(a) Angular momentum, $$mvr = {{nh} \over {2\pi }}$$

$$ \Rightarrow mvr \propto n$$ $$\propto$$ distance from the nucleus

(b) This statement is incorrect as size of an orbit $$\propto$$ Azimuthal quantum number (l) ($$\because$$ n = constant)

(c) This statement is incorrect as at ground state,

n = 1, l = 0

$$\Rightarrow$$ Orbital angular momentum (wave mechanics)

$$ = \sqrt {l(l + 1)} {h \over {2\pi }} = 0$$ [$$\because$$ $$l = 0$$]

(d) The plot of $$\psi $$ vs $$ r $$ for various azimuthal quantum numbers, shows peak shifting towards higher $$ r $$ value.
This statement is typically true. The radial distribution functions of atomic orbitals (which can be related to $$ |\psi|^2 $$ vs $$ r $$) for higher azimuthal quantum numbers generally have their peaks at larger values of the radial distance $$ r $$ from the nucleus. This is because with higher $$ l $$ values, the electrons are more likely to be found at greater distances due to the higher angular momentum.