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Standard XI

Physics

Archimedes' Principle

A rod of leng...

Question

A rod of length $6 m$ has specific gravity $ρ (= 25 / 36)$ . One end of the rod is tied to a $5 m$ long rope, which in turn is tied to the floor of a pool $10 m$ deep, as shown. Find the length (in $m$ ) of the part of rod which is out of water.

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Solution

Forces acting on rod are force of buoyancy, weight of rod and tension due to string.
Let Rod makes an angle $θ$ with horizontal

Balancing torque about O,
$F_{b} r_{⊥} = W r_{⊥}$
$ρ_{w} (6 - x) A g \frac{(6 - x)}{2} cos θ = ρ_{b} (6 A) (3) cos θ$

$(6 - x)^{2} = 36 ρ$
$6 - x = 5$
$x = 1$

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Similar questions

Q. A rod of length

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A rod of length 6 m has specific gravity ρ(=25/36). One end of the rod is tied to a 5 m long rope, which in turn is tied to the floor of a pool 10 m deep, as shown. Find the length (in m) of the part of rod which is out of water.

Forces acting on rod are force of buoyancy, weight of rod and tension due to string. Let Rod makes an angle θ with horizontal Balancing torque about O, Fbr⊥=Wr⊥ ρw(6−x)Ag(6−x)2cosθ=ρb(6A)(3)cosθ (6−x)2=36 ρ 6−x=5 x=1

A rod of length $6 m$ has specific gravity $ρ (= 25 / 36)$ . One end of the rod is tied to a $5 m$ long rope, which in turn is tied to the floor of a pool $10 m$ deep, as shown. Find the length (in $m$ ) of the part of rod which is out of water.

Forces acting on rod are force of buoyancy, weight of rod and tension due to string.
Let Rod makes an angle $θ$ with horizontal

Balancing torque about O,
$F_{b} r_{⊥} = W r_{⊥}$
$ρ_{w} (6 - x) A g \frac{(6 - x)}{2} cos θ = ρ_{b} (6 A) (3) cos θ$

$(6 - x)^{2} = 36 ρ$
$6 - x = 5$
$x = 1$