Title Varsity Weekly-1 (Set-B) Solution.pdf ✅

1 Varsity Weekly-01 [B (Solution)] wm‡jevm: †f±i + ivmvqwbK cwieZ©b-1 + g ̈vwUa· I wbY©vqK + cÖvYxi wewfbœZv I †kÖwYweb ̈vm + evsjv (AcwiwPZv, D”PviY, cvwifvwlK kã) + Bs‡iwR (Noun, Pronoun, determiners, Poem Summarizing) c~Y©gvb: 100 †b‡MwUf gvK©: 0.25 mgq: 1 NÈv 30 wgwbU MCQ c`v_©weÁvb (Physics) 1. |M|  = A + B Ges |N|  = A – B; M  I N  G‡`i ga ̈eZ©x †KvY 90 n‡j, M I N Gi jwäi gvb KZ? [|A| ]  = 3N; |B|  = 2N [|M|  = A + B and |N|  = A – B; M  and N  If the angle between M  and N  is 90, what is the value of resultant of the vectors M and N? [|A| ]  = 3N; |B|  = 2N 13 N 26 N 26 N 39 N DËi: 26 N e ̈vL ̈v: R = (A + B)2 + (A – B)2 + 2 (A + B) (A – B) cos90 = A2 + 2AB + B2 + A2 – 2AB + B2 = 2(A2 + B2 ) = 2 (9 + 4) = 26 N 2. `ywU e‡ji e„nËg jwä 10 N Ges ÿz`aZg jwä 4 N n‡j, ej `ywUi gvb KZ? [If the maximum resultant of two forces is 10 N and the minimum resultant is 4 N, what are the magnitudes of the two forces?] 6N, 4N 5N, 5N 7N, 3N 8N, 2N DËi: 7N, 3N e ̈vL ̈v: Rmax = P + Q  P + Q = 10 ..... (i)  Rmin = P – Q  P – Q = 4 .... (ii) (i) + (ii) = 2P = 14 N  P = 7N Avevi, (i) – (ii)  2Q = 6  Q = 3N 3. r  Q  P  wb‡Pi †KvbwU mwVK? [Which of the following is correct?] P  – Q  = r  P  + r  = Q  r  – Q  = – P  r  – Q  –P  = 0 DËi: r  – Q  –P  = 0 e ̈vL ̈v: P  + Q  = r   r  – Q  –P  = 0 4. A  = 2i  + 3j  + ak  †f±iwU z A‡ÿi mv‡_ sec–1 29 4 †KvY Drcbœ Ki‡j, a = ? [If the vector A  = 2i  + 3j  + ak  makes an angle sec–1 29 4 with the z axis, then a=?] 4 1 2 5 DËi: 4 e ̈vL ̈v:  = sec–1 29 4  = cos–1 4 29 Avgiv Rvwb, z A‡ÿi mv‡_ Drcbœ †KvY  = cos–1    H A‡ÿi mn‡Mi gvb †f±‡ii gvb   = cos–1 a 2 2 + 32 + a2 †hLv‡b, 4 n‡jv k  ev Z A‡ÿi mnM|  a = 4 5. A   B  = C  n‡j, wb‡Pi †KvbwU mwVK? [If A   B  = C  , which of the following is correct?] C  = – A   B  C  = – B   A  C  =(–A)   B  None of these DËi: C  = – B   A  e ̈vL ̈v: A   B  =   ABsin B   A  = –  ABsin C  = A   B  = – (–   ABsin) = – B   A  6. p  , q  , r  hw` GKK †f±i nq Ges †f±iÎq †Kvb wÎfz‡Ri evû eivei GKB μ‡g wμqvkxj n‡j, 9 2 (p )  . q  + q  .r  + p  .r  = ? [If p  , q  , r  are unit vectors and the vectors act in the same order along the sides of a triangle, 9 2 (p )  . q  + q  .r  + p  .r  =?] –27 4 – 3 2 4 9 – 2 3
2 DËi: –27 4 e ̈vL ̈v: p  + q  + r  = 0  (p )  + q  + r  2 = 0  p 2 + q2 + r2 + 2(p )  . q  + q  .r  + r  .p  = 0  2 (p )  . q  + q  .r  + r  .p  = – (p ) 2 + q2 + r2 [|P| ]  = 1; |q|  = 1; |r|  = 1  p  .q  + q  .r  + p  .r  = – 3 2 myZivs, 9 2 (p )  . q  + q  .r  + p  . r  = – 3 2  9 2 = – 27 4 7. m I p Gi gvb –8 3 I 3 2 n‡j, wb‡Pi †Kvb 2wU †f±i ci ̄úi mgvšÍivj n‡eÑ [If the values of m and p are –8 3 and 3 2 , which of the following 2 vectors will be parallel to each other?] (5i )  + 3j  – pk  I (2i )  + mj  – k  (5i )  – 3j  – pk  I (2i )  + mj  – k  (3i )  – 4j  + pk  I (2i )  + mj  + k  (i )  + j  + pk  I (2i )  + 6j  + 9k  DËi: (3i )  – 4j  + pk  I (2i )  + mj  + k  e ̈vL ̈v: (3i )  – 4j  + pk  I (2i )  + mj  + k  3 2 = –4 m = p 1 3 2 = P 3 2 = – 4 m m = – 4  2 3 = –8 3 8. S T P R 45 45 Q 2N 2N PRTS GKwU K ̈vbfvm‡K `yBwU myZvi mvnv‡h ̈ Q †c‡i‡K Szjv‡bv n‡jv| P I R we›`y‡Z myZv mgvb e‡ji wμqv Abyfe K‡i| Qwei IRb KZ? [PRTS canvas is hung on a Q nail with two threads. The yarn experiences equal force action at points P and R. How much does the photo weigh?] 2 2 N 1 2 N 4 2 N 3 2 N DËi: 4 2 N e ̈vL ̈v: S T P R 2N 2N F1 = 2sin 45 W F2 = 2sin 45 45 45 2cos 45 2cos 45 F1 + F2 = W  2sin45 + 2 sin45 = W  2.2sin45 = W  W = 4 2 N 9. cos2i  + sinj  + sink  †f±iwU GKwU- [The vector cos2i  + sinj  + sink  is a-] mgZjxq †f±i GKK †f±i k~b ̈ †f±i None of these DËi: GKK †f±i e ̈vL ̈v: ( cos2) 2 + sin2  + sin2  = cos2 + 2sin2  = 1 – 2sin2  + 2sin2  = 1 10. A  .B  = 0 n‡j, †KvbwU mZ ̈? [If A  .B  = 0, which one is true?] |A |  + B  = |A |  – B  A  + B  = 0 A  = 4i  + 9j  + 2k  , B  = 4i  + 2j  + 3k  †KvbwUB bq DËi: |A |  + B  = |A |  – B  e ̈vL ̈v: |A |  + B  = |A |  – B   A 2 + B2 + 2ABcos = A2 + B2 – 2ABcos  4ABcos = 0   = cos–1 (0)  = 90  A  . B  = ABcos90 11. †f±i wefvR‡bi D`vniY †KvbwU? [Which is an example of vector division?] i‡KU DÇqb †bŠKvi MwZ PjšÍ Mvwo‡Z cošÍ e„wó †bŠKvi ̧YUvbv DËi: †bŠKvi ̧YUvbv e ̈vL ̈v: GKwU †f±i‡K Zvi Dcvs‡k wef3 KivB n‡”Q †f±i wefvRb| †bŠKvi ̧YUvbv : †bŠKvi ̧YUvbvi †ÿ‡Î †f±iwU‡K `ywU Dcvs‡k wef3 Kiv †h‡Z cv‡i| GKwU Dcvsk †bŠKvi MwZi w`K wb‡`©k K‡i Ges Ab ̈ Dcvsk †bŠKvi w`K wb‡`©kbv wb‡`©k K‡i|
3 Fy F  Fx = F cos †bŠKvi ̧YUvbv †f±i wefvR‡bi me‡P‡q fv‡jv D`vniY| †f±i †hvRb : cvwLi DÇqb, PjšÍ Mvwo‡Z cošÍ e„wó| 12. GKwU †bŠKv 10 km/h †e‡M † ̄av‡Zi mv‡_ KZ wWwMÖ †Kv‡Y Pvjbv Kiv n‡j †bŠKvwU †mvRvmywR Acicv‡o †cuŠQv‡e? [b`x‡Z † ̄av‡Zi †eM 5 km/h.] [If a boat is driven at 10 km/h with the current, at what degree angle will the boat reach the other side straight? [The velocity of the current in the river is 5 km/h.]] 60 90 135 120 DËi: 120 e ̈vL ̈v: †mvRvmywR Acicv‡o/ cvo eivei/ me©wb¤œ `~i‡Z¡ ej‡j Angle 90 n‡e| cvivcv‡ii †ÿ‡Î jwä † ̄av‡Zi mv‡_ 90 †KvY K‡i _vK‡e| u = 5 km/h   v = 10 km/h jwä tan = vsin u + vcos  tan90 = 10sin 5 + 10cos u = 5 km/h v = 10 km/h  = ?  = 90  1 0 = 10sin 5 + 10cos  5 + 10 cos = 0  5(1 + 2 cos) = 0  1 + 2 cos = 0  cos = – 1 2  cos = cos 120   = 120 Shortcut : hw` †bŠKvi †eM † ̄av‡Zi †e‡Mi wØ ̧Y nq Z‡e †mvRvmywR Acicv‡o †cuŠQv‡bvi Rb ̈ † ̄av‡Zi mv‡_ 120 †Kv‡Y †bŠKv Pvjbv Ki‡Z n‡e| 13. GKwU jb †ivjvi‡K Uvbv I †Vjvi Rb ̈ Avbyf‚wg‡Ki mv‡_ 60 †Kv‡Y 10N ej cÖ‡qvM Kiv n‡j, †Vjv I Uvbvi gv‡S AvcvZ IR‡bi cv_©K ̈ KZ n‡e? [If a force of 10N is applied at an angle of 60° to the horizontal to pull and push a lawn roller, what is the apparent weight difference between the push and the pull?] 10 3 N 10 3 N 20 N 30 N DËi: 10 3 N e ̈vL ̈v: jb †ivjvi Uvbvi †ÿ‡Î,  Fsin R Fcos F mg Abyf‚Z IRb, R = (mg – Fsin) jb †ivjvi †Vjvi †ÿ‡Î Abyf‚Z IRb, R = mg + Fsin  Fsin R Fcos F mg  †Vjv I Uvbvi †ÿ‡Î IR‡bi cv_©K ̈ = mg + Fsin – mg + Fsin = 2 Fsin = 2  10  sin60 = 10 3 N 14. GKwU KYvi Dci wμqviZ `ywU e‡ji jwä GKwU e‡ji Dci j¤^ Ges Gi gvb AciwUi A‡a©‡Ki mgvb| ej؇qi ga ̈eZ©x †KvY KZ? [The resultant of two forces acting on a particle is perpendicular to one force and equal to half the value of the other. What is the angle between the forces?] 120 135 30 150 DËi: 150 e ̈vL ̈v: Q R P  g‡b Kwi, ci ̄úi  †Kv‡Y wμqviZ P I Q e‡ji jwä R, hv P Gi Dci j¤^| Zvn‡j R = Q 2 R e‡ji w`‡Ki j¤^vsk wb‡q cvB : Rcos0 = Pcos90 + Qcos( – 90)  Q 2 = Qcos( – 90)  1 2 = cos( – 90)  60 =  – 90   = 150 wKš‘ jwä R, P Gi Dci j¤^ n‡j ga ̈eZ©x †KvY 30 n‡e bv|  ej؇qi ga ̈eZ©x †KvY 150| 15. a  + b  + c  = 0; |a|  = 5, |b|  = 3, |c|  = 7 n‡j, a  I b  Gi ga ̈eZ©x †KvY KZ? [If a  + b  + c  = 0; |a|  = 5, |b|  = 3, |c|  = 7, then and what is the angle between a  and b  ?]   3 0  2 DËi:  3 e ̈vL ̈v: a  + b  = – c  GLb, a 2 + b2 + 2|a|  |b|  cos = |c|   5 2 + 32 + 2.5 3 cos = 72
4  cos = 7 2 – 5 2 – 3 2 2  5  3 = 1 2   =  3 imvqb (Chemistry) 1. CH2 = CH2 + Br2  CH2 Br – CH2 Br Gi Rb ̈ (28) (160) (188) Atom – Economy = ? [For the reaction CH2 = CH2 + Br2  CH2 Br – CH2 Br (28) (160) (188) Atom – Economy = ?] 20% 60% 100% 95% DËi: 100% e ̈vL ̈v: % AE = 188 (28 + 160)  100 = 100% 2. CH2 = CH2 + H2(g) ⇌ CH3 – CH3(g); H = – 137 kJ/mol CH2 = CH2 Gi NbgvÎv e„w× cv‡e hw`- [CH2 = CH2 + H2(g) ⇌ CH3 – CH3(g); H = – 137 kJ/mol. The concentration of CH2 = CH2 will increase if-] Pvc e„w× Kiv nq Drcbœ CH3 – CH3 †K mwi‡q wb‡q AwZwi3 H2 †hvM Ki‡j ZvcgvÎv evov‡j DËi: ZvcgvÎv evov‡j e ̈vL ̈v: jv-kv‡Zwjqvi bxwZ Abyhvqx, Zv‡cvrcv`x wewμqvq ZvcgvÎv evov‡j cðvrgyLx Zvcnvix wewμqvwU e„w× cvq| d‡j mvg ̈ve ̄’v Wvb †_‡K evg w`‡K m‡i wM‡q wewμqv CH2 = CH2 Gi NbgvÎv e„w× cvq| 3. Kp Gi GKK mwVK bq †KvbwU‡Z? [In which of the following cases is the unit of Kp incorrect?] 2SO2(g) + O2(g) ⇌ 2SO3(g); Kp = atm–1 N2(g) + 3H2(g) ⇌ 2NH3(g); Kp = (atm)–3 PCl5(g) ⇌ PCl3(g) + Cl2(g); Kp = atm H2(g) + I2(g) ⇌ 2HI(g); Kp Gi GKK †bB DËi: N2(g) + 3H2(g) ⇌ 2NH3; Kp = (atm)–3 e ̈vL ̈v: Dˇii ̄^c‡ÿ hyw3: Kp Gi GKK n‡jv (atm)n (K) wewμqvi †ÿ‡Î, n = 2 – (2 + 1) = – 1; myZivs (L) bs mgxKi‡Yi †ÿ‡Î Kp = (atm)–1 (L) wewμqvi †ÿ‡Î, n = 2 – (1 + 3) = – 2; myZivs (K) bs mgxKi‡Yi †ÿ‡Î Kp = (atm)–2 (M) wewμqvi †ÿ‡Î, n = (1 + 1) – 1 = + 1; myZivs (M) bs mgxKi‡Yi †ÿ‡Î Kp = (atm)1 (N) wewμqvi †ÿ‡Î, n = 2 – (1 + 1) = 0; myZivs (N) bs mgxKi‡Yi †ÿ‡Î Kp Gi †Kv‡bv GKK _vK‡e bv| 4. †Kvb wewμqvi †ÿ‡Î mvg ̈ve ̄’v m„wó m¤¢e bq? [In which of the following reactions is it not possible to establish equilibrium?] NH3 + H2O + NaCl ⇌ NH4Cl + NaOH SnCl4 + Hg2Cl2 ⇌ SnCl2 + 2HgCl2 Fe2O3 + 6HCl ⇌ 2FeCl3 + 3H2O H2 + I2 ⇌ 2HI DËi: Fe2O3 + 6HCl ⇌ FeCl3 + 3H2O e ̈vL ̈v: Dˇii ̄^c‡ÿ hyw3: (K) wewμqvwU mvg ̈ve ̄’v m„wó Ki‡e| G wewμqvi mvg ̈ve ̄’v- NH3 + H2O ⇌ NH+ 4 + OH– wewμqvi DfgyLx Ges mvg ̈ve ̄’v m„wó nq| (L) wewμqvwU‡Z †iW· mvg ̈ve ̄’v m„wó nq| Sn4+ + 2Hg+ ⇌ Sn2+ + 2Hg2+ (M) wewμqvwU GwmW-ÿvi cÖkgb wewμqv| d‡j G wewμqvwU m¤ú~Y©Zv jvf K‡i| GRb ̈ GwU GKgyLx wewμqv Ges G‡ÿ‡Î mvg ̈ve ̄’v m„wó Kiv m¤¢e bq| 5. †Kv‡bv wbw`©ó ZvcgvÎvq, NO2(g) ⇌ NO(g) + 1 2 O2(g) wewμqvq Kc Gi gvb 8.66 × 10–3 n‡j, H GKB ZvcgvÎvq 2NO2(g) ⇌ 2NO(g) + O2(g) Gi Rb ̈ Kc =? [If the value of Kc for the reaction NO2(g) ⇌ NO(g) + 1 2 O2(g) is 8.66 × 10–3 at a certain temperature, what is the value of Kc for the reaction 2NO2(g) ⇌ 2NO(g) + O2(g) at the same temperature?] 8.5 × 10–2 3.5 × 10–2 7.5 × 10–5 8.16 × 10–2 DËi: 7.5 × 10–5 e ̈vL ̈v: NO2(g) ⇌ NO(g) + 1 2 O2(g); Kc = 8.66 × 10–3 2NO2(g) ⇌ 2NO(g) + O2(g); Kc  = ?  Kc  = (Kc) 2 = (8.66 × 10–3 ) 2 = (0.866 × 10–2 ) 2 =    3  2 × 10–2 2 = 7.5 × 10–5 6. 600 K ZvcgvÎvq 2HI(g) ⇌ H2(g) + I2(g) Gi mvg ̈aaæe‡Ki gvb 1.5 H ZvcgvÎvq cðvrg~Lx wewμqvi nvi aaæeK 8.66 × 10–3 nq Z‡e m¤§yL wewμqvi nvi aaæeK n‡e- [At a temperature of 600 K the equilibrium constant for the reaction 2HI(g) ⇌ H2(g) + I2(g) is 1.5, If the rate of the backward reaction at this temperature is 8.66 × 10–3 , what will be the rate of the forward reaction?] 3 4 × 10–2 5 7 × 10–2 3 3 4 × 10–2 7 5 2 × 10–2 DËi: 3 3 4 × 10–2 e ̈vL ̈v: g‡b Kwi, m¤§yL I cðvrg~Lx wewμqvi nvi h_vμ‡g K1 I K2 mvg ̈aaæeK K = K1 K2 K1 = K × K2 = 1.5 × 8.66 × 10–3 = 3 2 × 3 2 × 10–2 = 3 3 4 × 10–2