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

1 Varsity Weekly-01 [A (Solution)] wm‡jevm: †f±i + ̧YMZ imvqb-1 (`ave ̈Zvi AvM ch©šÍ) + g ̈vwUa· I wbY©vqK + †Kvl I †Kv‡li MVb + evsjv-1 (AcwiwPZv, D”PviY, cvwifvwlK kã) + Bs‡iwR (Noun, Pronoun, determiners, Poem Summarizing) c~Y©gvb: 60 †b‡MwUf gvK©: 0.25 mgq: 45 wgwbU c`v_©weÁvb (Physics) 1. A   B  = 0 Ges A  = 2i  + 2j  + 2k  I B  = 3i  + mj  + nk  n‡j, m I n Gi gvb KZ? 4, 2 3, 3 3, 2 6, 6 DËi: 3, 3 e ̈vL ̈v:        i   2 3 j  2 m k  2 n = 0  (2n – 2m)i  + (6 – 2n)j  + (2m – 6)k  = 0 2n – 2m = 0 6 = 2n 2m – 6 = 0 m = n n = 3 m = 3 Trick : A   B  = 0 Abyiƒc mnM ̧‡jvi AbycvZ mgvb| 2 3 = 2 m = 2 n Zvn‡j, m I n Gi gvb 3, 3 n‡Z n‡e| 2. GKwU Mvwoi‡eM 5 ms–1 Ges e„wói †eM 4 ms–1 , evZv‡mi †eM 1ms–1 n‡j, e„wó †hw`‡K co‡Q e‡j g‡b nq Zvi mv‡_ Avbyf‚wgK w`‡K Drcbœ †KvY KZ? (Mvwoi MwZi w`‡K evZvm eB‡Z _v‡K|) tan–1 3 tan–1 4 cos–1 4 tan–1 1 DËi: tan–1 1 e ̈vL ̈v: Mvwoi MwZi w`‡K evZvm eB‡Z _vK‡j, tan = Q P – S = 4 5 – 1 = 1 3. `yBwU mggv‡bi †f±i GKwU we›`y‡Z wμqvkxj| G‡`i jwäi gvb †h‡Kv‡bv GKwU †f±‡ii gv‡bi mgvb n‡j, ga ̈eZ©x †KvY KZ? 90 120 45 0 DËi: 120 e ̈vL ̈v: P = Q = R n‡j, R = P 2 + Q2 + 2PQ cos  P = P 2 + P2 + 2P2 cos  P 2 = 2P2 + 2P2 cos  –2P2 cos = P2  cos = P 2 – 2P2 = – 1 2 = cos 120   = 120 4. A  = 3 i  + 2 6 j  – 4 k  n‡j, A  Gi mgvšÍivj GKK †f±i Kx n‡e? 3i  – 2 6j  – 4k  49 3i  + 2 6j  + 4k  7 3i  + 2 6j  – 4k  50 3i  + 2 6j  – 4k  49 DËi: 3i  + 2 6j  – 4k  49 e ̈vL ̈v: A  Gi mgvšÍivj GKK †f±i, a  = A  |A|  a  = 3i  + 2 6j  – 4k  49 |A|  = 3 2 +(2 6) 2 + (–4) 2 = 9 + 24 + 16 = 49 5. A  = 0.866i  + 3j  †f±iwU z A‡ÿi mv‡_ KZ †KvY ˆZwi K‡i? 0 90 45 †Kv‡bv †KvYB ˆZwi K‡i bv DËi: 90 e ̈vL ̈v:  = cos–1 0 (0.866) 2 + ( 3) 2 = 90 6. P  = 5i  + 4j  – 2k  n‡j, wb‡¤œi †Kvb gvbwU yz Z‡ji gvb bq? 2 5 20 16  5 None DËi: None e ̈vL ̈v: P  = 5i  + 4j  – 2k  yz Z‡j |P|  = 4 2 + (–2) 2 = 20 = 2 5 yz Z‡ji GKK †f±iØq j  I k  7. A  = x 2 i  + (–6yz)k  + 3yj  †ÿÎwU (3, 1, 0) we›`y‡Z KZUzKz diverge K‡i? 5 3 6 27 DËi: 3 e ̈vL ̈v: div(A)  =   .A  =  x x 2 +  y 3y –  z 6yz = 2x + 3 – 6y (3, 1, 0) we›`y‡Z = 6 + 3 – 6 = 3 8. `ywU GKK †f±i a  I b  Gi †hvMdj Aci GKwU GKK †f±i c  n‡j, a  . b  = ?
2 1 2 3 2 – 3 2 – 1 2 DËi: – 1 2 e ̈vL ̈v: awi, `ywU GKK †f±i a  I b  Gi mgwó Aci GKK †f±i a  + b  = c   (a )  + b  2 = (c)  2 [|a| ]  = |b|  = |c|   (a )  + b  . (a )  + b  = c  .c   a 2 + 2a  . b  + b2 = c2  1 + 2a  .b  + 1 = 1  2a  .b  = –1  a  .b  = – 1 2 9. GKwU b`xi † ̄av‡Zi †eM 6ms–1 |8ms–1 †e‡Mi GKwU †bŠKv H b`x‡Z Pj‡Q| b`xi cÖ ̄’ 20 m n‡j b~ ̈bZg mg‡q b`x cvi n‡Z †bŠKv‡K KZ c_ cvwo w`‡Z n‡e? 50 m 30 m 40 m 25 m DËi: 25 m e ̈vL ̈v: †gvU AwZμvšÍ `~iZ¡ = jwä †eM  cvivcv‡ii Rb ̈ b~ ̈bZg mgq = u 2 + v2  d v = 6 2 + 8 2  20 8 = 25 m 10. 135 10 N 10 2 N ej؇qi jwäi gvb KZ n‡e? 5 2 N 50 N 5 10 N 10 5 N DËi: 10 5 N e ̈vL ̈v: 10 2 N  = 45 10 N  = 180 – 135 = 45 jwä = (10 2) 2 + 102 + 2. 10 2. 10. cos45 = 200 + 100 + 2  10 2  10  1 2 = 500 = 10 5 N 11. wb‡Pi †Kvb `yBwU †f±i P  = (3i )  + 3j  – 3k  Gi mv‡_ ci ̄úi j¤^? B  = (2i )  + j  + k  , A  = (9i )  + 10j  + 4k  A  = (3i )  – 2j  + k  , B  = (2i )  – j  + k  A  = (2i )  + 2j  – 3k  , B  = (i )  + j  + k  A  = (i )  + j  + k  , B  = (2i )  + j  + k  DËi: A  = (3i )  – 2j  + k  , B  = (2i )  – j  + k  e ̈vL ̈v: 2q option n‡Z, A  .P  = 9 – 6 – 3 = 0 B  .P  = 6 – 3 – 3 = 0 myZivs, (L) G A  I B  †f±iØq DfqB P  Gi mv‡_ j¤^| 12. OA  = i  – 2j  + 5k  Ges OB  = 3i  – j  + 9k  n‡j, BA  Gi gvb wb‡Pi †KvbwU? [†hLv‡b, O g~jwe›`y|] 29 9.89 21 40 DËi: 21 e ̈vL ̈v: O A  B  OA  = i  – 2j  + 5k  ; OB  = 3i  – j  + 9k  BA  = OA  – OB  = – 2i  – j  – 4k  |BA|  = (–2) 2 + (–1) 2 + (–4) 2 = 4 + 1 + 16 = 21 13. A  .B  = 0 n‡j, †KvbwU mZ ̈? |A |  + B  = |A |  – B  A  + B  = 0 A  = 4i  + 9j  + 2k  , B  = 4i  + 2j  + 3k  †Kv‡bvwUB 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 = 0 14. A  = 3i  – 4j  , B  = –3i  + 4j  ; Ges R  Zv‡`i jwä n‡j, wb‡Pi †KvbwU mwVK? R  GKwU GKgvwÎK †f±i R  GKwU wØ-gvwÎK †f±i R  GKwU wÎgvwÎK †f±i R  GKwU bvj †f±i
3 DËi: R  GKwU bvj †f±i e ̈vL ̈v: R  = A  + B  = 3i  – 4j  – 3i  + 4j  = 0  15. GKwU e ̄`i Dci F  = (2i ) ^ + 3j ^ – 4 k ^ N ej cÖ‡qv‡Mi d‡j †mwU (3, 2, –1) we›`y n‡Z (7, 7, 3) we›`y‡Z ̄’vbvšÍwiZ nq| G‡ÿ‡Î K...Z KvR KZ n‡e? 3 Ryj 4 Ryj 2 Ryj 7 Ryj DËi: 7 Ryj e ̈vL ̈v: W = F  .r  = (2i ) ^ + 3j ^ – 4k ^ .(4i ) ^ + 5j ^ + 4k ^ = 8 + 15 – 16 = 7 Ryj imvqb (Chemistry) 1. †Kvb cigvYy ev Avq‡bi g‡a ̈ B‡jKUab, †cÖvUb I wbDUab msL ̈v me ̧‡jvB wfbœ? 27 13Al 35 17Cl – 32 16S 2– 39 19K + DËi: 39 19K + e ̈vL ̈v: 39 19K + Avq‡bi, B‡jKUab msL ̈v = 19 – 1 = 18 †cÖvUb msL ̈v = 19 wbDUab msL ̈v = 39 – 19 = 20 2. wb‡¤œv3 wewμqvi k~b ̈ ̄’v‡b Kx n‡Z cv‡i? 27 13Al + 4 2He  30 15P + 1 0 n 0 –1 e 1 1H 0 0  DËi: 1 0 n e ̈vL ̈v: †cÖvUb msL ̈vi cv_©K ̈ = (13 + 2) – 15 = 0 fi msL ̈vi cv_©K ̈ = (27 + 4) – 30 = 1  ms‡KZ 1 0 n 27 13Al + 4 2He  30 15P + 1 0 n 3. c ̈v‡ðb wmwi‡Ri me©wb¤œ Zi1⁄2 msL ̈v KZ? 7 144 RH 144 7 RH 16 225 RH 1 9 RH DËi: 7 144 RH e ̈vL ̈v: c ̈v‡ðb wmwi‡Ri Rb ̈ n1 = 3 me©wb¤œ Zi1⁄2msL ̈vi Rb ̈ n2 = 3  1 = 4  –  = RH     1 n1 2 – 1 n2 2 = RH     1 3 2 – 1 4 2 = 7 144 RH 4. wb‡Pi †KvbwU AbycÖfv m„wóKvix c`v_©? ZnS Na2S CaS K2S DËi: ZnS e ̈vL ̈v: iv`vi‡dvW©  KYv we‡ÿcY cixÿvq ZnS e ̈envi K‡ib| KviY GwU AbycÖfv m„wóKvix c`v_© Ges Gi gva ̈‡g -KYvi w`K cwieZ©b eySv hvq| 5. cigvYyi wØZxq Kÿc‡_i GKwU B‡jKUa‡bi Rb ̈ †KŠwYK fi‡e‡Mi gvb wbY©‡qi mgxKiY †KvbwU? mvr = h 2 mvr = h  mvr = 3h 2 mvr = 3h  DËi: mvr = h  e ̈vL ̈v: Avgiv Rvwb, †KŠwYK fi‡eM, L = mvr = nh 2 mvr = 2h 2 [n = 2]  mvr = h  6. Cr cigvYyi me©ewnt ̄’ ͇̄ii B‡jKUa‡bi Rb ̈ †Kvqv›Uvg msL ̈vi †mU †KvbwU? n = 4, l = 0, m = 0, s = – 1 2 n = 3, l = 0, m = 0, s = – 1 2 n = 3, l = 2, m = – 2, s = – 1 2 n = 4, l = 2, m = 2, s = – 1 2 DËi: n = 4, l = 0, m = 0, s = – 1 2 e ̈vL ̈v: 24Cr  1s2 2s2 2p6 3s2 3p6 3d5 4s1 GLv‡b, Cr cigvYyi me©ewnt ̄’ ͇̄ii B‡jKUabwU PZz_© kw3 ͇̄ii s Dckw3 ͇̄i Aew ̄’Z| 4s1 Gi †ÿ‡Î: n = 4, l = 0, m = 0, s = – 1 2 7. 4_© kw3 ͇̄i †gvU AiweUvj msL ̈v KqwU? 4 9 16 32 DËi: 16 e ̈vL ̈v: 4_© kw3 ͇̄i †gvU AiweUvj msL ̈v = n 2 = 42 = 16wU 8. _v‡qvmvj‡dU (S2O 2– 3 ) Avq‡b me©‡gvU †hvRb B‡jKUab msL ̈v KZ? 28 30 32 34 DËi: 32