c
(c) Maximum force by surface when friction works
$F = \sqrt {{f^2} + {R^2}} = \sqrt {{{(\mu R)}^2} + {R^2}} = R\sqrt {{\mu ^2} + 1} $
Minimum force $ = R$ when there is no friction
Hence ranging from $R$ to $R\sqrt {{\mu ^2} + 1} $
We get, $Mg \le F \le Mg\sqrt {{\mu ^2} + 1} $