A homogeneous solid cylinder of length L ( L

Question

A homogeneous solid cylinder of length L ( L < H / 2), cross-sectional area A is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L / 3 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure P0. Then, density D of solid is given by :- (A) (5/4) d (B) (4/5) d (C) 4 d (D) (4 d /3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e50cf6ca770b3f6e196535064f2aaa5a-.png" /> Bouyant force =( F b ) d +( F b )2 d = mg (d) A (2 L/3) g+(2 d ) ( AL /3) g = DALg D =(4 d /3)

$\frac{5}{4} d$

$\frac{4}{5} d$

$4 d$

$\frac{4 d}{3}$

Bouyant force $= (F_{b})_{d} + (F_{b})_{2 d} = m g$
(d) $A \frac{2 L}{3} g + (2 d) \frac{A L}{3} g = D A Lg$
$D = \frac{4 d}{3}$

45​d

54​d

4d

34d​

Bouyant force =(Fb​)d​+(Fb​)2d​=mg (d) A32L​g+(2d)3AL​g=DALg D=34d​

$\frac{5}{4} d$

$\frac{4}{5} d$

$4 d$

$\frac{4 d}{3}$

Bouyant force $= (F_{b})_{d} + (F_{b})_{2 d} = m g$
(d) $A \frac{2 L}{3} g + (2 d) \frac{A L}{3} g = D A Lg$
$D = \frac{4 d}{3}$