Given, radius of sphere =3 Now, In △OAB, by Pythagoras theorem (OA)2=(OB)2+(AB)2 
(3)2=(2h)2+r23=4h2+r2⇒r2=3−4h2 Now, volume of cylinder =πr2h V=π(3−4h2)hV=3πh−4πh3 Now, for largest possible right circular cylinder the volume must be maximum ∴ For maximum volume, dhdV=0 Now, Differentiating eq. (2) w.r.t. h V1=dhdV=3π−43πh2 or 3π−43πh2=0⇒3π=43πh2 ⇒h2=4⇒h=2 Now, volume (V) of the cylinder =π(3−4h2)h=π(6−2)=4π