Statement 1:Electric field intensity at surface of uniformly charged spehrical shell is `E` if the shell is puncurted at a point then intensity at punctured point becomed `E//2`.
Statement 2:Electric field intensity due to spherical charge distribution can be found out by using `Gauss Law`.

Answer

B

Explanation

`V_(P)=(1)/(4piepsilon_(0))[(q)/(r_(1))+(-Kq)/(r_(2))]=0 or r_(2)=Kr_(1)`
`V_(P)=(1)/(4piepsilon_(0))[(q)/(r_(1))+(-Kq)/(r_(2))]=0 or r_(2)=Kr_(1)`(B) Now , `PP_(1)=(sin(/_POP_(1)))/(sin(/_PAP_(1)))=(r_(2))/(r_(1))=(Kr_(1))/(r_(1))=K`
`(C ) OP_(1)=x,P_(1)A=(6-x)`
`r_(1)^(2)=x^(2)+y^(2) r_(2)^(2)=(6-x)^(2)+y^(2)=4(x^(2)+y^(2))`
`r_(2)=2r_(1)` (from `r_(2)=Kr_(1))(6-x^(2))+y^(2)=4(x^(2)+y^(2))` ltbr. On solving `(x+2)^(2)+(y=0)^(2)=(4)^(2)`
Locus of point `P` is a circle of radius `x_(1)=4m` and centre `x(2)=(-2,0)`
(d) the values of `OA` and `K` should be known.