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

Vidyamandir Classes VMC | Final Step| Part-B 1 Class XII | Physics Solutions to Final Step | Part - B | Physics Electrostatics 1.(A) tan Fe qE mg mg    ; 0 0 / 2 tan 2 q q mg mg        2.(C) v u at u    , 0 ; qE v t m  ; 2 2 2 1 2 2 2 q E t mv m  3.(D) Since, the surface densities are equal, hence 1 2 2 2 4 4 q q r R    (where 1 2 ( ) q q Q   Or 1 2 1 2 2 2 2 2 2 2 q q q q Q r R r R r R        2 1 2 2 Q q r r R    and 2 2 2 2 Q q R r R    So, potential at the common centre, 1 2 1 2 0 0 0 1 4 4 4 q q q q V r R r R              2 2 2 2 2 2 0 1 4 Q r Q R R r R r r R              2 2 0 1 ( ) 4 ( ) Q R r R r     4.(CD) Only option C and D are true 5.(B) In air, 2 1 2 0 1 4 q F r   ; In medium k , 2 1 2 2 0 2 1 4 1 q k F F k F r     6.(D) Let, the third charge q is placed at a distance of ‘x’ from the charge 2q . Then P.E. of the system is :            2 8 2 8 q q q q q q U k r x r x           U is minimum, when 2 2 2 8 q q x r x    is maximum ; Solve for 0 dy dx   2 2 3 r d y x , ve dx    i.e. at 3 r x , y  is minimum or U is maximum. So, there can’t be any point between them where P.E. is minimum. 7.(C) When the two are joined by a metal wire, they become a single conductor. As charge can reside only on the outer surface of a conductor, the entire charge Q must flow to the outer sphere. 8.(D) Let an element of length dx, charge dq, at distance x from point O dq dV k x  Where, Q dq dx L  2 2 L L L L Q dx kdq L V k x x           
Vidyamandir Classes VMC | Final Step| Part-B 2 Class XII | Physics = 2 2 0 0 1 4 4 L L e L L Q Q dx [log x]   L x L         = 0 [ 2 ] 4 e e Q log L log L  L  0 0 (2) 4 4 e Q L Q log ln L L L            9.(AB) Since V  0 , there must be charges on the surface or inside itself. 10.(C) With each rotation, charge Q crosses any fixed point P near the ring. Number of rotations per second    / 2 .  Charge crossing P per second current 2 Q  11.(C) W Q V Q V V Q V              f i i  0  , 2 0 0 1 2 2 . . 4 4 Q Q W Q a a            12.(B) 2 cos net f F    2 2 2 2 2 2 2 . net q kq y F y a y a                 [Negative sign indicate the net force is towards the mean position]   2 3/ 2 3 2 2 2 2 net q kq y kq y F y a y a              13.(A)   , 2 2 , x   x dv E E x dx     at  1 1 0 x x E    14.(BC) dv E dr   , since v is constant, E  0 Since E  0 , there is no charge inside the region 15.(C) 16.(B) Force on 1 q 1 2 1 3 2 2 0 0 1 1 [sin cos ] 4 4 q q q q F i i j b a        From above, x component of force is : 1 2 3 2 2 0 4 x q q q f sin b a           2 3 2 2 sin x q q F b a          17.(D) 18.(C) 19.(C) Complete the cube, adding five other faces. cube face 0 0 , 6 q q       20.(C) At P due to shell, potential 1 0 4 q V R   At P due to Q, potential 2 0 0 2 4 4 2 Q Q V R R      Net potential at P, 1 2 0 0 2 4 4 q Q V V V R R      
Vidyamandir Classes VMC | Final Step| Part-B 3 Class XII | Physics 21.(BC) Work done an equipotential surface is zero also B A W E.dl    22.(B) 0 0 1 2 1 2 4 2 4 A Q Q Q V R R R            0 0 1 3 , 4 2 8 B A B Q Q V V V R R      23.(A) Using Gaussls theorem for radius r 0 1 E ds Q q . ( )       2 0 1 E r Q q 4 ( )      . . . . .(i) q = charge enclosed between x = a and x = r. 2 4 4 r r a a A q x dx A xdx x       2 2 2 4 2 ( ) 2 r a x   A A r a            Putting the value of q in equation (i), we get 2 2 2 0 1 E r Q A r a 4 2 ( )            2 2 2 0 1 2 2 4 Q Aa E A r r             E will be constant if it is independent of r 2 2 2 2 2 or 2 Q Aa Q A r r a      24.(B) Consider a point in the emptied space whose position vector is r  from the centre of the given sphere. Let the position vector of this point from the centre of cavity be 2 r  . Let the position vector of the centre of cavity from the centre of the sphere be r1  . Clearly r1  is constant. Consider the uniformly charged sphere without cavity. Electric field at r  can be obtained using Gauss law. 2 3 13 E r r     4 1 3 E r   The direction of E1  vector is along r  , so we can write 1 2 3 E r     Superposition of the two charges from the emptied space. To get the net electric field due to actual charge distribution superimpose the above two fields. 1 2 1 3 E E E r         This implies that the electric field in the emptied region is non-zero and uniform.
Vidyamandir Classes VMC | Final Step| Part-B 4 Class XII | Physics 25.(C) We know dV E dr   Here, ΔV and Δr are same for any pair of surfaces. So, E = constant Now, electric field inside the spherical charge distribution, 0 3 E r    ; E would be constant if r  constant  1 ( )r r   26.(BD) According to gauss law option B and D are correct 27.(C) The electric field lines around the cylinder must resemble that due to a dipole. 28.(C) Electric field due to complete disc (R = 2a), 1 2 2 0 1 2 x E R x              2 2 0 1 2 4 h a h              0 1 ( ) 2 2 h h a a             Electric field due to disc (R = a), 2 0 1 2 h E a           Hence, electric field due to given disc, 1 2 0 0 4 4 h E E E C a a          29.(D) Let a particle of charge q having velocity v approaches Q upto a closest distance r and if the velocity becomes 2v, the closest distance will be r . The law of conservation of energy yields, kinetic energy of particle = electric potential energy between them at closest distance of approach. Or 2 0 1 1 2 4 Qq mv  r  Or 1 2 2 kQq mv r  . . . . .(i) 0 1 constant 4 k           and 1 2 2 2 kQq m( v ) r   Dividing equation (i) by equation (ii), we get : 2 2 1 2 1 2 2 kQq mv r kQq m( v ) r    1 4 4 r r r r     30.(C) On outer surface there will be no charge. So, 2 2 Q    0 On inner surface total charge will be zero but charge distribution will be there so 1 1 Q and   0 0  On inner surface total charge will be zero but charge distribution will be there so 1 1 Q   0 and 0 