JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 9)
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 $${{27} \over 8}$$ w.r.t water,
the value of r is :
$${{2} \over 3}$$R
$${{4} \over 9}$$R
$${{1} \over 3}$$R
$${{8} \over 9}$$R
Explanation
_5th_September_Morning_Slot_en_9_1.png)
$${4 \over 3}\pi \left( {{R^3} - {r^3}} \right){p_m}\,g = {4 \over 3}\pi {R^3}{p_w}\,g$$
$$1 - {\left( {{r \over R}} \right)^3} = {8 \over {27}}$$
$$ \Rightarrow {r \over R} = {\left( {{{19} \over {27}}} \right)^{1/3}} = {{{{19}^{1/3}}} \over 3}$$
$$ = 0.88 \simeq {8 \over 9}$$
Comments (0)
