Question

A hollow glass stopper of relative density 2.5 just sinks in water. The ratio of volume of cavity to that of stopper is

• A. 1:2
• B. 3:5
• C. 1:5
• D. 3:2

Hint:

For hollow objects, "just sinks" implies \( \text{Relative Density of material} \times V_{\text{material}} = V_{\text{total}} \). Here, \( 2.5 V_m = V_m + V_c \), which directly leads to \( 1.5 V_m = V_c \). \

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A "just sinking" object has an average density equal to the density of the fluid (water). We need to find the ratio of the cavity volume ($V_c$) to the material volume of the stopper ($V_m$). \

Step 2: Key Formula or Approach:
For an object to just sink, the total weight must equal the buoyancy force: \[ W = B \Rightarrow \rho_{\text{avg}} V_{\text{total}} g = \rho_w V_{\text{total}} g \Rightarrow \rho_{\text{avg}} = \rho_w \] \

Step 3: Detailed Explanation:
Let \( V_c \) be the volume of the cavity and \( V_m \) be the volume of the glass material. The total volume \( V_t = V_m + V_c \). The mass of the stopper is purely from the glass material: \( m = \rho_g V_m \). Given relative density of glass is 2.5, so \( \rho_g = 2.5 \rho_w \). \ For "just sinking": \[ \text{Weight of stopper} = \text{Buoyancy force} \] \[ \rho_g V_m g = \rho_w V_t g \] \[ (2.5 \rho_w) V_m = \rho_w (V_m + V_c) \] \[ 2.5 V_m = V_m + V_c \] \[ 1.5 V_m = V_c \] \[ \frac{V_c}{V_m} = \frac{1.5}{1} = \frac{3}{2} \] \

Step 4: Final Answer:
The ratio of the volume of the cavity to the material volume of the stopper is 3:2. \