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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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A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density sigma at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period T of the oscillations of the cylinder will be :

Knowledge Check

  • 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. Time period of vertical S.H.M of the cylinder is

    A
    `2pi((m)/(K+Apg))^(1//2)`
    B
    `2pi((m)/(K-Apg))^(1//2)`
    C
    `2pi((m)/(K-(1)/(4)Apg))^(1//2)`
    D
    `2pi((m)/(K+(1)/(4)Apg))^(1//2)`

  • 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
    `-([K+Apg]a)/(4)`
    B
    `-[K+(1)/(4)Apg]a`
    C
    `-[K-(1)/(4)Apg]a`
    D
    `-[K+Apg]a`

  • A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density sigma at equilibrium position. The extension x_0 of the spring when it is in equlibrium is:

    A
    (a) `(Mg)/(k)`
    B
    (b) `(Mg)/(k)(1-(LAsigma)/(M))`
    C
    (c) `(Mg)/(k)(1-(LAsigma)/(2M))`
    D
    (d) `(Mg)/(k)(1+(LAsigma)/(M))`

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