv`v‡bi Bqvs- Gi ¸Yv¼ nq Z‡e Zv‡ii cÖwZ GKK AvqZ‡b mwÂZ kw³: [CU. 18-19] A.2P2Y B. P 2 2Y C. 2Y p 2 D. P 2Y W = 1 2 cxob weK…wZ = 1 2 P P Y = P 2 2Y Y = P l L 1 L = P Y 02. GKwU Zv‡ii Dcv`v‡bi Bqs Gi ¸YvsK 2 1011Nm–2 ZviwUi •`N©¨ 15% e„w× Ki‡Z cÖhy³ cxob KZ? [CU. 10-11, RU. 14-15; CoU. 13-14, JnU. 15-16, MBSTU. 15-16] A. 3 1010Nm–2 B. 2 1010Nm–2 C. 4 1010Nm–2 D. 4.5 1010Nm–2 Y = cxob weK…wZ cxob = Y, weK…wZ = 2 1011N 0.15 = 3 1010Nm–2 03. †KvbwU B¯úv‡Zi ¸YvsK? [CU. 07-08] A.12.61010Nm–2 B. 111010 Nm–2 C. 201010 Nm–2 D.71010 Nm–2 Ans C 04. Bqs-Gi ¸YvsK wbY©‡qi cix¶vq †Kvb Dcv`v‡bi †¶‡Î w¯’wZ¯’vcK mxgvi g‡a¨ M- [fvi ebvg Avbylw½K •`N©¨ cÖmviY] †jLwPÎwU n‡e- [CU. 06-07] A. Awae„Ë B. Dce„Ë C. e„Ë D. g~jwe›`yMvgx mij †iLv Ans D STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. KZ Pv‡c 1000 Nb †mw›UwgUvi cvi‡`i 2 Nb †mw›UwgUvi ms‡KuvPb n‡e? (cvi‡`i AvqZb ¸YvsK = 2.5 1010 Nms-2 ) [KU. 06-07] A. 6 107 Nms-2 B. 5 107 Nms-2 C. 4 107 Nms-2 D. 3 107 Nms-2 AvqZb ¸YvsK, K = v PV cÖhy³ Pvc, P = Kv V = 1000 2.5 10 2 10 = 5107 Nms–2
280 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 BU 01. GKwU Zv‡ii Dcv`v‡bi ¸YvsK 21011Nm-2 ZviwUi •`©N¨ 25% e„wׇZ cÖhy³ cxob [BU. 11-12] A. 4108Nm-2 B. 51010Nm-2 C. 510Nm-2 D. 6106Nm-2 l PL Y ev, L Yl P = 100 2 10 25 11 =51010N/m2 PART B Analysis of Science & Technology Question SUST 01. GKwU bvBj‡bi `wo‡Z 70 kg f‡ii GKRb ce©Zv‡ivnx Szj‡j `woi •`N©¨ 1.5 m e„w× cvq| `woi g~j •`N©¨ 60 m Ges e¨vm 9 mm n‡j Dnvi Bqs Gi ¸Yv¼ KZ Pa? [SUST-B, 19-20] A. 6.4 × 107 B. 8.31 × 108 C. 4.31×108 D. 4.4×109 E. 1.35×106 Bqs Gi ¸Yv¼, Y = FL Al = mgL Al = mgL r 2 l = 70 9.8 60 1 4 3.14 (9 103 ) 2 1.5 Pa = 4.31 108 pa 02. 2m `xN© I 1mm e¨vmv‡a©i GKwU B¯úv‡Zi Zv‡ii Bqs Gi ¸YvsK 4m `xN© I 1mm e¨vmv‡a©i B¯úv‡Zi Zv‡ii Bqs Gi ¸YvsK [SUST. 01-02] A. wظY B. Pvi¸Y C. A‡a©K D. mgvb Dcv`vb GKB nIqv‡Z Bqs Gi ¸YvsK GKB n‡e| JUST 01. 2 10–4 m 2 cÖ¯’‡”Q‡`i †ÿÎdj wewkó GKwU B¯úv‡Zi Zv‡i KZ ej cÖ‡qvM Ki‡j Gi •`N©¨ wظY n‡e? (Y = 21011Nm–2 ) [JUST-C, 19-20; RU. 09-10] A. 3 107 N B. 4 107 N C. 5 107 N D. 6 107 N •`N©¨ n ¸b n‡j ej, F = YA (n – 1) = 2 × 1011 × 2 × 10–4 (2–1) = 4× 107 N MBSTU 01. B¯úv‡Zi NbZ¡ 7.8 103 kgm3 Ges Bqs Gi w¯’wZ¯’vcK ¸Yv¼ 2 1011 Nm2 | B¯úv‡Z k‡ãi †eM n‡e- [MBSTU-B, Set-2 19-20] A. 6050 ms1 B. 6000 ms1 C. 456 ms1 D. 5064 ms1 v = Y = 2 1011 7.8 103 = 0.5064 104 = 5064 ms1 02. 1mm2 cÖ¯’‡”Q` wewkó GKwU B¯úv‡Zi Zv‡ii •`N©¨ 6% e„w× Ki‡j KZ ej cÖ‡qvM Ki‡Z n‡e? (B¯úv‡Zi Y = 21011 Nm–2 ) [MBSTU. 15-16] A. 1104N B. 1.2104N C. 12104N D. 0.12104N Al FL Y ev, F = 100 2 10 10 6 11 6 L YAl = 12103 = 1.2104 NSTU 01. 2m `xN© SzjšÍ GKwU Zv‡ii wb‡Pi cÖv‡šÍ 8kg fi Szjv‡j Gi •`N©¨ 0.5mm ev‡o| Zv‡ii Dcv`v‡bi Bqs ¸Yv¼ 21011nm–2 n‡j Zv‡ii cÖ¯’‡”Q‡`i ‡¶Îdj wbY©q Ki| [NSTU. 14-15] A. 1.51210–5m 2 B. 1.003410–6m 2 C. 1.56810–6m 2 D. 1.56810–8m 2 6 11 3 1.568 10 2 10 0.5 10 8 9.8 2 Yl FL A Al FL Y DU Technology 01. 2m •`N©¨ Ges 1 mm2 cÖ¯’‡”Q‡`i †ÿÎdj wewkó GKB Zv‡i 20kg fi Szwj‡q †`Iqvi d‡j ZviwUi •`N©¨ 2mm e„w× †M‡j ZviwUi Bqs Gi ¸Yv‡½i gvb KZ? [DU-Tech. 2020-21] A. 5 × 1011N/m2 B. 1.96 × 1011N/m2 C. 1.96 × 1012N/m2 D. 5 × 1012N/m2 S B info Bqs Gi ¸YvsK, Y = FL Al = 20 9.8 2 1 10–6 2 10–3Nm-2 = 1.96 1011 Nm–2 STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. 1cm2 cÖ¯’‡”Q` wewkó Zvgvi Zvi‡K †U‡b wظY •`N©¨ Ki‡Z e‡ji cÖ‡qvRb n‡e- (Y=21011N/m2 ) [BUET. 07-08] A. 107N B. 2107N C. 1011N D. 21011N F=YA =2101110-4 = 2107 CKRuet. Combind 01. `yBwU wfbœ c`v‡_©i •Zix mgvb •`‡N©¨i (2m) Ges mgvb e¨v‡mi (1mm) `yBwU Zvi wmwi‡R mshy³ Av‡Q| †h cwigvY ej cÖ‡qvM Ki‡j G‡`i mw¤§wjZ •`N©¨ 0.9 mm e„w× cv‡e Zvi gvb †ei Ki| [Y1 = 2 × 1011N/m2 I Y2 = 7 × 1011 N/m2 ] [CKRUET. 2020-21] A. 219.8 N B. 635.85 N C. 2198 × 108 D. 109.9 N E. 6.36 N S Blank info wmwi‡R mshy³ _vK‡j Dfq Zv‡i mgvb ej cÖhy³ n‡e| Bqs Gi ¸YvsK, Y = FL Al FL A = Yl = constant Y1l1 = Y2l2 l1 l2 = Y2 Y1 = 7×1011 2×1011 = 7 2 2l1 – 7l2 = 0 ... ....... (i) ; l1 + l2 = 0.99 .... ... ... (ii) (i) I (ii) mgxKiYØq mgvavb K‡i cvB, l1 = 0.7 mm, l2 = 0.2mm F = Y1Al2 L = 2 × 1011 ××(0.5×10–3 ) 2 ×0.7×10–3 2 N = 54.98N KUET 01. 2m `xN© SzjšÍ GKwU Zv‡ii wb‡Pi cÖv‡šÍ 10kg fi Szjv‡j Gi ‣`N©¨ 0.6mm ev‡o| Zv‡ii Dcv`v‡bi Bqs ¸Yv¼ 2 1011Nm–2 n‡j Zv‡ii e¨vmva© †KvbwU? [KUET. 18-19] A. 1.28mm B. 0.52mm C. 0.06cm D. 0.84cm E. 0.72mm A = FL lY r 2 = mgL lY r = 0.72mm 02. 1.4m `xN© Ges 10–6m 2 cÖ¯’‡”Q‡`i GKwU mylg avZe Zvi †U‡b 410–3m cÖmvwiZ Ki‡Z m¤úvw`Z Kv‡Ri cwigvb KZ? [Y=21011N/m2 ] [KUET. 14-15] A. 1.066J B. 1.143J C. 1.15 N/m2 D. 1.143 N/m2 E. 1.066 N/m2 W = 1 2 YAl2 L = 2 1011 10–6 × = 1.143J 03. 10m j¤^v Ges 1mm e¨vm wewkó GKwU Zvi‡K 100N ej Øviv Uvbv n‡j ZviwUi •`N©¨ KZUzKz e„w× cvq (Y=21011N/m2 ) [KUET. 12-13] A. 6.410-3m B. 6.410-4m C. 6.410-5m D. 6.410-6m E. 6.410-2m Y = FL r 2 l ev, l = FL Yr2 = 10010 21011(0.510– 3) 2 = 6.410-3m 04. `ywU mgvb •`‡N©¨i Zvi A I B Gi e¨vm h_vµ‡g 110-3m I 410-3m Dfq‡K mgvb e‡j Uvb‡j A Gi •`N©¨ e„w× B Gi •`N©¨ e„w×i 4 ¸Y nq| A I B Gi Dcv`v‡bi Bqs Gi w¯’wZ¯’vcK ¸Yvs‡Ki Zzjbv Ki| [KUET. 08-09] A. 1:1 B. 1:2 C. 2:1 D. 4:1 E. 1:4 2 3 3 2 1 10 4 10 4 1 dA dB L L Y Y A B B A = 4:1 1 4 RUET 01. GKwU ÷x‡ji Zv‡ii •`N©¨ 2m Ges cÖ¯’‡”Q` 0.810-6m 2 . Av‡iK cÖvšÍ `„p fv‡e AvUKv‡bv Av‡Q| Ab¨ cÖv‡šÍ KZ ej cÖ‡qvM Ki‡j Zv‡ii •`N©¨ 0.5mm e„w× cv‡e| [÷x‡ji Bqs Gi ¸YvsK 21011N/m2 ] [RUET. 09-10] A. 8N B. 4N C. 16N D. 40N Al FL Y ev, L YAl F = 2 2 10 0.8 10 0.5 10 11 6 3 = 40
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 281 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. 4m ‣`N©¨ Ges 30.5mm e¨v‡mi GKwU w÷‡ji Zv‡ii Dci 5kg fi cÖ‡qvM Ki‡j •`N©¨ e„w× n‡e- [MAT. 14-15] A. 4.910–5m B. 4.9 10–4m C. 4.9 10–3m D. 4.9 10–6m S Blank info r l mgL Y 2 ev, 2 Y r mgL l 2 11 3 2 10 15.25 10 5 9.8 4 = 1.34110–6 STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. 1m `xN© GKwU Zv‡i 105Nm-2 ej cÖ‡qv‡M Gi •`N©¨ e„w× †cj 0.001m| ZviwUi Bqvs ¸bv¼ KZ? [P.‡ev. 2019] A. 10-7Nm-2 B. 10-3Nm-2 C. 107Nm-2 D. 108Nm-2 S D info Y = F A ℓ L = 105 0.001 1 = 108 Nm–2 02. w¯’wZ¯’v&cK mxgvi g‡a¨ AvKvi cxob I AvKvi weK…wZi AbycvZ n‡”QÑ [Xv. †ev. 2015] A. Bqs-Gi ¸Yv¼ B. AvqZb ¸Yv¼ C. `„pZvi ¸Yv¼ D. cqm‡bi AbycvZ Ans C 03. 1 eM© †m.wg. cÖ¯’‡”Q`wewkó GKwU Zv‡i KZ ej cÖ‡qvM Kiv n‡j Gi •`N©¨ e„w× Avw` •`‡N©¨i mgvb n‡e? [Y = 21011 N m 2 ] [iv. †ev. 2016] A. 2107N B. 4107N C. 2105N D. 4105N S A info Y = Fℓ AL 2 × 1011 = F × L 1 × 10–4 × L [∵ ℓ = L] F = 2 × 107 N wP‡Î GKwU avZe Zv‡ii Rb¨ •`N©¨ cxob •`N©¨ weK…wZ †jL †`Lv‡bv n‡jv: DÏxcK Abymv‡i wb‡Pi 04 I 05 bs cÖ‡kœi DËi `vI: 04. ZviwUi Dcv`v‡bi Bqvs Gi ¸bv¼ KZ? [w`.‡ev. 2019] A.7.2×109Nm-2 B.3.6×1010Nm-2 C.4.8×1010Nm-2 D.1.8×1011Nm-2 S D info Y = cxob weK…wZ = 3.6 × 1010 0.2 = 1.8 × 1011 Nm–2 05. ZviwUi Dci cxobÑ [w`.‡ev. 2019] i. 4.8 × 1010Nm-2 Gi †P‡q †ewk n‡j ZviwU wQ‡o hv‡e ii. 4.2 × 1010Nm-2 n‡j ZviwUi ¯ vqx weK…wZ n‡e iii. 3.6 × 1010Nm-2 Gi †P‡q Kg n‡j †Kv‡bv weK…wZ NU‡e bv wb‡Pi †KvbwU mwVKÑ A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans A 06. wP‡Î weK…wZ ebvg cxob †jLwP‡Îi OAB-Gi †ÿÎdj wb‡`©k K‡i [w`. †ev. 2016] A. Bqs Gi ¸YvsK B. me©‡gvU K…Z KvR C. cqm‡bi AbycvZ D. GKK AvqZ‡bi wefekw³ Ans D 07. 1m j¤^v I 1mm e¨v‡mi Zvi‡K ej cÖ‡qv‡M 0.025cm •`©N¨ e„w× Kiv n‡jv| e¨vm n«vm KZ? [h.‡ev. 2019] A. 1.5 × 105 mm B. 2.5 × 105 mm C. 3.5 × 105 mm D. 2.5 mm S B info d D = ℓ L d = ℓ L × D = 0.025 × 10–2 1 × 1 = 2.5 × 105 mm wb‡¤œv³ Z_¨ n‡Z 08bs cÖ‡kœi DËi `vI : [Kz. †ev. 2017] GKwU Zv‡i 0.01 •`N©¨ weK…wZ‡Z cvk¦© weK…wZ 0.0024 n‡jv|Ó 08. ‣`N©¨ weK…wZ ebvg cvk¦© weK…wZi †jLwP‡Îi cÖK…wZ †KvbwU? A. B. C. D. Ans D wb‡Pi DÏxcKwU co Ges 09bs cÖ‡kœi DËi `vI: D e¨vm I L •`‡N©¨i GKwU Zvi GK cÖv‡šÍ `„pfv‡e AvUKv‡bv Av‡Q| ZviwUi wb‡Pi cÖv‡šÍ GKwU fi Szjv‡bv‡Z Gi •`N©¨ x cwigvY e„w× †cj| X, L Gi A‡a©K| 09. Y = 2.01011 Nm2 n‡j cxob KZ? [Kz. †ev. 2016] A. 0.251011 Nm2 B. 0.51011 Nm2 C. 11011 Nm2 D. 41011 Nm2 S C info Y = F A ℓ L F A = Y × ℓ L = Y × L 2 L = Y 2 cxob = 2.0 × 1011 × 1 2 = 1 × 1011 Nm–2 10. wP‡Î cxob Ges weK…wZi ga¨Kvi †jLwP‡Î OA †iLvi Xvj wK wb‡`©k K‡i? [Kz. †ev. 2016] A. bwZ we›`y B. Bqs Gi ¸Yv¼ C. f½yi we›`y D. ¯ vqx weK…wZ Ans B 11. AvqZb ¸bv‡¼i Ab¨ bvg Kx? [e.‡ev. 2019] A. Amsbg¨Zv B. msbg¨Zv C. KvwV‡b¨i ¸bv¼ D. Bqvs Gi ¸bv¼ Ans A wb‡Pi DÏxcKwU c‡ov Ges cÖ‡kœi DËi `vI : [e. †ev. 2015] mgvb •`‡N©¨i wZbwU Zvi A, B Ges C G GKB gv‡bi cxob 51012 Nm2 cÖ‡qv‡Mi d‡j •`N©¨ e„w× h_vµ‡g 5%, 2% Ges 1% nj| 12. A, B Ges C Zv‡ii Bqs Gi ¸YvsK h_vµ‡g YA, YB I YC n‡j wb‡Pi †KvbwU mwVK? [e. †ev. 2015] A. YA > YC > YB B. YA < YB < YC C. YA > YB > YC D. YB < YA < YC Ans B
282 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 3 mv›`ªZvsK, mv›`ªej I cÖvwšÍK †eM msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. wbDU‡bi m~Î, F = A dx dv [ = mv›`ªZvsK, dx dv = †e‡Mi bwZ, F = mv›`ªej, A = ¯Íi؇qi †¶Îdj] 02. †÷vK‡mi m~Î, F = 6rv [v= cÖvšÍ†eM] 03. e¯‧i IRb, W= cøeZv (F1) + mv›`ªej (F2) 04. cÖvšÍ†eM, v = 9 2r ( )g 2 [ = †Mvj‡Ki NbZ¡, = cÖevnxi NbZ¡] 05. ZvcgvÎvi mv‡_ mv›`ªZv mn‡Mi m¤úK©, log = A + B T [Zi‡ji †ÿ‡Î] 06. 2 1 = 2 1 T T [M¨v‡mi †ÿ‡Î] CONCEPTUAL MATH MEx 01 210-3m e¨v‡mi GKwU ¶z`ª †MvjK Zi‡ji ga¨ w`‡q 4x10-2ms-1 cÖvšÍ †e‡M co‡Q| †Mvj‡Ki Dci wµqviZ mv›`ª ej 3106N Zi‡ji mv›`ªZvsK wbY©q Ki| F = 6vr ev, = F 6rv 2 4 10 3 6 3.14 1 1 10 6 3 10 = 3. 98109 kg m-1 s -1 MEx 02 2 10–2m e¨v‡mi GKwU M¨vm ey`ey` 1.75 103 kg.m3 Nb‡Z¡i GKwU `ªe‡Yi wfZi w`‡q 3.5 102ms1 w¯’i †e‡M D‡aŸ© DV‡Q| M¨v‡mi NbZ¡ AMÖvn¨ K‡i `ªe‡Yi mv›`ªZvsK wbY©q Ki| mv›`ªZvsK, = 2 9 r 2 g v = 2 9 (102 ) 2 1.75 103 9.8 3.5 102 = 1.09 103 kg.m1 s 1 NOW START PRACTICE 01. cvwbi wfZi w`‡q 105m e¨vmv‡ai GKwU evqy ey`ey` DV‡Q| cvwbi mv›`ªZv 3.5 × 10-3Nsm2 Ges NbZ¡ 103 kgm3 | cvwbi Nb‡Z¡i Zzjbvq evqyi NbZ¡ AMÖvn¨ K‡i ey`ey`wUi DaŸ©gyLx †ei Ki| 02. †Kvb cÖev‡ni ZvcgvÎv GK-PZz_©vsk Kiv n‡j Gi mv›`ªZv ¸YvsK KZ¸Y n‡e| NOW PRACTICE SOLVE : 01. DaŸ©gyLx †eM, v = 2 9 r 2 g = (105 ) 2 103 9.8 3.5 10–3 = 2.18 10ms1 02. T 1 4 1 2 A_©vr A‡a©K n‡e| REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. 0.01 m2 †¶Îdj wewkó GKwU cvZ 2mm cyiæ wMømvwi‡bi GKwU ¯Í†ii Dci ivLv i‡q‡Q| cvZwU 0.05 ms–1 †e‡M Pvjbv Ki‡Z 0.4 N Abyf‚wgK e‡ji cÖ‡qvRb n‡j mv›`ªZv ¸Yv‡¼i gvb KZ? [KU. 17-18] A. 1.6 Nm–2 B. 1.6 Nsm–2 C. 1.60 Nsm–2 D. 1600 Nm–2 AbyfzwgK ej, F = A dx dv Zi‡ji mv›`ªZvsK, = dv dx A F = 0.05 2 10 10 0.4 3 2 = 1.6 kgs1m 1 = 1.6 Nsm–2 02. GKwU 10–2m 2 cÖ¯’‡”Q‡`i †¶Îdj wewkó cv‡Zi cyiæZ¡ 210–3m| H †cøU‡K 0.03ms–1 †e‡M Pvjbv Ki‡Z 0.235N AbyfywgK e‡ji cÖ‡qvRb n‡j Zi‡ji mv›`ªZvsK KZ? [KU. 13-14] A. 3kgs–1m –1 B. 1.567kgs–1m –1 C. 1kgs–1m –1 D. 4kgs–1m –1 F = dx dv A 1.567 10 0.03 0.235 2 10 2 3 A dv F dx PART B Analysis of Science & Technology Question SUST 01. 100mm e¨vmv‡a©i GKwU ‡MvjK ‡Kv‡bv †Z‡ji g‡a¨ 210–2ms–1 cÖvšÍ‡eM wb‡q co‡Q| ‡Z‡ji mv›`ªZv¼ 210–3Nsm–2 n‡j mv›`ª ej KZ N? [SUST. 15-16] A. 7.53610–5 B. 8.41610–5 C. 9.43910–5 D. 5.15610–5 E. 4.83610–5 F = 6rv = 6210–3 0.1210–2 = 7.53610–5 02. GKwU •KwkK b‡ji e¨vm 0.22mm. G‡K 7.210–2Nm–1 c„ôUvb Ges 103 kgm–3 Nb‡Z¡i cvwb‡Z Wyev‡j b‡ji KZ m D”PZvq cvwb DV‡e? [SUST. 15-16] A. 0.1149 B. 0.1223 C. 0.1469 D. 0.1336 E. 0.1750 T = hrg 2 h = 2T rg = 2 7.2 10-2 0.11 10–3 103 9.8 = 0.13358 = 0.1336 03. 0.02 m2 †¶Îd‡ji GKwU avZe cvZ 5 mm cyiæ wMømvwi‡bi GKwU ¯Í‡ii Dci ivLv n‡q‡Q| cvZwU‡K 0.06 m/s †e‡M Pvjbv Ki‡Z 0.5 N Abyf~wgK e‡ji cÖ‡qvRb nq| wMømvwi‡bi mv›`ªZv ¸Yv‡¼i gvb KZ Nsm–2 ? [SUST. 14-15] A. 2.08 B. 1.04 C. 1.0410-2 D. 2.0810-2 E. 2.0810-3 F = dx dv A ev, Adv F.dx 0.02 0.06 0.5 5 10 3 = 2.0833
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 283 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 02 ANALYSIS OF ENGINEERING & BUTex QUESTION CKRuet. Combind 01. e„wói cvwbi GKwU †dvuUv evqyi ga¨ w`‡q cwZZ n‡”Q| †duvUwUi AšÍt‡eM 1.5 cms–1 ,evqyi mv›`ªZv mnM 1.8 × 10–4 Ges evqyi NbZ¡ 1.21 × 10–3 gm/cc n‡j cvwbi †duv‡Ui e¨vmva© KZ? [CKRUET. 2020-21] A. 1.1 × 10–3 cm B. 1.11 × 10–3 cm C. 11.1 × 10–6 cm D. 1.24 × 10–3 cm E. 1.11 × 10–5 cm S B info †duvUvi AšÍt‡eM, v = 2r2 (w –a)g 9 [meB C.G.S GK‡K] r 2 = 9v 2g(w – a) = 9 × 1.8 × 10–4 × 1.5 2 × 980 × (1 – 1.21 × 10–3 ) = 1.24 × 10–6 cm2 r = 1.11 × 10–3 cm KUET 01. 800kgm–3 NbZ¡ I 10–4m e¨vmv‡a©i GKwU †Zj †dvUv 1.5 kgm–3 NbZ¡ I 1.85 10–5NSm–2 mv›`ªZv ¸Yvs‡Ki evqyi ga¨ w`‡q co‡j Dnvi cÖvwšÍK †eM KZ? [KUET. 18-19] A. 1.14 10–4m/s B. 0.94m/s-1 C. 9.4 m/s D. 94 m/sE. 800 m/s Ve = 2r2 9 (s – f)g = 2 × (104 ) 2 9 × 1.85 × 105 (8001.5) = 0.94ms–1 STEP 03 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. GKwU ¶z`ª †MvjvKvi e¯‘ †Kvb Zi‡ji ga¨w`‡q cÖvšÍ †e‡M co‡Q| e¯‘i IRb 0.03 N n‡j Ges e¯‘i Dci wµqviZ cøeZv 0.01N n‡j, e¯‘i Dci wµqviZ mv›`ª ej n‡e? [MAT. 03-04] A. 0.01N B. 0.02N C. 0.03N D. 0.04N mv›`ª ej = 0.03 – 0.01 = 0.02N STEP 04 ANALYSIS OF HSC BOARD QUESTION 01. 100gm f‡ii GKwU e¯‘ cvwbi g‡a¨ covq Zvi Dci wµqviZ cøeZv 0.981N n‡j mv›`ª ej n‡e- [e. †ev. 2017] A. 981N B. 0.981N C. 0N D. 1.962N S C info W = u + F F = W – u = 0.1 × 9.8 – 0.98 = 0N wb‡Pi DÏxcKwU c‡o 02bs cÖ‡kœi DËi `vI: wP‡Î GKwU cÖevnxi ga¨ w`‡q gy³fv‡e cošÍ GKwU †Mvj‡Ki Dci wµqvkxj ej †`Lv‡bv nj Ges †MvjKwUi co‡bi gyn~Z© n‡Z †MvjKwUi MwZ‡eM-mgq MÖv‡di gva¨‡g †`Lv‡bv nj| 02. MÖvdwUi O we›`y‡Z Z¡iY KZ? [Kz. †ev. 2016] A. 0 ms2 B. 4.9 ms2 C. 9.8 ms2 D. 14.7 ms2 Ans A wb‡Pi DÏxcKwU c‡o 03bs cÖ‡kœi DËi `vI: [Kz. †ev. 2016] wP‡Î GKwU cÖevnxi ga¨ w`‡q gy³fv‡e cošÍ GKwU †Mvj‡Ki Dci wµqvkxj ej †`Lv‡bv nj Ges †MvjKwUi co‡bi gyn~Z© n‡Z †MvjKwUi MwZ‡eM-mgq MÖv‡di gva¨‡g †`Lv‡bv nj| 03. A Ges B we›`yi g‡a¨Ñ A. W = F B. W = u C. W = u + F D. W > u + F Ans D 04. wb‡Pi †Kvb m¤úK©wU †÷vKm-Gi m~Î? [Kz. †ev. 2015] A. F r B. F r C. F rv D. F Ans C 05. cÖvšÍ‡eM mv›`ªZv ¸Yv‡¼i- [mKj †evW© 2014] A. mgvbycvwZK B. e¨¯ÍvbycvwZK C. eM©g~‡ji mgvbycvwZK D. eM©g~‡ji e¨¯ÍvbycvwZK Ans B Concept 4 cqm‡bi AbycvZ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. cqm‡bi AbycvZ: = cvk¦© weK…wZ ‣`N©¨ weK…wZ = l L d D l L r R / / / / , (–1 < < 1 2 ) CONCEPTUAL MATH MEx 01 GKwU 1m Zv‡ii e¨vmva© 5m| H Zv‡i ej cÖ‡qvM Ki‡j 0.02m •`N©¨ e„w× cvq, wKš‘ e¨vmva© 0.05m K‡g hvq| cqm‡bi AbycvZ n‡ecqm‡bi AbycvZ, = cvk^© weK…wZ •`N©¨ weK…wZ = Lr lR = 1 0.05 0.02 0.5 = 5 MEx 02 GKwU Zv‡i •`N©¨ weK…wZ Ges cvk¦© weK…wZ h_vµ‡g 0.01m Ges 0.25 cm n‡j Zv‡ii Dcv`v‡bi cqm‡bi AbycvZ KZ? = cvk¦© weK…wZ ‣`N©¨ weK…wZ 0.25 1 0.25 NOW START PRACTICE 01. ej cÖ‡qv‡Mi d‡j GKwU Zv‡ii •`N©¨ 1%cwieZ©b n‡j Gi e¨vm kZKiv KZ fvM cwiewZ©Z n‡e? cqm‡bi AbycvZ 0.2| 02. GKwU Zv‡i 0.01m •`N©¨ weK…wZ‡Z cvk^© weK…wZ 0.0024m n‡j Zv‡ii Dcv`v‡bi cqm‡bi AbycvZ wbY©q Ki| NOW PRACTICE SOLVE : 01. d = cvk¦© weK…wZ •`N©¨ weK…wZ ev, 0.2 = 0.01 d ev, d = 0.002 = 0.2% 02. cqm‡bi AbycvZ, = cvk^© weK…wZ •`N©¨ weK…wZ = 0.0024 0.01 = 0.24
284 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 RU QUESTION 01. GKwU Zv‡i 0.01m •`N©¨weK…wZ‡Z cvk¦©weK…wZ 0.0024 m n‡j Zv‡ii Dcv`v‡bi cqm‡bi AbycvZ KZ? [RU. 12-13] A. 0.20 B. 0.30 C. 0.24 D. 0.28 cqm‡bi AbycvZ, = cvk¦© weK…wZ •`N©¨ weK…wZ = 0.01 0.0024 = 0.24 STEP 02 ANALYSIS OF CU QUESTION 01. cqmb (Poisson) Abycv‡Zi mxgv Kx? [CU. 18-19; mKj †evW© 2015] A. – 1 2 < 1 2 B. – 1 2 < < 1 C.1 < < 2 D. –1 < < 1 2 Ans D 02. GKwU Zv‡i •`N©¨ weK…wZ Ges cvk¦© weK…wZ h_vµ‡g 0.01m Ges 0.25 cm n‡j Zv‡ii Dcv`v‡bi cqm‡bi AbycvZ KZ? [CU. 12-13] A. 0.25 B. 0.025 C. 2.5 D. 25 E. 0.0025 = cvk¦© weK…wZ ‣`N©¨ weK…wZ 0.25 1 0.25 STEP 03 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. 2m `xN© SzjšÍ GKwU Zv‡ii wb‡Pi cÖv‡šÍ 10kg fi Szjv‡j Gi ‣`N©¨ 0.6mm ev‡o| Zv‡ii Dcv`v‡bi Bqs ¸Yv¼ 2 1011Nm–2 n‡j Zv‡ii e¨vmva© †KvbwU? [KUET. 18-19] A. 1.28mm B. 0.52mm C. 0.06cm D. 0.84cm E. 0.72mm A = FL lY r 2 = mgL lY = r = 0.72mm 02. ej cÖ‡qv‡Mi d‡j GKwU Zv‡ii •`N©¨ 1%cwieZ©b n‡j Gi e¨vm kZKiv KZ fvM cwiewZ©Z n‡e? cqm‡bi AbycvZ 0.2. [BUET. 07-08] A. 1% B. 2% C. 0.2% D. 5% d = cvk¦© weK…wZ •`N©¨ weK…wZ ev, 0.2 = 0.01 d ev, d = 0.002 = 0.2% CUET 01. 3m `xN© Ges 1mm e¨vm wewkó GKwU avZe Zvi‡K 10kg IRb Øviv Uvbv n‡jv| hw` Bnvi Dcv`v‡bi Bqs Gi ¸YvsK I cqm‡bi AbycvZ h_vµ‡g 12.51011 dyne/cm2 I 0.26 nq Zvn‡j Gi cvk¦©xq ms‡KuvPb ‡ei Ki| [CUET. 11-12] A. 2.610-5 cm B. 2.610-8 cm C. 2.610-2 cm D. None of them Al FL Y ev, 2 r mg YA F L l = 11 2 2 12.5 10 (0.5 10 ) 10 1000 981 = 0.1 cqm‡bi AbycvZ = cvk¦© weK…wZ ‣`N©¨ weK…wZ ev, cvk¦© weK…wZ = ‣`N©¨ weK…wZ = 0.260.1 =0.026 =2.610-2 cm STEP 04 ANALYSIS OF HSC BOARD QUESTION 01. GKwU Zv‡ii •`N©¨ weK…wZ 0.02 Ges cvk^© weK…wZ 0.002 n‡j Gi cqm‡bi AbycvZ KZ? [iv. †ev. 2017] A. 0.004 B. 0.1 C. 10 D. 0.0004 S B info = cvk¦© weK…wZ •`N©¨ weK…wZ = 0.002 0.02 = 0.1 02. GKwU Zv‡ii •`N©¨ eivei ej cÖ‡qvM Kiv n‡j Gi •`N©¨ 1m n‡Z 1.02m nq Ges e¨vm 5 mm n‡Z 4.99 mm nq| cqm‡bi AbycvZ KZ? [P. †ev. 2015] A. 0.01 B. 0.1 C. 1 D. 10 Ans B 03. hLb cvwb‡Z wKQz wWUvi‡R›U †gkv‡bv nq ZLb Gi c„ôUvb- [w`. †ev. 2015] A. e„w× cvq B. n«vm cvq C. AcwiewZ©Z _v‡K D. k~b¨ nq Ans B 04. †Kvb Zi‡ji c„ôkw³ msL¨vMZfv‡e c„ôUv‡bi- [w`. †ev. 2015] A. A‡a©K B. mgvb C. wظY D. wZb¸Y Ans B 05. 2m j¤^v I 2mm e¨vmva©wewkó GKwU Zv‡ii •`N©¨ e„w× 0.25mm n‡j ZviwUi e¨vmva© KZ n«vm cv‡e? (g = 9.8ms-2 ) [Kz.‡ev. 2019] A. 5×10-3m B. 2.5×10-3m C. 5×10-8m D. 2.5×10-8m S C info = r × L r × L r = × r × L L = 0.2 × 2 × 10–3 × 0.25 × 10–3 2 = 5 × 10–8 m GKwU Zv‡i 0.01 •`N©¨ weK…wZ‡Z cvk¦© weK…wZ 0.0024 n‡jv|Ó [Kz. †ev. 2017] Dc‡iv³ Z_¨ n‡Z 06bs cÖ‡kœi DËi `vI: 06. cqm‡bi Abycv‡Zi gvb A. 0.2 B. 0.24 C. 2 D. 2.4 S B info = cvk¦© weK…wZ •`N©¨ weK…wZ = 0.24 wb‡Pi DÏxcKwU co Ges 07bs cÖ‡kœi DËi `vI: [Kz. †ev. 2016] D e¨vm I L •`‡N©¨i GKwU Zvi GK cÖv‡šÍ `„pfv‡e AvUKv‡bv Av‡Q| ZviwUi wb‡Pi cÖv‡šÍ GKwU fi Szjv‡bv‡Z Gi •`N©¨ x cwigvY e„w× †cj| X, L Gi A‡a©K| 07. GKB Dcv`v‡bi 2D e¨vm Ges 3L •`‡N©¨i Aci GKwU Zv‡i mgcwigvY fi Szj‡jÑ i. cqm‡bi AbycvZ AcwiewZ©Z _vK‡e ii. ‣`N©¨ e„w× n‡e 3x 4 iii. cxo‡bi cwieZ©b n‡e wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans D 08. cqm‡bi AbycvZ- [wm. †ev. 2016] A. •`N©¨ weK…wZ I cvk^© weK…wZi AbycvZ B. ‡Kv‡bv GKK †bB C. Gi gvb –1 †_‡K 0.5 ch©šÍ D. B I C DfqB Ans D 09. cqm‡bi Abycv‡Zi †ÿ‡Î- [wm. †ev. 2016] A. = – L0r rL B. cqm‡bi Abycv‡Zi gvb –1 †_‡K 1 2 Gi g‡a¨ C. cqmb Abycv‡Zi †Kv‡bv gvÎv I GKK †bB D. me¸‡jv mwVK Ans D
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 285 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 5 c„ôUvb wbY©q msµvšÍ MvwYwZK cÖ‡qvM FORMULA c„ôUvb: 01. cvwbi †dvUvi †ÿ‡Î, T = F L 02. †K․wkK b‡ji mv‡_ Zi‡ji ¯úk©‡Ki Gi †ÿ‡Î, T = hgr 2cos 03. cvwbi ey`ey‡`i †ÿ‡Î, T = F 2L 04. myuP: T = F 2l 05. mvev‡bi †ÿ‡Î, T = F 4L 06. AvqZvKvi e¯‧, T = F 2(a + b) hLb, L = 2(a + b) 07. e„ËvKvi e¯‧, T = F 2r hLb, L = 2r 08. ZvcgvÎvi mv‡_ c„ôUv‡bi cwieZ©b, Tt = T0(1 – t) ◈ T = hrpg 2cos ; 0 90 Av‡ivnb; ◈ 90 180 Aebgb; 90 n‡j Av‡ivnb n‡e bv, AebgbI n‡e bv| (Av‡ivnb n‡j h (+) ve Aebgb n‡j, h(–) ve) CONCEPTUAL MATH MEx 01 GKwU •KwkK b‡ji e¨vm 0.2mm G‡K 7210–3Nm–1 c„ôUvb Ges 103 kgm–3 Nb‡Z¡i cvwb‡Z Wyev‡j b‡ji KZ D”PZvq cvwb DV‡e2 hr g T ev, rpg T h 2 0.1 10 10 9.8 2 72 10 3 3 3 h 0.1469m MEx 02 cvwbi DcwiZj n‡Z 0.05g j¤^v GKwU Abyf~wgK Zvi‡K †U‡b Zzj‡Z †h me©vwaK e‡ji cÖ‡qvRb Zvi gvb 7.28 10-2Nmcvwbi c„ôvb †ei Ki| cvwbi c„ôUvb, T = F 2L = 7.28 10–3 2 0.05 = 7.28 10–2Nm1 MEx 03 GKwU ‣KwkK bj cvwb‡Z AvswkK †Wvev‡bv| Gi A‡a©K e¨vmv‡a©i Avi GKwU ‣KwkK bj cvwb‡Z AvswkK Wzev‡bv n‡j Zvi g‡a¨Kvi cvwbi ¯Í‡¤¢i D”PZv cÖ_gwUi g‡a¨Kvi cvwbi ¯Í‡¤¢i D”PZvi Zzjbvq KZ n‡e? h1r1 = h2 r2 h2 h1 = r1 r2 h2 h1 = 1 1 2 = 2 (wظb) NOW START PRACTICE 01. cvwbi DcwiZj n‡Z 0.05m j¤^v GKwU Avbyf~wgK Zvi‡K †U‡b Zzj‡Z Zv‡ii IRbmn me©vwaK 7.2810–3N e‡ji cÖ‡qvRb| cvwbi c„ô Uvb KZ? 02. `yBwU †K․wkK b‡ji g‡a¨ GKwU AciwU A‡c¶v †ekx miæ| `yBwU‡KB Lvovfv‡e cvwbi g‡a¨ AvswkK Wzev‡bv n‡j †ekx miæwUi wfZi cvwbi ¯Í‡¤¢i D”PZv †ekx nq| Gi KviY03. GKwU ‣KwkK bj cvwb‡Z AvswkK †Wvev‡bv| Gi GK Z…Zxqvsk e¨vmv‡a©i Avi GKwU ‣KwkK bj cvwb‡Z AvswkK Wzev‡bv n‡j Zvi g‡a¨Kvi cvwbi ¯Í‡¤¢i D”PZv cÖ_gwUi g‡a¨Kvi cvwbi ¯Í‡¤¢i D”PZvi Zzjbvq KZ n‡e? NOW PRACTICE SOLVE : 01. T = F 2L = 7.28 10–3 2 0.05 = 7.28 10–2 Nm-1 02. GLv‡b cvwbi c„ôUvb b‡ji e¨v‡mi Dci wbf©ikxj wKš‧ cvwbi ¯Í‡¤¢i IRb e¨v‡mi e‡M©i Dci wbf©ikxj bq| (cÖvq) 03. h1r1 = h2 r2 h2 h1 = r1 r2 h2 h1 = 1 1 3 = 3 (wZb¸b) REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU •KwkK b‡ji e¨vm 0.0410–4m. Gi GK cÖvšÍ cvwb‡Z Wzev‡j cvwb b‡ji wfZi 0.082 m Dc‡i D‡V| cvwbi Zj Uvb KZ? †`qv Av‡Q, ¯úk© †KvY = 0 Ges cvwbi NbZ¡ = 1.0103 kg/m3 . [DU. 12-13] A. 8.510–4 N/m B. 7.510–4 N/m C. 9.010–4 N/m D. 8.010–4 N/m b‡ji e¨vm, d = 0.04 m 4 10 b‡ji e¨vmva© r = 0.02 m 4 10 , D”PZv, h=0.082m ZjUvb , T =?, ¯úk© †Kvb, = 00 cvwbi NbZ¡ = 1.0 3 3 10 kg / m T = 2 cos hr g = 2 0.082 0.02 10 10 9.8 4 3 = 8 4 10 N/m 02. cvwbi c„ôUvb 7210-3 N/m | 0.2 mm e¨v‡mi b‡j cvwbi Av‡ivnY n‡e- [DU.04-05, 02-03; RU. 06-07] A. 14.694m B. 14.694 10-2m C. 10.0cm D. 7.347 cm c„ôUvb, T = 2 rhg Zi‡ji Av‡ivnY, h = gr 2T = 3 3 3 10 9.8 0.1 10 2 72 10 =14.694 10–2 m STEP 02 ANALYSIS OF JU QUESTION 01. cvwbi DcwiZ‡j ivLv 3cm `xN© myB‡K †U‡b Zzj‡Z me©vwaK KZ ej `iKvi? [T = 72 10–3Nm–1 ] [JU. 18-19] A. 4.9 10–3N B. 4.32 10–3N C. 2.16 10–3N. D. 3.3 10–3N F = 2T1 = 2 72 10-3 3 100 = 4.32 10–3N
286 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. 20C ZvcgvÎvq cvwbi DcwiZj n‡Z 0.05m j¤^v GKwU Abyf‚wgK Zvi‡K †U‡b Zzj‡Z †h me©vwaK Z‡ji cÖ‡qvRb Zvi gvb 7.2810-3N, cvwbi c„ôUvb †ei Ki| [Zv‡ii IRb bMY¨] [JU. 16-17] A. 7.210-2Nm-2 B. 2.310-2Nm-1 C. 7.210-3Nm-1 D. 7.210-1Nm-1 T = F 2L = 2.28 10–3 2 0.05 = 2.3 10–2Nm–1 03. cvwbi c„ô Uvb 7210-3 Nm-1 n‡j 4mm. e¨v‡mi †Kvb cvwbi we›`yi wfZ‡ii I evwn‡ii Pv‡ci cv_©K¨ n‡e- [JU. 12-13] A. 144Nm-2 B. 100Nm-2 C. 180Nm-2 D. 110Nm-2 P = 4T R = 47210–3 210–3 = 144Nm–2 04. Zi‡ji IR‡bi d‡j GKwU †K․wkK b‡ji cvwbi D”PZv †e‡o GKwU wbw`©ó D”PZvq †c․‡Qb †hLv‡b c„ô Uv‡bi d‡j D™¢~Z DaŸ©gyLx 7510-4N ej Øviv mvg¨ve¯’vq _v‡K| hw` cvwbi c„ô Uvb 610–2N/m nq Zvn‡j †K․wkK b‡ji wfZi c„‡ôi cwiwa KZ? [JU. 12-13] A. 1.2510-2m B. 0.5010-2m C. 6.510-2m D. 12.510-2m Avgiv Rvwb, Zi‡ji Dci †gvU DaŸ©gyLx ej=2rTcos (cos = 1) = 2rT wfZ‡ii c„‡ôi cwiwa= F T = 75 10–4 6 10–2 = 12.5 10–2m STEP 03 ANALYSIS OF RU QUESTION 01. cvwbi DcwiZ‡j Avj‡Zvfv‡e ivLv 3cm `xN© GKwU m~P‡K †U‡b Zzj‡Z me©vwaK KZ e‡ji cÖ‡qvRb? [RU. 13-14, 06-07] (cvwbi c„ôUvb 72 10-3 Nm-1 ) A. 4.32 10-3 N B. 4.52 10-2 N C. 4.44 10-3N D. 310-3 N ej, F = T 2L = 7210–3 20.03 = 4.3210–3 N Ans. 02. 2mm e¨v‡mi GKwU †MvjK‡K 106 ‡QvU †QvU we›`y‡Z †¯cÖ Kiv n‡j e¨wqZ kw³i cwigvb KZ? c„ôUvb=7210-3N/m [RU. 08-09] A. 4.510-5 J B. 910-5 J C. 110-4 J D. 910-4 J 1 3 R nr ev, 3 3 6 1 1 10 10 1 r = 3 5 2 10 10 10 1 kw³ = 8(Nr2 – R 2 )T= 8{106 (10-5 ) 2 –(10–3 ) 2 }7210–3 =110-4N STEP 04 ANALYSIS OF CU QUESTION 01. M¨v‡mi Abyi Mo MwZkw³ Dnvi- [CU. 06-07] A. ZvcgvÎvi e¨¯ÍvbycvwZK B. Pv‡ci mgvbycvwZK C. ZvcgvÎvi mgvbycvwZK D. Pv‡ci e¨¯ÍvbycvwZK E. AvqZ‡bi mgvbycvwZK Ans C STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. †Kvb Zv‡ii •`N©¨ 3m Ges fi 200kg Uv‡b H Zv‡ii •`N©¨ 1mm ev‡o| Zv‡ii NbZ¡ 7.510–3 kg/m3 n‡j, Gi Dcv`v‡bi Bqs Gi ¸Yv¼ wbY©q Ki| [JnU. 08-09] A. 2.691011N/m2 B. 2.351011N/m2 C. 41011N/m2 D. 541011N/m2 Zv‡ii †¶Îdj A = 7.5 10–3 3 = 2.5 10–3m 2 Y = mgL Al = 200 9.8 3 2.5 10–3 10–3 = 2.35 1011 N/m2 02. 0.35m `xN© Ges 0.20mm e¨vmv‡a©i GKwU A¨vjywgwbqvg Zv‡ii •`N©¨ 1.4mm e„w× Kiv n‡jv| Gjywgwbqv‡gi Bqs Gi ¸YvsK 71010N/m2 n‡j ZviwUi cxob KZ? [JnU. 06-07] A. 2.8107N/m2 B. 28108N/m2 C. 2.8108N/m2 D. 2.81011N/m2 Y = PL l P= Yl L = 7 1010 1.4 10–3 0.35 = 28107 2.8108 N/m2 CoU 01. 20000C ZvcgvÎvq cvwbi DcwiZj n‡Z 0.50m j¤^v GKwU Avbyf~wgK Zvi‡K †U‡b Zzj‡Z me©vwaK †h e‡ji cÖ‡qvRb nq Zvi gvb 7.2810-3N. cvwbi c„ôUv‡bi cwigvb †ei Ki? [CoU. 12-13] A. 7.2810-2Nm-1 B. 7.0010-2Nm-1 C. 7.7710-5Nm-1 D. 7.9910-1Nm-1 T = F 2l= 7.2810–3 2×0.5 3 1 3 7.28 10 2 0.05 7.28 10 2 Nm l F T PART B Analysis of Science & Technology Question MBSTU 01. 1 mm e¨v‡mi GKwU •KwkK bj 4C ZvcgvÎvi weï× cvwbi g‡a¨ Wzev‡bv n‡j b‡ji wfZi cvwbi D”PZv nq 3 cm| cvwbi c„ôUvb KZ? [MBSTU-C, Set- 19-20] A. 14.70 102 Nm1 B. 0.1740 Nm1 C. 7.35 102 Nm1 D. 0.1407 Nm1 T = hrg 2 = 3 102 0.5 103 1000 9.8 2 = 7.35 102 Nm1 02. cvwbi c„ôUvb- 7210–3N/m| 0.2mm e¨v‡mi b‡j cvwbi Av‡ivnY n‡e- [MBSTU. 15-16] A. 14.694 m B. 14.69410-2m C. 10.0cm D. 7.347 cm T= hrg 2 ev, h = 2T rg = 0.14693 0.1 10 10 9.8 2 72 10 3 3 3 =14.6910-2m BSMRSTU 01. cvwb n‡Z 5 cm myuB‡K Zz‡j Avb‡Z cÖ‡qvRbxq b~¨bZg ejÑ [BSMRSTU-B, 19-20] A. 7.2 10–3N B. 3.6 10–3N C. 1.4 10–3N D. 2.2 10–3N Avgiv Rvwb, T = F 2L F = 2L T = 20.05 7.2 10–2 = 0.7210–2 N = 7.210–3 N STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. cvwbi DcwiZ‡ji ivLv 0.05m `xN© GKwU m~P‡K †U‡b Zzj‡j me©vwaK †h e‡ji cÖ‡qvRb- (c„ôUvb = 72 10-3Nm-1 ) [BUET. 08-09] A. 7.210-3N B. 3.610-3N C. 1.410-3N D. 7.210-4N L F T 2 ev, F=T2L =7210-3 20.05 = 7.210-3N KUET 01. 10cm e¨vmv‡a©i GKwU cvi` †dvUv‡K 106 mg AvqZb †duvUvq wef³ Kiv n‡jv| G‡Z wK cwigvY KvR m¤úvw`Z n‡jv? cvi‡`ic„ôUvb = 0.55Nm–1 [KUET. 18-19] A. 10–3 J B. 6.84 10–3 J C.68.39 10–3 J D. 0.684J E. 6.84J W = 4(Nr2 – R 2 )T = 6.84J; r = R 3 106 02. GKwU mvev‡bi ey`ey`‡K 1cm e¨vm n‡Z ax‡i ax‡i AvK…wZ e„w× K‡i 10cm e¨v‡m cwiYZ Kiv n‡jv| K…Z Kv‡Ri cwigvb wbY©q Ki| (mvev‡b cvwbi c„ôUvb = 2510-3 ) [KUET. 11-12] A. 1.5510-3 J B. 1.5510-4 J C. 1.5510-3 J D. 1.65510-3 J E. 1.55010-2 J KvR W= 8{(0.05)2 – (0.005)2 }T =8{(0.05)2 (0.005)2 2510-3 =1.5510-3 J
ASPECT PHYSICS cÖ_g cÎ c`v‡_©i MvVwbK ag© 287 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 03. GKwU mvev‡bi ey`ey‡`i e¨vmva© 1cm Ges mvev‡bi `ªe‡bi c„ôUvb 3.2 10-2 N/m ey`ey‡`i evB‡ii I wfZ‡ii g‡a¨ AwZwi³ Pv‡ci cwigvb wbY©q Ki| [KUET. 04-05] A.12.8N/m2 B. 6.4N/m2 C. 4.8Nm2 D. 3.2N/m2 E. 1.6N/m2 P = 4T R = 4 3.2 10–2 0.01 = 12.8N/m2 STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. †h a‡g©i d‡j Zij Zvi wewfbœ ¯Í‡ii Av‡cwÿK MwZi we‡ivwaZv K‡i Zv‡K Zi‡ji Kx e‡j? [iv. †ev. 2017] A. mv›`ªZv ej B. mv›`ªZv C. mv›`ªZv¼ D. c„ôUvb Ans B 02. cvwbi DcwiZj n‡Z 0.5m j¤^v GKwU Abyf~wgK Zvi‡K †U‡b Zzj‡Z Zv‡ii IRbmn me©vwaK 72.810 3N e‡ji cÖ‡qvRb nq| cvwbi c„ôUvb KZ? [h. †ev. 2016] A. 145.6 103Nm1 B. 72.8 103Nm1 C. 14.56 103Nm1 D. 7.28 103Nm1 S B info T = F 2L = 72.8 × 10–3 2 × 0.5 = 72.8 × 10–3 Nm–1 03. KwV‡bi NbZ¡ s , Zi‡ji NbZ¡ 1, Ges ¯úk© †KvY n‡j †KvbwU mwVK? [iv. †ev. 2016] A. S > pL, = 90 B. S < L, = 90 C. S > pL, > 90 D. S > pL, < 90 Ans D 04. L evûwewkó eM©vKvi †d«g Zi‡j wbgw¾Z K‡i †Zvjv n‡jv| Gi GK evû x `~iZ¡ miv‡Z K…ZKvR KZ? [‡hLv‡b T c„ôUvb wb‡`©k K‡i] [h.‡ev. 2019] A. W = 2LT B. W = T C. W = 2LTx D. W = LTx Ans C 05. ¯úk©‡Kvb n‡jÑ [Kz.‡ev. 2019] i. KvP I cvwbi †ÿ‡Î 090 ii. cvi` I Kv‡Pi †ÿ‡Î 0180 iii. KvP I ‡K‡ivwm‡bi †ÿ‡Î 090 wb‡Pi †KvbwU mwVKÑ A. i B. ii C. i I iii D. i, ii I iii Ans D 06. cvwbi c„ôUvb 0.06N/m n‡j Zvi c„ôkw³Ñ [wm.‡ev. 2019] A. 60 N/m B. 6 N/m C. 0.6 N/m D. 0.06 N/m S D info E = W A = A × T A E = T GKwU •KwkK b‡ji e¨vm 0.4mm| G‡K 72×103N/m c„ôUvb Ges 103 kg m3 Nb‡Z¡i cvwb‡Z Wzev‡bv n‡jv| Z‡_¨i Av‡jv‡K 07bs cÖ‡kœi DËi `vI: [wm.‡ev. 2019] 07. e¨vmva© wظb n‡j b‡ji KZ D”PZvq cvwb DV‡e? A. 0.00367 m B. 0.03673 m C. 0.3673 m D. 3.6734 m S B info h1 h2 = r2 r1 h2 = r1 r2 × h1 = r1 2r1 × 0.0734 = 0.03673 m DÏxc‡Ki Av‡jv‡K 08bs cÖ‡kœi DËi `vI: [gv. †ev. 2018] h T sin T cos wP‡Î r e¨vmv‡a©i GKwU Kuv‡Pi •KwkK bj weï× cvwb‡Z Wzev‡j Zv cvwb‡Z h D”PZvq D‡V| c‡i 2r e¨vmv‡a©i Dci GKwU •KwkK bj cvwb‡Z Wzev‡bv n‡jv| 08. †h D”PZvq wØZxq †ÿ‡Î cvwb DV‡e Zv n‡jvA. h 2 B. h C. 2h D. 3h S A info h1 h2 = r2 r1 h2 = r1 r2 × h1 = r 2r × h = h 2 Concept 6 KvR wbY©q msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. v = 2r2 ()g 9 ; F = 6rv, F = A dx dv , w = AT = 4(Nr2 R 2 )T CONCEPTUAL MATH MEx 01 mvevb cvwb Øviv 0.1m e¨vmva© wewkó GKwU ey`ey` dzjv‡Z KZ KvR Ki‡Z n‡e| mvev‡bi cvwbi c„ô Uvb 0.030Nm–1 W = 8r 2T 8 3.14 0.1 0.030 2 = 7.53910–3 j NOW START PRACTICE 01. GKwU mvev‡bi ey`ey‡`i e¨vmva© wZb¸Y Ki‡Z KZ¸Y KvR Ki‡Z n‡e? NOW PRACTICE SOLVE : 01. W= 8r 2T W r 2 W (3)2 KvR 9 ¸Y Ki‡Z n‡e| REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU 1m Zv‡ii e¨vmva© 0.5 m H Zv‡i ej cÖ‡qvM Ki‡j 0.02m e„w× cvq| wKš‘ e¨vmva© 0.05 m K‡g hvq cqm‡bi AbycvZ n‡e- [DU. 08-09] A. 0.2 B. 0.1 C. 0.01 D. 5 cqm‡bi AbycvZ, = cvk^© weK…wZ •`N©¨ weK„wZ dL Dl = 0.05 1 0.5 0.02 = 5 STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. 0.17 10-2 e¨vmva© wewkó 25wU cvwbi ¶z`ª †dvUv wg‡j GKwU eo †dvUv •Zix Kij| G‡Z wbM©Z kw³i cwigvY wbY©q Ki| [KU. 06-07] A. 43.62410-7 J B. 40.34510-7 J C. 42.34510-5 J D. None R r 3 25 r 2 2.924 49.7 10 2 0.17 10 wbM©Z kw³ E = 4AT = 4(25r2 R 2 ) 72 103 = 4 3.14 {25 (0.17 102 ) 2 – (49.7 102 ) 2 }72 10–3 = 4 3.1410–4 {25 (0.17)2 – 49.7)2 } 72 10–3 = 43.624 10–7 BRUR 01. cÖwZwU 110-4m e¨vmv‡a©i 1000000 wU Wªc wgwjZ n‡q GKwU eo Wª‡c cwiYZ n‡jv| eo Wª‡ci e¨vmva© †KvbwU? [BRU. 13-14] A. 110-2m B. 110-3m C. 110-4m D. 110-5m R = 3 3 6 4 2 10 10 1 10 n r
288 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 STEP 03 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. 2m ‣`N©¨ Ges 1mm2 cÖ¯’‡”Q‡`i †¶Îdj wewkó Zvi‡K †U‡b 0.1mm cÖmvwiZ Ki‡Z Kv‡Ri cwigvb KZ? (Y=21011N/m2 ) [KUET. 12-13] A. 510-3 J B. 510-4 J C. 210-3 J D. 10-4 J E. 2.510-4 J L YAl W 2 2 2 2 1 2 10 10 (0.1 10 ) 11 6 3 2 = 5 10–4 =510-4 J CUET 02. 310-3m e¨vmv‡a©i GKwU mvev‡bi ey`ey‡`i c„ô kw³ wbY©q Ki? (wgkª‡Yi c„ôUvb 2010-3N.) [CUET. 11-12] A. 4.2510-6 J B. 2.2610-6 J C. 1.1310-6 J D. None of these E=8r 2T= 8(310-3 ) 2 2010-3 = 4.5210-6 J STEP 04 ANALYSIS OF MEDICAL & DENTAL QUESTION DAT 01. cÖwZwU 10–4 m e¨vmwewkó cvwbi 1000 wU ÿz`ª †duvUv wg‡j GKwU e„nr †duvUv •Zwi Kij| e„nr †duvUvi e¨vmva© KZ? [DAT; 19-20] A. 5 10–4 m B. 1/10 m C. 10–2 m D. 1010–4 m e„nr †duvUvi AvqZb = 1000 wU ÿz`ª †duvUvi AvqZb 4 3 R 3 = 1000 4 3 r 3 R 3 = 1000 r 3 R = 10 r = 10 d 2 = 5 10–4 m Shortcut: R = 3 N r = 3 1000 10–4 2 = 5 10–4 STEP 05 ANALYSIS OF HSC BOARD QUESTION 01. GKwU ey`ey` n«‡`i Zj‡`k †_‡K Dcwic„‡ô D‡V Avmvq Gi AvqZb AvU¸Y nq| evqygÛ‡ji Pvc H wgUvi D”PZvi cvwb¯Í‡¤¢i Pv‡ci mgvb n‡j n«‡`i MfxiZv n‡e [h. †ev. 2016] A. H B. 3H C. 5H D. 7H S D info h = (n – 1) × H = (8 – 1) × H = 7H 02. †Kv‡bv Zi‡ji c„‡ôi †ÿÎdj GK GKK e„w× Ki‡Z K…Z KvR‡K ejv nq [wm. †ev. 2016] A. c„ôUvb B. mv›`ªZv C. c„ôkw³ D. AvqZb cxob Ans C 03. 5 cm e¨vmv‡a©i ey`ey` m„wó Ki‡Z K…Z KvR KZ? (T = 3 10–2 Nm–1 ) [e. †ev. 2016] A. 0.88 10–3 J B. 0.98 10–3 J C. 1.88 10–3 J D. 2.88 10–3 J S C info W = A × T = 4r 2 × T = 4 × 3.14 × (0.05)2 × 2 × 3 × 10–3 = 1.88 × 10–3 J SAQ Short Answer Questions WRITTEN PART BAQ Broad Answer Questions 01. GKwU Zv‡ii •`N¨© 3m, cÖ¯’‡”Q‡`i †ÿÎdj 2mm2 Ges Amn cxob 2.45 108Nm–2 | ZviwUi Amn IRb I Amn fi wbY©q Ki| Solve Avgiv Rvwb, Amn IRb = Amn cxob cÖ¯ ‡”Q‡`i †ÿÎdj = 2.45 108 2 10–6 = 4.90 102N Amn fi = Amn IRb AwfKl©xq Z¡iY = 4.90102 9.8 = 50kg GLv‡b, •`N¨© L = 3m cÖ¯ ‡”Q‡`i †ÿÎdj, A = 2mm2 = 2 10–6m 2 Amn cxob = 2.45 108Nm–2 Imn IRb = ? Amn fi = ? 02. 1mm2 cÖ¯’‡”Q`wewkó GKwU B¯úvZ Zv‡ii •`N¨© 5% e„w× Ki‡j KZ ej cÖ‡qvM Ki‡Z n‡e? [B¯úv‡Zi Y = 2 1011Nm–2 ] Solve Y = F A L l ev, F = YAl L = 2 1011110–6 0.05L L = 1 104 N (Ans.) awi, Avw` ‣`N¨© = L •`N¨© e„w×, l = 5L 100 = 0.05L †ÿÎdj, A = 1mm2 = 1 10–6m 2 Y = 2 1011 Nm–2 ej, F = ? 03. 150cm •`‡N¨©i GKwU avZe Zv‡ii GK cÖvšÍ AvU‡K †i‡L Aci cÖv‡šÍ Zvi Szjv‡j 2mm •`N¨© cÖmviY nq| Zv‡ii e¨vm 1mm Ges Zv‡ii Dcv`v‡bi cqm‡bi AbycvZ 0.24 n‡j cÖmvwiZ Ae¯’vq ZviwUi e¨v‡mi cwieZ©b wbY©q Ki| Solve Avgiv Rvwb, = d/d l/L Zv‡ii e¨v‡mi cwieZ©b, d = ld 1 = 0.240.20.1 150 = 3.2 10–5 cm (Ans.) GLv‡b, Zv‡ii •`N¨© l = 150cm •`N¨© e„w×, l = 2mm = 0.2cm Zv‡ii e¨vm, d = 1mm = 0.1cm = 0.24 d = ? 04. 1 m 2 ‡ÿÎd‡ji GKwU P¨vÞv †cøU Aci GKwU eo †cøU n‡Z 0.1cm cyiæ wMømvwib ¯Íi Øviv c„_K Kiv Av‡Q| H †cøU‡K 1 10–2ms–1 †e‡M Pvjbv Ki‡Z 1.510–5N e‡ji cÖ‡qvRb n‡j wMømvwi‡bi mv›`ªZv¼ wbY©q Ki| Solve Avgiv Rvwb, F = A dv dy ev, = Fdy Adv = 1.510–5 110–3 110–3 110–2 = 1.5 10–3Ns m–2 (Ans.) GLv‡b, A = 1 10–3m 2 F = 1.5 10–5N dv = 1 10–2ms–1 dy = 0.1cm = 110–3m 05. 3m `xN© I 3 mm e¨vmv‡a©i GKwU Zvi‡K Uvb‡j Zv‡ii •`N©¨ e„w× nq 0.2 mm| ZviwUi e¨vm KZUzKz Kg‡e? (cqmb AbycvZ = 0.2) [BUTex. 2020-21] Solve Avgiv Rvwb, cqm‡bi AbycvZ = •`N©¨ weK…wZ cvk¦© weK…wZ = l L d D = lD Ld 0.2 = 0.2 10–3 610–3 3d d = 2 10–6m GLv‡b, = 0.2 Avw` •`N©¨, L = 3m •`‡N©¨i cwieZ©b l = 0.2 10–3m Avw` e¨vmva© = 3mm = 3 10–3m Avw` e¨vm, D = 2310–3m = 610–3m e¨v‡mi cwieZ©b ev e¨vm n«vm cv‡e, d = ? 06. GKwU mvev‡bi ey`ey‡`i e¨vmva© 1 †m.wg. n‡Z evwo‡q 10 †m.wg. Ki‡Z KZ KvR m¤úbœ Ki‡Z n‡e? [mvevb cvwbi c„ôUvb = 2.6 10–2 wbDUb/wgUvi] Solve c„‡ôi †ÿÎdj e„w×, A = 2 (4R 2 – 4r 2 ) [∵ mvev‡bi 2wU c„ô _v‡K] = 2 4 (0.12 – 0.012 ) = 0.2488m2 K…ZKvR W = AT = 0.2488 2.6 10 –2 = 6.469 10–3 J