Q. 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) 9/10
Explanation
(a) Let $x$ be the portion of exposed height of the body of length$l$, area of cross-section $A$. As the body is floating, so
$A\ l\ \rho\ g = A(l - x)3\rho\ g$
or $l = 3l - 3x$
or $x = 2l/3$
or $\frac{x}{l} = \frac{2}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.