JEE Advanced 2021 Paper 2 · Q07 · Buoyancy & Archimedes

Correct answer:0.30

Solution

At the sinking threshold the test tube is in neutral equilibrium: weight equals buoyant force. Weight of test tube $= 5$ g (ignoring trapped air mass). Volume submerged = volume of glass + volume of trapped air $= m_{\text{glass}}/\rho_{\text{glass}} + v_{\text{air}} = 5/2.5 + v_{\text{air}} = 2 + v_{\text{air}}$ cc. Buoyancy = $(2 + v_{\text{air}}) \cdot \rho_w \cdot g$ with $\rho_w = 1$ g/cc. Setting weight $=$ buoyancy: $5 = 2 + v_{\text{air}}$, so $v_{\text{air}} = 3$ cc. Initially $v_0 = 3.3$ cc, so $\Delta v = v_0 - v_{\text{air}} = 0.3$ cc, hence $X = 0.30$.

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