Question

Problem: If the density of the liquid is 835kg/m3, find its specific weight,and specific gravity(S.G)

Answer 1

Answer on Question #52995, Physics / Mechanics | Kinematics | Dynamics

If the density of the liquid is $835\,\mathrm{kg/m^3}$, find its specific weight and specific gravity (S.G)

Solution:

We know that the density of a substance is that quantity of matter contained in unit volume of the substance. According to the condition of the task we have the density of the liquid is $835\,\frac{\mathrm{kg}}{\mathrm{m}^3}$.

$$\text{Specific gravity of a substance} = \frac{\text{Weight of substance}}{\text{Weight of equal volume of water}} = \frac{\text{Density of substance}}{\text{Density of water}}$$ $$\text{Density}, \rho = \frac{\gamma}{g}$$

Specific weight can be calculated from the noted above formula, $\gamma = \rho \cdot g = 835\,\frac{\mathrm{kg}}{\mathrm{m}^3} \cdot 9.81\,\frac{\mathrm{m}}{\mathrm{s}^2} \approx 8.20\,\frac{\mathrm{kN}}{\mathrm{m}^3}$

Now, we can determine the specific gravity (S.G) of the liquid, which is equal to

$$\text{Specific gravity of a substance} = \frac{8.20\,\frac{\mathrm{kN}}{\mathrm{m}^3}}{9.79\,\frac{\mathrm{kN}}{\mathrm{m}^3}} = 0.838$$

Thus, the specific weight is equal to $8.20\,\frac{\mathrm{kN}}{\mathrm{m}^3}$ and the specific gravity (S.G) is equal to 0.838.

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