Tiêu đề Vector Varsity Daily-2 (Set-A) Solution.pdf ✅

3 10. `ywU †f±‡ii jwäi m‡e©v”P gvb 40 N Ges GKwU †f±‡ii gvb 15 N n‡j, wb‡¤œi †Kvb gvbwU 2q †f±‡ii Rb ̈ cÖ‡hvR ̈ bq? [The maximum magnitude of the resultant of two vectors is 40 N and the magnitude of one of the vectors is 15 N. Which of the following values is not possible for the magnitude of the second vector?] 5 N 15 N 30 N All DËi: All e ̈vL ̈v: Rvwb, `ywU †f±i A  I B  n‡j Zv‡`i jwäi m‡e©v”P gvb, Rmax = |A|  + |B|   40 = 15 + |B|  Rmax = 40 N |A|  = 15 N  |B|  = 25 N B  Gi gvb 25 N n‡ZB n‡e Zv bv-n‡j Ab ̈wKQz cÖ‡hvR ̈ bq| 11. ABC GKwU wÎfzR| GLb, P I Q e‡ji jwä KZ? [ABC is a triangle. What is the resultant of forces P and Q?] 5N = Q P = 10 N A B C 60 175 N 60 N 75 N 80 N DËi: 75 N e ̈vL ̈v: Q = 5N P = 10 N A B C 60 R = P 2 + Q2 + 2PQ cos120 = 102 + 52 + 2  10  5      –1 2 = 100 + 25 – 50 = 75 N 12. A  = 2i  + 4j  , B  = 2j  + 3k  `ywU †f±i| GLb, C  Ggb GKwU GKK †f±i wbY©q Ki hv Dc‡ii `ywU †f±‡ii mv‡_ j¤^fv‡e Aew ̄’Z? [Given vectors A  = 2i  + 4j  and B  = 2j  + 3k  find a unit vector C  that is perpendicular to both A and B.] 6 7 i  – 3 7 j  + 3 7 k  5 7 i  – 4 7 j  + 2 7 k  6 7 i  + 2 7 k  – 3 7 j  4i  + 3j  + 7k  DËi: 6 7 i  + 2 7 k  – 3 7 j  e ̈vL ̈v: A   B  =        i   2 0 j  4 2 k  0 3 = i  (12) – j  (6) + k  (4) = 12i  – 6j  + 4k  C  =   =  A   B  |A |   B  =  12i  – 6j  + 4k  122 +(–6) 2 + 42 =  12i  – 6j  + 4k  196 =      12 14 i  – 6 14 j  + 4 14 k  =      6 7 i  – 3 7 j  + 2 7 k  13. j   i  + i   k  + k   j  + j   k  = ? – i  – j  – k  –j  – k  i  + j  9i  DËi: –j  – k  e ̈vL ̈v: j   i  = – k  , i   k  = –j  , k   j  = –i  , j   k  = i  = –k  –j  – i  + i  = –j  – k  14. evwn ̈K evZvm _vK‡j Mvwoi KvP †fRvi kZ© hw` evZv‡mi w`K Mvwoi †e‡Mi w`‡K n‡j, †KvbwU mwVK? [If the speed of the wind is greater than the speed of the car, which part of the car's windshield will get wet?] VevZvm > VMvwo  †cQ‡bi KuvP wfR‡e (Vair > Vcar  Back windshield) VevZvm = VMvwo  †Kvb KuvP wfR‡e bv (Vair = Vcar  Neither windshield) VevZvm < VMvwo  †cQ‡bi KuvP wfR‡e (Vair < Vcar  Back windshield) †KvbwUB bq (None of the above) DËi: VevZvm > VMvwo  †cQ‡bi KuvP wfR‡e e ̈vL ̈v: (i) evZvm I Mvwo/ ch©‡eÿK wecixZ w`‡K _vK‡j  me©`vB mvg‡bi KuvP wfR‡e (ii) evZvm I Mvwo/ ch©‡eÿK GKB w`‡K _vK‡j  Vc > Va n‡j mvg‡bi KuvP wfR‡e  Vc < Va n‡j wcQ‡bi KuvP wfR‡e  Vc = Va n‡j Dfq KuvP wfR‡e