Question

A hollow glass stopper of relative density $2 \cdot 5$ just floats in water. The ratio of volume of cavity to that of the stopper is:
A. 1:2
B. 3:5
C. 1:5
D. 3:2

Answer 1

Hint- It is given that a hollow stopper with relative density $2 \cdot 5$ just floats in water. This means the weight of the stopper is balanced by the buoyant force of the water. So, by equating weight with buoyant force we can find the ratio of volume of cavity to the volume of stopper.

Complete step-by-step answer:
Let the volume of the stopper including the cavity be V and the volume of the cavity alone be v.
We need to find the ratio of volume of cavity to that of the stopper.
It is given that the stopper just floats in water. This means the weight of the stopper is balanced by the buoyant force of the water.
Let the mass of th3e glass stopper be M. Then the weight of the stopper will be $M\,g$ .
Buoyant force is given as ${V_w}{\rho _w}g$
Here the volume of water ${V_w}$ is equal to the volume of the stopper with a cavity. Thus, we can replace ${V_w}$ with V.
Thus on equating the weight and the buoyant force we get ,
$M\,g = V{\rho _w}g$
$\Rightarrow M\, = V{\rho _w}$................(1)
Now the mass of the glass M can be written as the product of volume of glass and density of glass. The volume of glass is the total volume of the glass stopper minus the volume of the cavity. That is, V-v
Therefore, equation (1) can be written as,
$(V - v){\rho _g}\, = V{\rho _w}$
$\dfrac{{(V - v)}}{V}\, = \dfrac{{{\rho _w}}}{{{\rho _g}}}$
On simplifying we get,
$1 - \dfrac{v}{V}\, = \dfrac{{{\rho _w}}}{{{\rho _g}}}$
$\Rightarrow \dfrac{v}{V}\, = 1 - \dfrac{{{\rho _w}}}{{{\rho _g}}}$................(2)
It is already given that the relative density of glass is $2 \cdot 5$.
We know that relative density of a substance is the ratio of density of substance to the density of water .
That is,
$\rho = \dfrac{{{\rho _g}}}{{{\rho _w}}}$
$\Rightarrow \dfrac{{{\rho _g}}}{{{\rho _w}}} = 2.5$
$\therefore \dfrac{{{\rho _w}}}{{{\rho _g}}} = \dfrac{1}{{2 \cdot 5}}$
Let us substitute this value in equation (2) .
$\Rightarrow \dfrac{v}{V}\, = 1 - \dfrac{1}{{2 \cdot 5}}$
$\therefore \dfrac{v}{V}\, = \dfrac{{1 \cdot 5}}{{2 \cdot 5}} = \dfrac{3}{5}$
Therefore the ratio of volume of cavity to the volume of the stopper is 3:5.

So the correct answer is option B.

Note: While calculating the weight of the stopper, remember that the stopper is hollow, so we need to consider the volume to be V-v. That is the volume of the stopper minus the volume of the cavity. Since, mass is distributed only in that much volume. If solid glass is given instead of hollow glass then we will take volume to be V.