Fluid Mechanics

JEE Main level, single correct, medium

14 of 192

Question 14 of 192

A hollow spherical shell at outer radius RR floats just submerged under the water surface. The inner radius of the shell is r.r. If the specific gravity of the shell material is 278\dfrac{27}{8} with respect to water, the value of rr is:

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Answer: (A) 89R\dfrac{8}{9}R

In equilibrium mg=FBmg={F}_{B}
43π(R3−r3)ρ0g=43πR3ρwg\dfrac{4}{3}\pi \left({R}^{3}-{r}^{3}\right){\rho }_{0}g=\dfrac{4}{3}\pi {R}^{3}{\rho }_{w}g
[1−(rR)3]278ρw=ρw\left[1-{\left(\dfrac{r}{R}\right)}^{3}\right]\dfrac{27}{8}{\rho }_{w}={\rho }_{w}
1−r3R3=9271-\dfrac{{r}^{3}}{{R}^{3}}=\dfrac{9}{27}
1−13=r3R31-\dfrac{1}{3}=\dfrac{{r}^{3}}{{R}^{3}}
23=r3R3\dfrac{2}{3}=\dfrac{{r}^{3}}{{R}^{3}}
rR=(23)1/3\dfrac{r}{R}={\left(\dfrac{2}{3}\right)}^{1/3}
1−r3R3=8271-\dfrac{{r}^{3}}{{R}^{3}}=\dfrac{8}{27}
r3R3=1−827=1927\dfrac{{r}^{3}}{{R}^{3}}=1-\dfrac{8}{27}=\dfrac{19}{27}
r=0.89r=0.89

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