Question

A hydrometer has a bulb of volume 10.5 cm³ and the area of the cross-section of its stem is 0.3 cm². The hydrometer weighs 13.5 gf. Find the density of a liquid in which only the bulb of the hydrometer gets immersed.

Answer 1

Step 1: Given data

Volume of the bulb of the hydrometer, $V=10.5c{m}^3$

Area of the cross-section of the stem of the hydrometer, $A=0.3c{m}^2$

Weight of hydrometer, $W=13.5gf$

Weight of liquid displaced $=W_l$

Density of liquid, ${\rho }_l=?$

Step 2: Formula used

${\rho }_l=\frac{m_l}{v_l}$ ………………………………(a)

Where $m_l$ is the mass of the liquid displaced and $v_l$ is the volume of the liquid displaced

Step 3: Calculation of the density of the liquid

From the basic crux of the law of floatation, it follows that the volume of the liquid displaced will be equal to the volume of the bulb of the hydrometer.

$v_l=V$

$v_l=10.5c{m}^3$ …………………….(b)

From the basic crux of the law of floatation, it follows that the weight of the liquid displaced will be equal to the weight of the hydrometer.

$$\begin{gathered} W_l=W \\ {W}_l=13.5gf \end{gathered}$$

Converting the weight of the liquid displaced into the mass of the liquid displaced, we get

$m_l=\frac{W_l}{g}$

$m_l=13.5g$ ……………………(c)

Substituting the values obtained from equation (b) and equation (c) in equation (a), we get

$$\begin{gathered} {\rho }_l=\frac{13.5g}{10.5c{m}^3} \\ {\rho }_l=1.28g/c{m}^3 \end{gathered}$$

Hence, the density of the liquid is equal to 1.28 kg/m³.