A uniform cylinder of mass m, length l and cross sectional area A is suspended with its vertical from a fixed point, with the help of a massless spirng of force constant (K), The cylinder is then submerged in a liquid of density p. At equilibrium the cylinder remains submerged in the liquid with 1//4^(th) of its volume outside the liquid,as shown in the figure. The elongation of hte spring at this point is x_(0). WHen the cylinder is given a slight downward push into the liquid and then released, its starts oscillating with vertical S.H.M. of small amplitude. If the cyleinder is displced by a small downward displacement a from its equilibrium postion and then released , the restoring force on it will be

A uniform cylinder of mass m, length l and cross sectional area A is s

1.	A uniform cylinder of mass m, length l and cross sectional area A is suspended with its vertical from a fixed point, with the help of a massless spirng of force constant (K), The cylinder is then submerged in a liquid of density p. At equilibrium the cylinder remains submerged in the liquid with 1//4^(th) of its volume outside the liquid,as shown in the figure. The elongation of hte spring at this point is x_(0). WHen the cylinder is given a slight downward push into the liquid and then released, its starts oscillating with vertical S.H.M. of small amplitude. If the cyleinder is displced by a small downward displacement a from its equilibrium postion and then released , the restoring force on it will be
Answer» `-([K+Apg]a)/(4)` `-[K+(1)/(4)Apg]a` `-[K-(1)/(4)Apg]a` `-[K+Apg]a` Solution :When a small downward displacement a is given to the cyliner, then the submerged length of the cylinder becomes (3L/4 + a). So, the extension PRODUCED in the SPRING is now `(x_(0) + a)`. The upthrust acting on the cylinder will become `U. = ((3l)/(4) + a) Arhog ` Force in the spring ` = k(x_(0) + a)` So, the restoring force on the cylinder is `F = -[k(x_(0) + a) - {mg -((3l)/(4) + a)} Arhog] ...(II)` Using (i) and (ii), we get `F = -[k + Arhog]a`