Question

Problem: If the specific gravity of agiven oil is 0.750, find its density and specific weight. It is given that the density and specific weight of the water at 4 0C is 1000kg/m3 and 9810N/m3 respectively

Answer 1

Answer on Question #52996 - Physics - Mechanics - Kinematics - Dynamics

If the specific gravity of given oil is $SG = 0.750$, find its density $\rho$ and specific weight $\gamma$. It is given that the density and specific weight of the water at $4{}^{\circ}\mathrm{C}$ is $\rho_{H_2O} = 1000\frac{\mathrm{kg}}{\mathrm{m}^3}$ and $\gamma_{H_2O} = 9810\frac{\mathrm{N}}{\mathrm{m}^3}$ respectively.

Solution:

The specific gravity is given by

$$SG = \frac{\rho}{\rho_{H_2O}}$$

Since $SG = 0.750$, and $\rho_{H_2O} = 1000\frac{\mathrm{kg}}{\mathrm{m}^3}$, we obtain

$$\rho = SG \cdot \rho_{H_2O} = 0.750 \cdot 1000\frac{\mathrm{kg}}{\mathrm{m}^3} = 750\frac{\mathrm{kg}}{\mathrm{m}^3}$$

The specific weight is given by

$$\gamma = g \cdot \rho,$$

where $g$ – is the acceleration due to gravity. Therefore,

$$g = \frac{\gamma_{H_2O}}{\rho_{H_2O}}$$

and

$$\gamma = \frac{\gamma_{H_2O}}{\rho_{H_2O}} \cdot \rho = \frac{9810\frac{\mathrm{N}}{\mathrm{m}^3}}{1000\frac{\mathrm{kg}}{\mathrm{m}^3}} \cdot 750\frac{\mathrm{kg}}{\mathrm{m}^3} = 7357.5\frac{\mathrm{N}}{\mathrm{m}^3}$$

Answer:

$\rho = 750\frac{\mathrm{kg}}{\mathrm{m}^3}, \gamma = 7357.5\frac{\mathrm{N}}{\mathrm{m}^3}.$

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