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A uniform cylinder of mass m, length l a...

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.

The elongation `X_(0)` produced in the springs when the syste is in equilibrium is

A

`(g)/(K)(m-(1)/(4)lAp)`

B

`(g)/(K)(m+(1)/(4)lAp)`

C

`(g)/(K)(m-(3)/(4)lAp)`

D

`(g)/(K)(m+(3)/(4)lAp)`

Text Solution

Verified by Experts

The correct Answer is:
C
With `3//4^(th)` of its length submerged, the upthrust acting on the cylinder is
U = weight of the liquid displaced by the length 3/4 of the cylinder
` = A xx (3l)/(4) xx rho xx g = (3l)/(4) (Arhog)`
Extension produced in the spring at equilibrium is `x_(0)`. So,
`Kx_(0) = mg - (3l)/(4)(Arhog)" "...(i)`
`rArr x_(0) = (mg)/(k) - (3l)/(4)((Arhog)/(k)) = (g)/(k)(m-(3)/(4)lArho)`
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