JEE PYQ: Mechanical Properties Of Fluids Question 15
Question 15 - 2020 (05 Sep 2020 Shift 1)
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:
(1) $\frac{8}{9}R$
(2) $\frac{4}{9}R$
(3) $\frac{2}{3}R$
(4) $\frac{1}{3}R$
Show Answer
Answer: (1) $\frac{8}{9}R$
Solution
In equilibrium, $mg = F_b$ $\frac{4}{3}\pi(R^3 - r^3)\rho_0 g = \frac{4}{3}\pi R^3 \rho_w g$ Given, relative density $\frac{\rho_0}{\rho_w} = \frac{27}{8}$ $\Rightarrow \left[1 - \left(\frac{r}{R}\right)^3\right] \frac{27}{8} = \rho_w$ $\Rightarrow 1 - \frac{r^3}{R^3} = \frac{8}{27} \Rightarrow 1 - \frac{8}{27} = \frac{r^3}{R^3} = \frac{19}{27}$ $\Rightarrow \frac{r}{R} = \left(\frac{2}{3}\right)^{1/3} \Rightarrow 1 - \frac{r^3}{R^3} = \frac{8}{27} \Rightarrow \frac{r^3}{R^3} = \frac{19}{27}$ $\therefore r = \frac{8}{9}R$
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