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.

Option 1 - <p>2/3 R<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 2 - <p>4/9 R<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>1/3 R<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>8/9 R<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
7 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 4
Detailed Solution:

As it just floats B = mg
(4πR³/3) (ρ‎? ) (g) = (4πR³/3 - 4πr³/3) (ρ‎? ) (g)
R³ = (R³ - r³) (27/8)
On solving we get,
r = 8/9 R (approx)

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Physics Mechanical Properties of Solids 2025

Physics Mechanical Properties of Solids 2025

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