i m also waiting for d solution of 2nd part. somebody reply yaar!!!!!!!!!!!!

Bhaskarya! Why didnt u try solving the question. U r such a genius.

Can't it be in -ve Z direction......Akshay?

1) Find v_x by the given KE by --- ( 1/2 mv_x² ) = 10 * 10³ * 1.6 * 10^~19   

2) now 1m[ X ] = v_xt  i.e.    t = ( 1 / v_x)..............(1)

3) v_z =  ( qv_xB sin / m ) t                        [  = 90^o  ; subst t frm (1) ]

4) Now tan  = ( Z / X )    OR    v_z / v_x = tan

5) Equate them ull get Z = 1.4967 cm ~ 1.5cm [ I ve checked frm book u ve copied it incoorectly by mistake ]

ya, i got it. thanx...