A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be:

Question

A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be: (A) (2/3) (B) (5/6) (C) (1/6) (D) (1/3)

Tardigrade Admin · Accepted Answer

For floating body W=Th=V textin ρ Lg AH(ρ )Bg=(.Ah.)× 3(ρ )Bg (h/H)=(1/3) Exposed height =(H - h/H)=(3 - 1/3)=(2/3)

Q. A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be:

$\frac{2}{3}$

$\frac{5}{6}$

$\frac{1}{6}$

$\frac{1}{3}$

For floating body $W = T h = V_{in} ρ_{L} g$
$A H (ρ)_{B} g = (A h) \times 3 (ρ)_{B} g$
$\frac{h}{H} = \frac{1}{3}$
Exposed height $= \frac{H - h}{H} = \frac{3 - 1}{3} = \frac{2}{3}$