$A$ spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$.
$(a)$ $A$ charge $q$ is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
$(b)$ Is the electric field inside a cavity (with no charge) zero,even if the shell is not spherical,but has any irregular shape? Explain.

(N/A) The charge placed at the centre of the shell is $+q$. Due to electrostatic induction,a charge of magnitude $-q$ is induced on the inner surface of the shell. Therefore,the total charge on the inner surface is $-q$.
The surface charge density on the inner surface is given by:
$\sigma_{1} = \frac{\text{Total charge}}{\text{Inner surface area}} = \frac{-q}{4 \pi r_{1}^{2}}$
$A$ charge of $+q$ is induced on the outer surface of the shell. Since the shell already has a charge $Q$,the total charge on the outer surface becomes $Q + q$. The surface charge density on the outer surface is:
$\sigma_{2} = \frac{\text{Total charge}}{\text{Outer surface area}} = \frac{Q + q}{4 \pi r_{2}^{2}}$
$(b)$ Yes,the electric field inside a cavity (with no charge) is zero,even if the shell is not spherical and has an irregular shape.
This is because if we consider a closed loop such that a part of it is inside the cavity along a field line and the rest is inside the conductor,the net work done by the electric field in moving a test charge along this closed loop must be zero. Since the electric field inside the conducting material is zero,the field inside the cavity must also be zero to satisfy the conservative nature of the electrostatic field.