A prism shaped styrofoam of density ρstyrofoam < ρwater < is held completely submerged in water. It lies with its base horizontal. The base of foam is at a depth h0 below water surface and atmospheric pressure is P0. Surface is open to atmosphere. Styrofoam prism is held in equilibrium by the string attached symmetrically. (Take : ρstyrofoam = ρf ; ρwater = ρw).

A prism shaped styrofoam of density ρ_styrofoam < ρ_water < is held completely submerged in water. It lies with its base horizontal. The base of foam is at a depth h₀ below water surface and atmospheric pressure is P₀. Surface is open to atmosphere. Styrofoam prism is held in equilibrium by the string attached symmetrically. (Take : ρ_styrofoam = ρ_f ; ρ_water = ρ_w).

Linked Question 1

Magnitude of force on any one of the slant face of styrofoam is

$(P_{0} + ρ_{w} g (h_{0} - \frac{ℓ}{\sqrt{2}})) Lℓ$

$(P_{0} + ρ_{w} g (h_{0} - \frac{ℓ}{2 \sqrt{2}})) Lℓ$

$(P_{0} - ρ_{w} g (h_{0} - \frac{ℓ}{\sqrt{2}})) Lℓ$

$(P_{0} + ρ_{w} g (h_{0} - \sqrt{2} ℓ)) Lℓ$

Arrange pressure on slant surface

$P_{avg} = \frac{(P_{0} + ρ_{w} {gh}_{0}) + (P_{0} + ρ_{w} {gh}_{0} - \frac{ℓ}{\sqrt{2}})}{2} = (P_{0} + ρ_{w} {gh}_{0} - \frac{ℓ}{2 \sqrt{2}})$

Force on any one of the slant face

$= (P_{0} + ρ_{w} g (h_{0} - \frac{ℓ}{2 \sqrt{2}})) Lℓ$

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Linked Question 2

Net force exerted by liquid on the styrofoam is

$ρ_{w} {gℓ}^{2} L$

$\sqrt{2} ρ_{w} {gℓ}^{2} L$

$2 ρ_{w} {gℓ}^{2} L$

$ρ_{w} g \frac{ℓ^{2} L}{2}$

Net force exerted by liquid on styrofoam is buoyout force = $ρ_{w} g \frac{ℓ^{2} L}{2}$

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Linked Question 3

Now string is cut and styrofoam is allowed to come to surface. A point mass is to be placed symmetrically on the upper surface of styrofoam such that it is in equilibrium with its base in horizontal plane. In equilibrium position styrofoam has half of its slant length submerged. Surface tension of water is T, contact angle is 0°. Determine mass m to achieve equilibrium.

$\frac{ρ_{f} {gLℓ}^{2}}{2} + \frac{3}{4} ρ_{w} {gLℓ}^{2} - 2 T [L + ℓ]$

$\frac{- ρ_{f} {gLℓ}^{2}}{2} + \frac{3}{8} ρ_{w} {gLℓ}^{2} - \frac{T [L + ℓ]}{\sqrt{2}}$

None of these

$\frac{- ρ_{f} {gLℓ}^{2}}{2} + \frac{3}{8} ρ_{w} {gLℓ}^{2} - \sqrt{2} T [L + ℓ]$

Resultant downward force due to surface tension $= \sqrt{2} T [L + ℓ]$

Buoyent force $= \frac{3}{8} ρ_{w} {gLℓ}^{2}$

Weight $= \frac{ρ_{f} {gLℓ}^{2}}{2}$

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