A body floats with one-third of its volume outside water and 3/4 of its volume outside another liquid. The density of the other liquid is

Question

A body floats with one-third of its volume outside water and 3/4 of its volume outside another liquid. The density of the other liquid is (A) (9/4)g cc-1 (B) (4/0)g cc-1 (C) (8/3) gcc-1 (D) (3/8)g cc-1

Tardigrade Admin · Accepted Answer

Let V be volume of the body. In water weight of body = weight of water displaced =(2/3)V×1× g In another liquid, weight of body = (2/3)Vg=(1/4)Vρ g or ρ=(8/3)g cc-1

Q. A body floats with one-third of its volume outside water and 3/4 of its volume outside another liquid. The density of the other liquid is

$\frac{9}{4} g c c^{- 1}$

$\frac{4}{0} g c c^{- 1}$

$\frac{8}{3} g c c^{- 1}$

$\frac{3}{8} g c c^{- 1}$

Let V be volume of the body.
In water weight of body = weight of water displaced
$= \frac{2}{3} V \times 1 \times g$
In another liquid, weight of body
= $\frac{2}{3} V g = \frac{1}{4} V ρ g$
or $ρ = \frac{8}{3} g c c^{- 1}$

Q. A body floats with one-third of its volume outside water and 3/4 of its volume outside another liquid. The density of the other liquid is

49​gcc−1

04​gcc−1

38​gcc−1

83​gcc−1

Let V be volume of the body. In water weight of body = weight of water displaced =32​V×1×g In another liquid, weight of body = 32​Vg=41​Vρg or ρ=38​gcc−1

$\frac{9}{4} g c c^{- 1}$

$\frac{4}{0} g c c^{- 1}$

$\frac{8}{3} g c c^{- 1}$

$\frac{3}{8} g c c^{- 1}$

Let V be volume of the body.
In water weight of body = weight of water displaced
$= \frac{2}{3} V \times 1 \times g$
In another liquid, weight of body
= $\frac{2}{3} V g = \frac{1}{4} V ρ g$
or $ρ = \frac{8}{3} g c c^{- 1}$