Oct 30, 2016

$1. \dot{6}$

#### Explanation:

Operation of specific gravity meter is based on Archimedes' principle.
It states that when a solid body is suspended in a fluid it experiences an upward force which is equal to the weight of the fluid displaced by the submerged part of the body.

The depth is read off the scale, as shown in the figure below:

When submerged to the marking $1.0$, it corresponds to fluid being water at ${4}^{\circ} \text{C}$ and $1$ atm of pressure.

The position of the marking of $1.0$ is not clear from the statement of the problem. As such if the tube sinks in the sample to a point above the mark, the specific gravity of the sample is lower than the that of water and vice-versa.

Assuming that the $1.0$ mark is placed at the $10 c m$ length of tube.
Let $A$ be the area of cross-section of the tube.
Its volume which is also Volume of water displaced $V = 0.1 A$
As the tube is in equilibrium (upwards thrust$=$downwards weight)
Weight of tube $=$Weight of the water displaced
Weight of tube $= 0.1 \times A \times 1000 g$
where $g$ is acceleration due to gravity.

Now the meter floats with $6 c m$ of its length submerged in the liquid whose density is $d$ and is in equilibrium,
Weight of the liquid displaced $=$ Weight of tube
$0.06 \times A \times \mathrm{dg} = 0.1 \times A \times 1000 g$
$\implies 0.06 \times \frac{d}{1000} = 0.1$
We know that SG$= \frac{d}{d} _ \left(w a t e r\right)$, therefore we get
${\text{SG}}_{l i q u i d} = \frac{0.1}{0.06} = 1. \dot{6}$

Oct 30, 2016

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#### Explanation:

We know that the density of water at ${4}^{\circ} C$ is $1000 \frac{k g}{m} ^ 3 \mathmr{and} 1 g c {m}^{-} 3$.So specific gravity of water will be 1.0.

The given uniform tube of length 10cm has also sp.gravity 1.0.

If the area of cross section of the tube be $A c {m}^{2}$ then its volume will be $10 A c {m}^{3}$

And its weight becomes $= 10 A \times 1 \times g$ dyne.
$\text{where " g->"Acceleration due to gravity}$

If it floats in water immersing x cm length in water,then weight of water displaced by it will be $= x \cdot A \cdot 1 \cdot g \text{ }$dyne

By Archimedes principle

$x \cdot A \cdot g = 10 \cdot A \cdot g$

So $x = 10 c m$ Hence it floats in water submerged.

Again it floats in the given liquid keeping 6cm immersed.

So the weight of displaced liquid will be$= 6 \cdot A \cdot \rho \cdot g$ dyne
,where $\rho$ is the sp.gravity of the liquid.

Now applying Archimedes principle we have

$\to 6 \cdot A \cdot \rho \cdot g = 10 A \cdot 1 \cdot g$

$\implies \rho = \frac{10}{6} \approx 1.67$