E and V respectively represent magnitudes of electrostatic field and potential inside a charged metallic shell. We have:

$E=0, V≠0$

The correct answer is Option (2) → $E=0, V≠0$

For a charged, metallic shell, the electrostatic field is zero due to the symmetry of charge distribution.

$∴E=0$

and, The electric potential V inside the shell, however, is constant and non-zero. This is because the shell forms a equipotential surface.

$∴V=K≠0$  [K = Constant]