Tiêu đề Motion of Particles Practice Sheet Solution HSC FRB-25.pdf ✅

mgZ‡j e ̄‘KYvi MwZ  Final Revision Batch '25 1 Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 1 2 1 1 1 1 1 1 2 2022 1 1 1 1 1 1 1 1 1 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 5 3 3 4 4 4 3 4 4 2022 4 4 4 4 5 4 3 3 4 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcK-1: [XvKv †evW©- Õ23] C B A f1 f2 u = 0 DÏxcK-2: B a C 45 D b A (K) † ̄avZ bv _vK‡j GK e ̈w3 240 wgUvi cÖk ̄Í GKwU b`x muvZvi w`‡q 6 wgwb‡U †mvRvmywRfv‡e cvi nq| wKš‘ † ̄avZ _vK‡j H GKB c_ 10 wgwb‡U cvi n‡Z cv‡i| muvZviæi MwZ‡eM wbY©q Ki| (L) DÏxcK-1 G e ̄‘wUi mgZ¡iY f1, mgg›`b f2 Ges t mgq c‡i A n‡Z S `~i‡Z¡ C we›`y‡Z †_‡g hvq, cÖgvY Ki †h, S = t 2 f1f2 2(f1 + f2) | (M) DÏxcK-2 G, A we›`y n‡Z cÖwÿß e ̄‘wUi C we›`y‡Z †e‡Mi w`K Avbyf~wg‡Ki mgvšÍivj n‡j, cÖgvY Ki †h, CD †`Iqv‡ji D”PZv ab a + b| mgvavb: (K) GLv‡b, b`xi cÖ ̄’, d = 240 wgUvi † ̄avZ bv _vK‡j b`x †mvRvmywRfv‡e cvi n‡Z mgq jv‡M, t = 6 wgwbU  muvZviæi MwZ‡eM, v = d t = 240 6 = 40 wgUvi/wgwbU (Ans.) (L) awi, KYvwU A we›`y n‡Z f1 mylg Z¡i‡Y t1 mgq ci B we›`y‡Z †cuŠ‡Q| Gici B we›`y n‡Z C we›`y‡Z f2 mylg g›`‡b t2 mgq ci _v‡g| u = 0 A f1 t1 v = v B f2 t2 C v = 0 †gvU mgq, t = t1 + t2 myZivs †gvU AwZμvšÍ `~iZ¡,S = AB + BC mylg Z¡i‡Yi †ÿ‡Î, v = u + f1t1  v = 0 + f1t1  t1 = v f1 mylg g›`‡bi †ÿ‡Î, 0 = v – f2t2  t2 = v f2 myZivs, t1 + t2 = t = v f1 + v f2  t = v    1 f1 + 1 f2 ......(i) Avevi, AB = 0 + v 2 .t1  AB = vt1 2 Ges BC = v + 0 2 .t2 = vt2 2  AB + BC = S = v 2 (t1 + t2)  S = v 2 t  v = 2S t
2  Higher Math 2nd Paper Chapter-9 mgxKiY (i) G, v = 2S t ewm‡q cvB, t = 2S t     1 f1 + 1 f2  t = 2S t     f1 + f2 f1f2  S = t 2 f1f2 2(f1 + f2) (Proved) (M) GLv‡b, wb‡ÿcY †KvY,  = 45 Avgiv Rvwb, evqyk~b ̈ ̄’v‡b cÖwÿß KYvi MwZc‡_i mgxKiY, y = x tan     1 – x R CD †`Iqv‡ji D”PZv h n‡j, x = a Ges y = h Avevi, DÏxc‡Ki cÖvmwUi Avbyf~wgK cvjøv, R = a + b  h = atan     1 – a R = a tan45     1 – a a + b = a    a + b – a a + b  h = ab a + b  CD †`Iqv‡ji D”PZv, h = ab a + b (Proved) 2| `„k ̈Kí-1: w ̄’ive ̄’v n‡Z mij‡iLvq Pjgvb GKwU e ̄‘KYv cÖ_‡g y mgZ¡i‡Y Ges c‡i z mgg›`‡b P‡j| `„k ̈Kí-2: GKwU ͇̄¤¢i kxl© †_‡K 98 wg./†m‡KÛ †e‡M A e ̄‘‡K Lvov Dc‡ii w`‡K wb‡ÿc Kiv n‡jv| 2 †m‡KÛ c‡i GKB we›`y n‡Z Aci GKwU B e ̄‘‡K †Q‡o †`Iqv n‡jv| [ivRkvnx †evW©- Õ23] (K) 64 wgUvi DuPz `vjv‡bi Qv` †_‡K GKwU cv_i †Q‡o w`‡j f~wg‡Z co‡Z KZ mgq jvM‡e? (L) `„k ̈Kí-1 G KYvwU hw` t mg‡q d `~iZ¡ AwZμg K‡i Z‡e †`LvI †h, t = 2d     y + z yz (M) `„k ̈Kí-2 G e ̄‘ `ywU f~wg n‡Z KZ D”PZvq wgwjZ n‡e Zv wbY©q Ki| mgvavb: (K) GLv‡b, `vjv‡bi D”PZv, h = 64 m AwfKl©R Z¡iY, g = 9.8 ms–2 Avw`‡eM, u = 0 ms–1 GLb mgq t n‡j, Avgiv Rvwb, h = ut + 1 2 gt2  64 = 0  t + 1 2  9.8  t 2  t 2 = 64  2 9.8  t = 13.06  t = 3.61 †m‡KÛ (Ans.) (L) ̄’vb `yBwUi ga ̈eZx© †gvU `~iZ¡ = d awi, 1g As‡ki Rb ̈ `~iZ¡ = a, mgq = t1, mgZ¡iY = y Ges 2q As‡ki Rb ̈ `~iZ¡ = d – a, mgq = t2, mgg›`b = z mgq, t = t1 + t2 mgZ¡i‡Yi †ÿ‡Î: v = yt1  v y = t1 .......(i) Ges a = 1 2 yt 2 1 = 1 2 yt1.t1 = 1 2 vt1 .......(ii) mgg›`‡bi †ÿ‡Î: 0 = v – zt2  v = zt2  v z = t2 .......(iii) Ges d – a = vt2 – 1 2 zt 2 2 = vt2 – 1 2 vt2 = 1 2 vt2 ......(iv) (i) + (iii) K‡i cvB, t1 + t2 = v y + v z  t = v    1 y + 1 z ......(v) (ii) + (iv) K‡i cvB, d = 1 2 vt1 + 1 2 v2t2 = 1 2 v(t1 + t2)  d = 1 2 vt [⸪ t = t1 + t2]  v = 2d t ......(vi) mgxKiY (v) G v Gi gvb ewm‡q cvB, t = 2d t     y + z yz  t 2 2d = y + z yz  t 2 = 2d(y + z) yz  t = 2d     y + z yz (Showed) (M) B uA = 98ms–1 h A g‡b Kwi, B e ̄‘wU †d‡j †`Iqvi t †m‡KÛ c‡i ͇̄¤¢i kxl© n‡Z h wgUvi wb‡P A e ̄‘wU B e ̄‘i mv‡_ wgwjZ nq| A e ̄‘i †ÿ‡Î, h = – ut + 1 2 gt2  h = – 98(t + 2) + 1 2  9.8  (t + 2)2 .....(i)