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 Concept:
When an object "just sinks" (floats totally submerged), its average density is equal to the density of the fluid (water). The presence of a cavity reduces the overall average density of the object.
Step 2: Key Formula or Approach:
1. Average density $\rho_{avg} = \frac{\text{Total Mass}}{\text{Total Volume}} = \rho_{water}$
2. Mass of stopper = $\rho_{glass} \times V_{glass}$
3. Total Volume ($V_{total}$) = $V_{glass} + V_{cavity}$
Step 3: Detailed Explanation:
1. Let $V_s$ be the volume of the glass material and $V_c$ be the volume of the cavity.
2. Total volume $V = V_s + V_c$.
3. Density of water $\rho_w = 1 \, \text{g/cm}^3$. Relative density of glass = 2.5, so $\rho_g = 2.5 \, \text{g/cm}^3$.
4. For the object to just sink: \[ \text{Weight} = \text{Buoyant Force} \] \[ \rho_g V_s g = \rho_w (V_s + V_c) g \]
5. Substitute densities: \[ 2.5 V_s = 1(V_s + V_c) \] \[ 2.5 V_s - V_s = V_c \] \[ 1.5 V_s = V_c \]
6. The ratio of cavity volume to stopper (glass) volume is: \[ \frac{V_c}{V_s} = 1.5 = \frac{3}{2} \]
Step 4: Final Answer
The ratio of the volume of the cavity to that of the stopper is 3:2.