Note on Specific Gravity and Hydrometer

• Note
• Things to remember

Specific Gravity

Specific gravity of a substance is defined as the ratio of the mass of certain volume of it to the mass of same volume of water at 40C. So,

\begin{align*} \text {Specific gravity of a body} &= \frac {\text {mass of certain volume of substance}}{\text {mass of equal volume of water at} 4^o C} \\ &= \frac {\text {mass of 1 c.c. of a substance}}{\text {mass of 1c.c. of water at} 4^o C} \\ &= \frac {\rho }{\rho _w at 4^o C} \\ &= \frac {\rho }{1 gm/c.c.} \end{align*}

Since ρw = 1 gm/cm3 at 4oC, specific gravity of a substance is numerically equal to its density in C.G.S. system.

Again,

Specific gravity of a body $$= \frac{mass \;of \;certain \;volume \;of \;substance}{mass \;of equal\; volume\; of \;water \;at \;4^o C}$$

$$= \frac {weight \;of \;certain \;volume \;of \;substance} {weight\; of \;equal\; volume\; of \;water\; at \;4^o C}$$

\begin{align*}= \frac {\text {weight of V c.c. of a substance}}{\text {weight of Vc.c. of water at room temperature} } \times \frac {\text {weight of V c.c. of water at room temperature}}{\text {weight of Vc.c. of water at} 4^o C \text {temperature} } \\ \\ \end{align*}

\begin{align*}= \frac {\text {weight of V c.c. of a substance}} {\text {loss in weight of substance in water}} \times \text {specific gravity of water at room temperature} \\ \end{align*}

Measurement of Specific Gravity of Substances
1. An object heavier than water (ρ>ρw)
an object is weighted in air, and than in water. Let W1 and W2 be the weights in air and in water respectively. Loss in weight = W1 – W2. Then,
\begin{align*} \text {Specific gravity of object} &= \frac {\text {weight of object in air}}{\text {loss in weight in water}} \\ &= \frac {W_1}{W_1 – W_2} \times \text {specific gravity of water at room temperature} \\ \end{align*}
2. An object that floats in water (ρ<ρw)
Suppose an object such as cork which floats in water. Let weight of object in air = W1. Another object which can sink in water is tied on the first object, and with sinker in water and the object in air, weight of both system is taken. Let weight of the object in air and sinker in water = W2. Then, weight of the object and sinker both in water is taken. Let weight of object and sinker both in water = W3.
\begin{align*} \text {loss in weight of object in water} &= W_2 - W_3 \\ \text {Specific gravity of object} &= \frac {W_1}{W_2 – W_3} \times \text {specific gravity of water at room temperature} \\ \end{align*}
3. Specific gravity of a liquid

A body which can sink in the liquid and in water is taken. Let its weight in air be W1.
\begin{align*} \text {Let, weight of the solid in water} = W_2 \\ \text {weight of the solid in liquid} =W_3 \\ \text {weight of V.c.c. of liquid} = W_1 – W_2 \\ \text {weight of V.c.c. of water} = W_1 – W_3 \\ \end{align*}

Specific gravity of liquid =\begin{align*} \frac {\text {weight of certain volume of liquid}} {\text {weight of equal volume of water}} \\\times \text {specific gravity of water at room temperature} \text {Specific gravity} &= \frac {W_1 – W_3}{W_1 – W_2} \\\times \text {specific gravity of water at room temperature} \end{align*}

4. Specific gravity of an object soluble in water (specific gravity of salt)
\begin{align*} \text {Let weight of salt in air} = W_1 \\ \text {weight of the salt in kerosene} = W_2 \\ \text {weight of displaced kerosene} = W_1 – W_2 \\ \text {Then, take a solid sinker which is insoluble in water and in kerosene} \\ \text {Let weight of solid in air} = W_3 \\ \text {weight of solid in water} = W_4 \\ \text {weight of solid in kerosene} = W_5 \\ \text {weight of certain volume of water} = W_3 - W_4 \\\text {weight of equal volume of kerosene} = W_3 - W_5 \\ \text {Specific gravity of salt} = \end{align*} sp. gravity of salt with respect to kerosene $$\times$$ sp. gravity of kerosene with respect to water $$\times$$sp. gravity of water at room temperature.\begin{align*} \\ = \frac {W_1}{W_1 – W_2} \times \frac {W_3 –W_5}{W_3 – W_4} \times \text {specific gravity of water at room temperature} \end{align*}

Hydrometer

A hydrometer is a device used to measure the specific gravity of a solid or liquid. It floats in a liquid and so works in the principle of floatation; the weight of the floating body is equal to the weight of the displace liquid.

There are two types of hydrometer: the variable immersion type and constant immersion type. In both types, the hydrometer consists of a hollow cylinder that floats in the liquid with a stem at upper part.

1. To measure specific gravity of solid
The hydrometer floats in water with the index mark well above the water level. Let W1 be the weight on the upper pan needed to sink the meter up to the index mark. Some weight and small solid on upper pan needed to sink the meter up to the index. Let W3 be weight on upper pan and solid lower conical bowl that immerse the meter to index mark
\begin{align*} \text {So, weight solid in air} = W_1 – W_2 \\ \text {The weight of solid in water} = W_1 – W_3 \\ \text {The loss in weight of solid in water} = W_1 – W_2 –(W_1 –W_3) = W_3 –W_2 \\ \text {Specific gravity} &= \frac {W_1 – W_2}{W_3 – W_2} \times \text {specific gravity of water at room temperature} \end{align*}
2. To measure the specific gravity of liquid
Let, W1 = weight of hydrometer.
W2 = weight on pan needed to sink the meter in water to the index.
To find the specific gravity of a liquid, the hydrometer is immersed in it and some weights are kept on the pan as shown in figure. Let W3 be the weight on pan needed to sink the meter in liquid to the index.
\begin{align*} \text {So, weight of displaced liquid} &= \text {weight of floating body} \\ &= W_1 + W_3 \\ \text {Similarly, weight of the displaced water} &= \text {weight of floating body} \\ &= W_1 + W_2 \\ \text {Since the both liquids have same volume,} \\ \text {Specific gravity of the liquid} &= \frac {W_1 + W_2}{W_3 + W_2} \times \text {specific gravity of water at room temperature} \end{align*}

Specific Gravity Bottle

Specific gravity bottle is a device used to measure specific gravity of a liquid as well as a solid.

1. Specific gravity of a liquid
The specific gravity bottle is first cleaned with water and dried, and its weight is taken. Let W1 be the weight of the bottle.
The bottle is completely filled with water and with stopper, the weight is taken, let it would be W2. The bottle is completely filled with the liquid and is weighted again. Let W3 be the weight of the bottle and the liquid.
\begin{align*} \text {Therefore, weight of water} = W_2 – W_1 \\ \text {Weight of liquid} = W_3 – W_1 \\ \text {Specific gravity of liquid} &= \frac {\text {weight of liquid}} {\text {weight of equal volume of water}} \times \text {specific gravity of water at room temperature} \\ &= \frac {W_3 – W_1}{W_2 – W_1} \times \text {specific gravity of water at room temperature} \\ \end{align*}
2. Specific gravity of a solid
To find the specific gravity of a solid, weigh the clean empty bottle. Let W1 be this weight. Put the solid about 1/3 of volume of bottle in it and weigh again. Let it would be W2. Then fill the remaining volume of the bottle with water (if the solid is in water, use other liquid in which it does not melt) and weigh the bottle. Let W3 be the weight of the system.
Pour out all content of the bottle, and fill it completely with water and weight. Let W4 with the new weight.
\begin{align*} \text {Now, Weight of solid} = W_2 – W_1 \\ \text {Weight of water alone in bottle} = W_4 – W_1 \\ \text {Weight of water partially filled in bottle} = W_3 – W_2 \\ \text {Weight of water whose volume is equal to that of solid} = (W_4 –W_1) – (W_3 – W_2) \\ \text {Specific gravity} = \frac { W_3 – W_1}{(W_4 –W_1) – (W_3 – W_2)} \\ \end{align*}

Specific gravity of a substance is defined as the ratio of the mass of certain volume of it to the mass of same volume of water at 40C. So,

\begin{align*} \text {Specific gravity of a body} &= \frac {\text {mass of certain volume of substance}}{\text {mass of equal volume of water at} 4^o C} \\ &= \frac {\text {mass of 1 c.c. of a substance}}{\text {mass of 1c.c. of water at} 4^o C} \\ &= \frac {\rho }{\rho _w at 4^o C} \\ &= \frac {\rho }{1 gm/c.c.} \end{align*}

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