Question

if a depth of liquid of 1.0meter causes pressure of 7 kpa,what is the specific gravity of the liquid

Answer 1

$p=\rho gh; \rho=\frac{p}{gh}=\frac{7\times10^3}{9.8\times 1}=714.29\frac{kg}{m^3}\\or\ 0.71429\frac{g}{cm^3}.\\Answer:the\ specific\ gravity\ of\ the\ liquid\\ is\;714.29\frac{kg}{m^3}\;or\ 0.71429\frac{g}{cm^3}.$