Tiêu đề Physics ÔÇó Final Step-A ÔÇó SOL.pdf ✅

Vidyamandir Classes VMC | Final Step | Part - A 1 Class XI | Physics Solutions to Final Step | Part - A | Physics Introduction to Vectors & Forces 1.(C) A B    lies in plane  to A B C , ,    ( ) A B C      will be in plane containing A, B and C 2.(C) actual velocity of the train velocity of train along east velocity of train along north T T E N E N V V V V V V              First Condition 10 1 E N V V   Second Condition 30 0 E N V V    VE = 30 VN = 20 ; V0 (my velocity) = 50 3.(B)    R F N mg ,     R N F mg 0, sin cos R F N mg F       4.(D) A B and   may be anti-parallel with magnitude of A twice of B. 2 2 2 A B AB A     cos     B B( 2cos ) 0 0 or cos 2 B B A      2 cos 0 2 B        A B AB   5.(C) a  andb  must be | | or anti-parallel with | a | | b |    6.(B) a c sin pˆ b c sin pˆ           pˆ is unit vector prep to plane of incidence (outward) F N R mg (i) F N R mg (ii) F N R mg (iii)
Vidyamandir Classes VMC | Final Step | Part - A 2 Class XI | Physics 7.(C) sin 75 sin 45 A C    and sin 75 sin 30 , sin 75 sin(30 45) B C      1 3 3 1 sin 30 cos 45 cos30 sin 45 2 2 2 2 2 2          1 ( 3 1) 2 ( 3 1) 2 2 2 3 1 2 B C C B C         8.(BD) | A B | | A B |        2 2 2 2 A B AB cos A AB cos      2 2    AB cos  0 9.(BD) a  is anti-parallel to b  and | a | | b | a b     0      10.(A)   1 N F mg m M g R N Mg F F R                  11.(ABD) If   0 5. fmax = 20 N which will not be needed so block will remain at rest at f = 10N If   0 2. , f = 8N hence block will move down. 12.(D) DB a b      DB . AB a b . a          2    a b a   . . . .(i) Also a b c      2 2 2 a b a . b c    2   . . . .(ii) From (i) and (ii) 2 2 2 3 2 a b c DB . AB      13.(D) Q sin tan P Q cos       3 2 2 2 P sin P P cos     14.(ABD) If F , F , F 1 2 3    2 3 and 4    then maximum value of F F F 1 2 3      can be 9 and minimum value can be 0 and particle can take accelerations in between 2 2 9 0 4 5 and 0 . m / s m / s m m   N F mg R R Mg F N
Vidyamandir Classes VMC | Final Step | Part - A 3 Class XI | Physics 15.(AB) 10 and 0 R N R A B   in case 1 Case - II 60 60 10 60 60 A B A B A B R sin R sin R R N R cos R cos mg          2 60 20 3 3 60 3 10 3 B A A A B mg R R cos R / R sin mg R mg / /           16.(C) ˆ ˆ ˆ b xi yj zk     a b x y z       1 1   ˆ ˆ a b j k              z y ; x z ; y x 0 1 1 17.(AB) As AB BC V cos V sin    SR SR      5 1 5 cos sin       5 2 45 1 sin  1 1 45 53 5 2  sin        0 4 0 1 6 5 . km time . hr min sin km / hr     18.(B) b c b a b a b                  ; c a a b a a b                   19.(CD) Let velocity components (horizontal and vertical) be V each as for stationary observer it is falling at angle of 45 , let u ; velocity of man in the direction of horizontal component of rain.    v m / s i.e. cos 2 2 2 45   Now relative horizontal components relative velocity of rain in opposite direction if u v  and | v u |   2 (i.e.  2 2 45 sin    u m / s 4 20.(D)     2 2 2 2 a b c a b c a .b b c c a           2          21.(D) Magnitude of vector should not change 22.(ABCD) (i) Maximum value of a b c      in case 1 will be 14 and minimum value will be zero because a . b c and    can form a triangle. (ii) Maximum value of a b c      in Case II will be 12 and minimum value will be | | 1 4 7 2    23.(C) Sum of magnitude of any two vector is greater or equal to magnitude of the third, if three vector adding to zero. 24.(C) For Parallel RA RB 10N 60 60 RA 60 Mg RB 45  B C VSR 45 V V | V u |  45 V
Vidyamandir Classes VMC | Final Step | Part - A 4 Class XI | Physics 2 2 8 2 t t t   ( 3 t  16 (Not possible as t  0 ) For perpendicular 3 2 a . b t t    0 8 2   t = 1 4 25.(B) 26.(B)     2 2 2 x t t     1500 10 1800 15 , for x to be minimum, its first derivative should be zero. Thus 0 dx dt  .      20 1500 10 30 1800 15  t t    or t . s 129 23 27.(D) 28.(A) 3 2 6 A . B cos , A B           29.(A) 2 3 1 4 5 1 1 1 ˆ ˆ ˆ i j k A B ˆ ˆ ˆ n , A B i j k ˆ A B                ,   1 42 4 5 42 ˆ ˆ ˆ A B , n i j k      ˆ   30.(C) 5 4 10 f i av V V ˆ ˆ j i a t     or 1 2 aav  North west. 31.(B)  2 3 8 4 4 0    ˆ ˆ ˆ ˆ ˆ ˆ i j k . j i k       or 1 8 12 8 0 2 ,         32.(D) s river v sin v   1 30 2 sin     33.(D) 3 2 1 4 3 2 ˆ ˆ ˆ i j k A B      10 17 390 ˆ ˆ ˆ A B i j k nˆ A B           ; 29 10 17   390 ˆ ˆ ˆ i j k x B nˆ       34.(C) rm r m 3 4 V V V k j    ˆ ˆ    ; 4 53 3 tan       North vertical 35.(A) A regular hexagon will be formed if we continue. After 5th turn displacement = OE = 500 m After 3rd turn displacement = OC = 1000 m (OC is diameter of the circle circumscribing regular hexagon). 36.(D) Use parallelogram law. 37.(A) cm  rx m ry  V V V i V j   ˆ ˆ 