Question

A shell of relative density 27/9 w.r.t. water, is just submerged in water. If it's inner and outer radius is r and R then r/R will be :

Answer 1

It has given that, relative density of shell = 27/9 w.r.t water. shell is just submerged in water. it's inner and outer radius are r and R respectively.

To find : The ratio of their inner and outer radius is r/R

solution : In equilibrium,

weight of shell = force of buoyancy

⇒4/3π(R³ - r³)ρ_sg = 4/3πR³ρ_wg

⇒(R³ - r³)ρ_s = R³ρ_w

here density of shell = 27/8 × density of water

⇒ρ_s = 27/8 ρ_w

⇒(R³ - r³) × 27/9 ρ_w = R³ × ρ_w

⇒(R³ - r³)/R³ = 9/27 = 1/3

⇒1 - (r/R)³ = 1/3

⇒2/3 = (r/R)³

⇒r/R = (2/3)⅓

Therefore the ratio of their inner to outer radius is (2/3)⅓ .i.e., correct option is (2)

Answer 2

So the upthrust acting on the shell is equal to its weight.

The shell is hollow so that its volume will be $\sf{\dfrac{4}{3}\,\pi(R^3-r^3).}$

Let $\sf{\rho_B}$ and $\sf{\rho_w}$ be densities of the shell and water respectively.

We know upthrust is the weight of the water displaced, whose volume is equal to that of a solid sphere of radius R.

Thus,

$\sf{\longrightarrow U=W}$

$\sf{\longrightarrow\rho_B\cdot\dfrac{4}{3}\,\pi(R^3-r^3)\,g=\rho_w\cdot\dfrac{4}{3}\,\pi R^3\,g}$

$\sf{\longrightarrow\dfrac{\rho_B}{\rho_w}=\dfrac{R^3}{R^3-r^3}}$

Here $\sf{\dfrac{\rho_B}{\rho_w}=\sigma_B}$ is the relative density of the shell.

$\sf{\longrightarrow\sigma_B=\dfrac{R^3}{R^3-r^3}}$

By rule of dividendo,

$\sf{\longrightarrow\dfrac{r^3}{R^3}=\dfrac{\sigma_B-1}{\sigma_B}}$

$\sf{\longrightarrow\dfrac{r}{R}=\left(\dfrac{\sigma_B-1}{\sigma_B}\right)^{\frac{1}{3}}}$

In the question, $\sf{\sigma_B=\dfrac{27}{9}=3.}$ Then,

$\sf{\longrightarrow\dfrac{r}{R}=\left(\dfrac{3-1}{3}\right)^{\frac{1}{3}}}$

$\sf{\longrightarrow\underline{\underline{\dfrac{r}{R}=\left(\dfrac{2}{3}\right)^{\frac{1}{3}}}}}$

Hence (2) is the answer.