A U tube with both ends open to the atmosphere,is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is 

1. $650$ $kg$ $m^{-3}$

2. $425$ $kg$ $m^{-3}$

3. $800$ $kg$ $m^{-3}$

4. $928$ $kg$ $m^{-3}$

Equal masses of water and a liquid of relative density \(2\) are mixed together, then the mixture has a density of:
1. \(\dfrac{2}{3}\)

2. \(\dfrac{4}{3}\)

3. \(\dfrac{3}{2}\)

4. \(3\)

If two liquids of same masses but densities ${\mathrm{\rho }}_1$ and ${\mathrm{\rho }}_2$ respectively are mixed, then density of mixture is given by

1. $\mathrm{\rho }=\frac{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}{2}$                             

2. $\mathrm{\rho }=\frac{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}{2{\mathrm{\rho }}_1{\mathrm{\rho }}_2}$

3. $\mathrm{\rho }=\frac{2{\mathrm{\rho }}_1{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}$                             

4. $\mathrm{\rho }=\frac{{\mathrm{\rho }}_1{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}$

The density $\mathrm{\rho }$ of water of bulk modulus B at a depth y in the ocean is related to the density at surface ${\mathrm{\rho }}_0$ by the relation

1. $\mathrm{\rho }={\mathrm{\rho }}_0\left[1-\frac{{\mathrm{\rho }}_0\mathrm{gy}}{\mathrm{B}}\right]$                           

2. $\mathrm{\rho }={\mathrm{\rho }}_0\left[1+\frac{{\mathrm{\rho }}_0\mathrm{gy}}{\mathrm{B}}\right]$

3. $\mathrm{\rho }={\mathrm{\rho }}_0\left[1+\frac{\mathrm{B}}{{\mathrm{\rho }}_0\mathrm{gy}}\right]$                             

4. $\mathrm{\rho }={\mathrm{\rho }}_0\left[1-\frac{\mathrm{B}}{{\mathrm{\rho }}_0\mathrm{gy}}\right]$

With rise in temperature, density of a given body changes according to one of the following relations

(a) $\mathrm{\rho }={\mathrm{\rho }}_0\left[1+\mathrm{\gamma } \mathrm{d\theta }\right]$                        (b) $\mathrm{\rho }={\mathrm{\rho }}_0\left[1-\mathrm{\gamma } \mathrm{d\theta }\right]$

(c) $\mathrm{\rho }={\mathrm{\rho }}_0\mathrm{\gamma } \mathrm{d\theta }$                                (d) $\mathrm{\rho }={\mathrm{\rho }}_0/\mathrm{\gamma } \mathrm{d\theta }$