When water flows out of a hole in a tank, it is in contact with the edges of the hole over a very small area, and therefore viscosity effects are small

In this case the terminal velocity is upward. Since $\sigma >\rho$, the density of the air

inside the bubble may be neglected completely.

From equation 

$v=\frac{2a^2(\sigma -\rho )g}{g\eta }=\frac{2a^2\rho g}{9\eta }\Rightarrow \eta =\frac{2a^2\rho g}{9v}$

where, a = 5mm, $\rho =1.4\times {10}^{3}kg/m^3,v=4mm/s$

$\therefore \eta =\frac{2\times 25\times {10}^{-6}\times 1.4\times {10}^3\times 9.8}{9\times 4\times {10}^{-3}}kg/m-s=19.05kg/m-s$

Hence, (D) is correct. 

When water flows out of a hole in a tank, it is in contact with the edges of the hole over a very small area, and therefore viscosity effects are small

$A=(0.1{)}^2=0.01m^2$, $\eta =0.01$ Poise = 0.001 decapoise (M.K.S. unit),

$dv=0.1m/s$ and F = 0.002 N

$F=\eta A\frac{dv}{dx}$   $\therefore dx=\frac{\eta Adv}{F}=\frac{0.001\times 0.01\times 0.1}{0.002}=0.0005m$

Apparent weight

$=V(\rho -\sigma )g=l\times b\times h\times (5-1)\times g$

$=(5\times 5\times 5\times 4\times g)$ dyne = $(4\times 5\times 5\times 5)gf$

Volume of log of wood $V=\frac{mass}{density}=\frac{120}{600}=0.2m^3$

Let x weight that can be put on the log of wood.

So weight of the body = $(120+x)\times 10N$

Weight of displaced liquid = $V\rho g=0.2\times {10}^3\times 10N$

The body will just sink in liquid if the weight of the body will be equal to the weight of displaced liquid.

$(120+x)\times 10N=0.2\times {10}^3\times 10N$
$\Rightarrow x=80kg$

If two liquids of equal masses and different densities are mixed together then density of mixture 

$\rho =\frac{2{\rho }_1{\rho }_2}{{\rho }_1+{\rho }_2}=\frac{2\times 1\times 2}{1+2}=\frac{4}{3}$

Apparent weight = $V(\rho -\sigma )g=\frac{M}{\rho }(\rho -\sigma )g$

$=M⟮1-\frac{\sigma }{\rho }⟯g=2.1⟮1-\frac{0.8}{10.5}⟯g=1.94gN$

$=1.94Kg-wt$

The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second

$=Av=A\sqrt{2gh}$                ......(i)

Volume of water flowing in per second

=70 $c{m}^3/sec$                     ......(ii)

From (i) and (ii) we get

$A\sqrt{2gh}=70\Rightarrow 1\times \sqrt{2gh}=70$

$\Rightarrow 1\times \sqrt{2\times 980\times h}=70$

$\therefore h=\frac{4900}{1960}=2.5cm$

If the liquid is incompressible then mass of liquid entering through left end, should be equal to mass of liquid coming out from the right end.

$\therefore M=m_1+m_2\Rightarrow A{v}_1=A{v}_2+1.5A.v$

$\Rightarrow A\times 3=A\times 1.5+1.5A.v\Rightarrow v=1m/s$