Question
Download Solution PDF2 litres of an oil weigh 16 N. Calculate the specific gravity of the oil. Consider the density of water to be 998 kg/m3.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Relative density/specific gravity = \({Density\:of\:fluid \over Density\:of\:standard\:fluid}\) = \( {\rho_{f} \over \rho_{w}}\)
Density of fluid = ρ = \({Weight \over Volume}\)
The specific gravity of water (Standard Fluid) is 1.0 at 4 °C
Given:
Weight of Oil =16N, Volume of Oil = 2L = 0.002 m3
Density of water = ρw = 998 Kg/m3
Calculation:
Density of Oil =\({Weight \over Volume}\) = \({16 \over 0.002}\) =8KN/m3
ρoil = \({8000 \over 9.81}\) = 815.49 Kg/m3
Specific gravity = \( {\rho_{oil} \over \rho_{water}}\) = \({815.49 \over 998}\) = 0.816
G = 0.816 (ans)
Last updated on Jul 1, 2025
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