A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is (27/8) w.r.t. water, the value of r is :

Question

A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is (27/8) w.r.t. water, the value of r is : (A) (4/9) R (B) (8/9) R (C) (1/3) R (D) (2/3) R

Tardigrade Admin · Accepted Answer

(4/3) π( R 3- r 3) ρ m g =(4/3) π R 3 ρ w g 1-(( r / R ))3=(8/27) ⇒ ( r / R )=((19/27))1 / 3=(191 / 3/3) =0.88 ≃ (8/9)

Q. A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$ . If the specific gravity of the shell material is $\frac{27}{8}$ w.r.t. water, the value of $r$ is :

$\frac{4}{9} R$

$\frac{8}{9} R$

$\frac{1}{3} R$

$\frac{2}{3} R$

$\frac{4}{3} π (R^{3} - r^{3}) ρ_{m} g = \frac{4}{3} π R^{3} ρ_{w} g$
$1 - (\frac{r}{R})^{3} = \frac{8}{27}$
$\Rightarrow \frac{r}{R} = (\frac{19}{27})^{1/3} = \frac{1 9 ^{1/3}}{3}$
$= 0.88 ≃ \frac{8}{9}$

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94​R

98​R

31​R

32​R