Tiêu đề DU Standerd Model Test-03 (Set-A) - Solution.pdf ✅

1 DU Standard Model Test-03 [Set-A (Solution)] c~Y©gvb: 100 †b‡MwUf gvK©: 0.25 mgq: 1 NÈv 30 wgwbU MCQ c`v_©weÁvb (Physics) 1. NÈvq 40 km †e‡M DËi w`‡K Pjgvb GKwU Mvwoi PvjK NÈvq 30 km †e‡M GKwU UavK‡K cwðg w`‡K Pj‡Z †`Lj| UavKwUi cÖK...Z †eM KZ? [A car is moving north at a speed of 40 km/h. The driver sees a truck moving west at a speed of 30 km/h. What is the actual velocity of the truck?] 70 kmh–1 75 kmh–1 50 kmh–1 78 kmh–1 DËi: 50 kmh–1 e ̈vL ̈v: N W E S vT  vc  vT  v  TC vT = v 2 TC + v2 C = 402 + 302 = 50 kmh–1 GLv‡b, Mvwoi cÖK...Z †eM, vC = 40 kmh–1 Mvwoi mv‡c‡ÿ Uav‡Ki †eM, vTC = 30 kmh–1 Uav‡Ki cÖK...Z †eM, vT = ? 2. 5 kg I 3 kg f‡ii `ywU e ̄‘ GKB mij‡iLv eivei wKš‘ wecixZ w`‡K Pjv Ae ̄’vq G‡K Aci‡K av°v w`j| av°vi c~‡e© Df‡qi †eM 3 ms–1 wQj| av°vi ci wØZxq e ̄‘ 2 ms–1 †e‡M wcwQ‡q †M‡j, 1g e ̄‘i †eM KZ? [Two objects, one with mass of 5 kg and 3 kg respectively, are moving in the same straight line but in opposite directions. They collide with each other. Before the collision,both of them has same velocity of 3 m/s, After the collision, the second object moves with a velocity of 2 m/s in the opposite direction. What is the velocity of the first object after the collision?] – 3 ms–1 + 2 3 ms–1 – 2 ms–1 – 1 3 ms–1 DËi: + 2 3 ms–1 e ̈vL ̈v: 5kg (– ve) 3kg (+ ve) 3ms–1 v 2ms–1 m1u1 + m2u2 = m1v1 + m2v2  5  3 + 3  (–3) = 3  v1 + 2  (2)  15 – 9 = 6v1 + 4  6 – 4 = 3v1  v1 = – 2 3 ms–1 3. B A h 60 50m e‡ei fi 5 kg n‡j, wefekw3 KZ? (B we›`y‡Z) [If the mass of the bob is 5 kg, What is the potential energy at the point B?] 1225 J 2450 J 250 3 J 250 3 2 DËi: 1225 J e ̈vL ̈v: Ep = mgl (1 – cos) = 5  9.8  50     1 – 1 2 = 5  9.8  50  1 2 = 1225 J 4. 500 gm Gi GKwU eid LÐ h D”PZv †_‡K GKwU Rjvk‡q coj Ges 50% M‡j †Mj| h = ? [A 500 gm piece of ice falls from a height of h into water and melts 50%. h = ?] 5  105 3.36  9.8 m 9.8  3.36 2  105 3.36  105 2  9.8 m 3.36  2 98  105 DËi: 3.36  105 2  9.8 m e ̈vL ̈v: h D”PZvq wefekw3 = mgh Rjvk‡q covi ci fi M‡j n‡jv, m1 = 50m 100 = m 2  mgh = m 2 lf  h = lf 2g = 3.36  105 2  9.8
2 5. f‚-c„ô n‡Z KZ Mfx‡i AwfKl©R Z¡iY f‚-c„‡ôi AwfKl©R Z¡i‡Yi gv‡bi `yB-Z...Zxqvsk n‡e? [At what depth below the Earth's surface will the gravitational acceleration be two-third the value of the gravitational acceleration at the surface?] 4266.67 km 2133.33 km 426.667 km 213.33 km DËi: 2133.33 km e ̈vL ̈v: g = g     1 – h R  g g = 1 – h R  2 3 = 1 – h R  h = 2133.33 km 6. GKwU c`v‡_©i AvqZb ̧Yv1⁄4 3  1011 Pa n‡j, msbg ̈Zv KZ? [If the bulk modulus of a substance is 3 × 1011 Pa, what is its compressibility?] 33  10–12 3  1012 0.33  1010 3.3  10–12 DËi: 3.3  10–12 e ̈vL ̈v: msbg ̈Zv = 1 3  1011 = 0.33  10–11 = 3.3  10–12 7. mij Qw›`Z MwZm¤úbœ †Kv‡bv KYvi e ̈eKjbxq mgxKiY : 5 d 2 x dt2 + 80x = 0 n‡j, Gi †KŠwYK K¤úv1⁄4 KZ n‡e? [Differential equation of a particle with simple harmonic motion 5 d 2 x dt2 + 80x = 0. What is its angular frequency?] 4 rads1 5 rads–1 2 rads1 25 rads1 DËi: 4 rads1 e ̈vL ̈v: 5 d 2 x dt2 + 80x = 0  d 2 x dt2 + 16x = 0   2 = 16   = 4 rads–1 8. 900nm Zi1⁄2‰`‡N© ̈i `ywU Av‡jvK Zi‡1⁄2i g‡a ̈ c_ cv_©K ̈ 3  10–9 m n‡j Zv‡`i g‡a ̈ `kv cv_©K ̈ n‡eÑ [If the path difference between two light waves of wavelength 900nm is 3  10–9 m, the phase difference between them will be-] π 2 2π 3π 2 π 150 DËi: π 150 e ̈vL ̈v: λ = 900 nm = 900  10–9 m c_ cv_©K ̈ = 3  10–9 m `kv cv_©K ̈ = 2π λ  c_ cv_©K ̈ = 2π 900  10–9  3  10–9 = π 150 9. †Kv‡bv n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvi mgq evZv‡mi ey`ey` AvqZ‡b wZb ̧Y nq| evqygЇji Pvc 105 Nm–2 n‡j, n«‡`i MfxiZv KZ? [Air bubbles tripple in volume as they rise from the bottom of a lake to the water surface. If the atmospheric pressure is 105 Nm–2 , what is the depth of the lake?] 10.2 cm 30.6 m 20.4 m 10.2 m DËi: 10.2 m e ̈vL ̈v: Shortcut: h = (n – 1)  P   g = (3 –1)  105 1000  9.8 n = AvqZb hZ ̧Y n‡q‡Q = 3 P = 105 Nm–2  = 1000 kgm–3 g = 9.8 ms–2 = 2  102 1000  9.8 = 2  102 9.8 = 200 9.8  h = 20.4 m 10. GKwU ÎæwUc~Y© _v‡g©vwgUv‡i eidwe›`y 7C Ges w÷g we›`y 117C, _v‡g©vwgUv‡i Gi cvV 73C n‡j, e ̄‘i ZvcgvÎv KZ? [If a faulty thermometer has a freezing point of 7°C and a steam point of 117°C, the thermometer reads 73°C, what is the temperature of the substance?] 60C 50C 70C 40C DËi: 60C e ̈vL ̈v: 73 – 7 117 – 7 = C – 0 100 – 0  C = 66  100 110  C = 66 11  10 = 6  10 = 60C 11. wZbwU avi‡Ki aviKZ¡ h_vμ‡g 4 F, 5 F Ges 2 F G‡`i cÖ_g I Z...ZxqwU‡K †kÖwY‡Z mshy3 K‡i, wØZxqwUi mv‡_ mgvšÍiv‡j ivLv n‡j, Zzj ̈ aviKZ¡ KZ? [Three capacitors of capacitance 4 μF, 5 μF and 2 μF respectively, first and third connected in series, placed in parallel with the second, what is the equivalent capacitance?] 7 F 10.83 F 9 F 6.33 F DËi: 6.33 F e ̈vL ̈v: C13 = 4   = 4 3 F
3 †gvU Zzj ̈ aviKZ¡, C123 =     4 3 + 5 F = 4 +15 3 = 19 3 = 6.33 F 12. V 20A cÖwZwU †iv‡ai gvb 5| g~j Zwor cÖevn 20A n‡j, wefe cv_©K ̈ KZ? [Each resistor has a value of 5. If the original current is 20A, what is the potential difference?] 1V 40V 20V 25.5V DËi: 20V 5 5 5 5 5 5 I = 32A V e ̈vL ̈v: R =     1 5 + 1 5 + 1 5 + 1 5 + 1 5 –1 =     5 5  Rp = 1 V = 20  1 = 20V 13. GKwU B‡jKUab I †dvU‡bi wW-eaMwj Zi1⁄2‰`N© ̈ GKB| hw` Ee I Ep Øviv h_vμ‡g B‡jKUab I †dvU‡bi kw3‡K cÖKvk Kiv nq, Z‡e Ee Ep = ? [v I c h_vμ‡g B‡jKUab I †dvU‡bi †eM] [An electron and a photon have the same de-Broglie wavelength. If Ee and Ep denote the energy of electron and photon respectively, then Ee Ep =? [v and c are velocities of electron and photon respectively]] v 2c v 4c 2c v v c DËi: v 2c e ̈vL ̈v:  = h 2mEe [B‡jKUa‡bi †ÿ‡Î]  Ee = h 2 2m 2 Avevi,  = hc Ep [†dvU‡bi †ÿ‡Î]  Ep = hc  GLb, Ee Ep = h 2 2m 2   hc = h 2mc = h 2m  h mv  c = v 2c 14. 1wU B‡jKUab hw` E2 D”P kw3 ̄Íi †_‡K E1 wb¤œkw3 ͇̄i Mgb K‡i Z‡e Zi1⁄2 ˆ`N© ̈ n‡e? [If an electron moves from E2: higher energy level to E1 lower energy level, what will be the wave number?] hc E2 – E1 E2 – E1 hc E2 – E1 hc c h(E2 – E1) DËi: hc E2 – E1 e ̈vL ̈v: E = hc   1  = E hc   = hc E2 – E1 15. †n•v‡Wwmgvj 7FF Gi cieZ©x msL ̈v Kx? [What is the next number of 7FF in hexadecimal?] 800 7FE 7 GF 701 DËi: 800 e ̈vL ̈v: (7FF)16 = (2047)10 2047 Gi cieZ©x msL ̈v 2047 + 1 = 2048 (2048)10 = (800)16 imvqb (Chemistry) 1. H cigvYyi B‡jKUab n kw3 ̄Íi n‡Z f‚wg ͇̄i wd‡i Avm‡j 6wU eY©vwj †iLv cvIqv †M‡j, n = ? [6 spectral lines are obtained when the electron of H atom returns from the n energy level to the ground energy level, n=?] 6 3 4 2 DËi: 4 e ̈vL ̈v: eY©vwj †iLvi msL ̈v = (n2 – n1) (n2 – n1 + 1) 2  6 = (n – 1) (n – 1 + 1) 2  12 = n2 – n  n 2 – n – 12 = 0  (n – 4) (n + 3) = 0  n – 4 = 0 n = 4 2. H-Gi 4_© Kÿc_ †_‡K GKwU B‡jKUab weZvwoZ Ki‡Z cÖ‡qvRbxq kw3 Ges H Gi AvqwbKiY wef‡ei Zzjbv KZ? [What is the comparison between the energy required to eject an electron from the 4th orbit of H and the ionization potential of H?] 1 : 4 4 : 1
4 1 : 16 16 : 1 DËi: 1 : 16 e ̈vL ̈v: H Gi 5g Kÿc_ n‡Z B‡jKUab weZvwoZ Ki‡Z cÖ‡qvRbxq kw3, E = E – E5 =       0 –     – 13.6  1 16 eV = 0.85 eV  H Gi AvqbxKiY kw3 = 13.6 eV  Zzjbv = 0.85 13.6 = 1 16  1 : 16 3. AXn †K›`axq †gŠj A Gi msKivqb sp3 d Ges A †Z GKwU gy3‡Rvo we` ̈gvb Zvn‡j n = ? [Hybridization of center elements A of AXn is SP3 d and a lone pair exists in A then n = ?] 2 4 5 6 DËi: 4 e ̈vL ̈v: †K›`axq †gŠj A Gi msKivqb sp3 d  nvBweaW AiweUvj = 5 wU gy3‡Rvo = 1 wU  eÜb‡Rvo = 5 – 1 = 4  hy3 GK‡hvRx (X) cigvYyi msL ̈v n = 4 4. SF4 AYyi †ÿ‡Î- [For SF4 molecule-] (i) AvK...wZ K-shape/see saw shape (ii) T-Shape/weK...Z PZz ̄ÍjKxq (iii) msKivqY sp3 d (i) I (ii) (i) I (iii) (ii) I (iii) (i), (ii), (iii) DËi: (i) I (iii) e ̈vL ̈v: SF4 -G x = 1 2 [6 + 4 – 0 + 0] = 5 = sp3 d GLv‡b, lp = 1 S F F < 90 < 120 F F AvK...wZ: weK...Z PZz ̄ÍjKxq/K-Shape/see saw shape 5. gvby‡li i‡3i P H = 7.4 n‡j, [H+ ] = ? [If the pH of human blood is 7.4, then [H+ ] =?] 1.4  10–8 mol/L 4  10–7m ol/L 1  10–7 mol/L 3.98  10–8 mol/L DËi: 3.98  10–8 mol/L e ̈vL ̈v: pH = – log[H+ ]  [H+ ] = 10–pH = 10–7.4 = 3.98  10–8 mol/L 6. HNO3 GwmW, H3PO4 Gwm‡Wi Zzjbvq kw3kvjx KviYÑ [HNO3 acid is stronger than H3PO4 acid, because-] N Gi RviY msL ̈v, P Gi Zzjbvq †ekx HNO3 Gwm‡Wi Zzjbvq H3PO4 Gwm‡W H Gi msL ̈v †ekx N cigvYy‡Z PvR© NbZ¡ †ekx †KvbwUB bq DËi: N cigvYy‡Z PvR© NbZ¡ †ekx 7. wbw`©ó Pv‡c †Kv‡bv M ̈vm AYyi RMS †eMÑ [The RMS velocity of a gas molecule at a certain pressure is-] AvbweK f‡ii e ̈ ̄ÍvbycvwZK AvbweK f‡ii mgvYycvwZK Nb‡Z¡i eM©g~‡ji e ̈ ̄ÍvbycvwZK f‡ii eM©g~‡ji mgvYycvwZK DËi: Nb‡Z¡i eM©g~‡ji e ̈ ̄ÍvbycvwZK e ̈vL ̈v: Crms = 3RT M = 3P  ;   NbZ¡ I P  Pvc  p  aaæeK n‡j, Crms  1  8. wMÖMbvW© weKviK w`‡q GKwU cÖvBgvwi A ̈vj‡Kvnj ˆZwii Rb ̈ †h e ̄‘wU e ̈envi Kiv nq Zv njÑ [The substance used to prepare a primary alcohol with Grignard reagent is-] CH3MgBr HCHO RCHO DËi: HCHO e ̈vL ̈v: HCHO + RMgX ﮋ B_vi  H – OMgX | C – H | R R – CH2 – OH + Mg(OH)X cÖvBgvwi A ̈vj‡Kvnj 9. wb‡Pi †KvbwU †_‡K A ̈vgvBW cÖ ̄‘Z Kiv hvq? [Amide can be prepared from which of the following?] ˆRe Gwm‡Wi Amonium jeY †_‡K Kve©w•wjK GwmW I A ̈vwgb †_‡K ÔKÕ I ÔLÕ DfqB †Kv‡bvwUB bq DËi: ÔKÕ I ÔLÕ DfqB e ̈vL ̈v: ˆRe GwmW A ̈v‡gvwbqvg jeY †_‡K- R – CO – ONH4   230 R – CO – NH2 + H2O Kve©w·wjK GwmW I A ̈vwgb †_‡K, CH3 – O || C – OH + H – NH – CH3 CH3 – O || C – NH – CH3 + H2O 10. k~b ̈ ̄’v‡b Kx e ̈eüZ n‡e? [What should be used in the blank?] R C = O R1 H2O/H+