Fluid Mechanics

The bulk modulus B is defined as B = -delta_p / (delta_V / V). Rearranging gives delta_V / V = -delta_p / B. Plugging in values: -(1.6 * 10^7) / (5.0 * 10^9) = -0.0032, which is -0.32%.

We know,
P1V1 = P2V2
At the top of the lake P1 = $P _o$ (say)
If h = height of the lake then,
($P _o$ + $\rho g h ) (4/3) \pi r^3$= $P _o$ $(4/3) 8 \pi r^3$
Solving we get, h = 7$P _0$/ $\rho$ g
Now, $P _o$ = 1 X 9.8 X 11
Considering density of water = 1,
we get, h = 11 x 7 = 77 m

In this case, (4/3) $\pi r^3(p _0 +p _1)$ = (4/3) $\pi r^3 (5^3/4^3)P _0$
where, $P _0$ = atm pressure, $P _1$ = water pressure
Rearranging, we get, $P _1 = [(5^3/4^3) - 1] P _0$
Or, $\rho g H = (61/64) P _0$
Or, H = $(61/64) P _0/\rho g$
$P _0 $= 10g and $\rho$ of water = 1
Putting these values we get, H = 9.53m

At a depth of h water pressure will be
${P} _{h} = {P} _{a} + \rho gh$                eq(1)
where 
$\rho  =1000 kg/{m}^{3}$
$g  = 9.81 m/{s}^{2} $
$h = \text{depth  of  water  bubble}$
${P} _{a} = {10}^{5}Pa$

initially bubble is below water at h, so
${P} _{1} = {P} _{h}$
finally it rises to surface of lake so
${P} _{2} = {P} _{a}$

we assume that temperature is constant when it is rising from bottom to surface
by ideal gas equation
${P} _{1}{V} _{1} = {P} _{2}{V} _{2}$

volume can be written as $V = \dfrac{4}{3}\pi {r}^{3}$
where r is radius.
${P} _{h} \times \dfrac{4}{3}\pi {r} _{h}^{3} = {P} _{a} \times \dfrac{4}{3}\pi {r} _{s}^{3}$
${r} _{h} = {(\dfrac{{P} _{a}}{{P} _{h}})}^{1/3} r$
${r} _{h} = \dfrac{r}{{(\frac{{P} _{h}}{{P} _{a}})}^{1/3}}$
by eq(1)
${r} _{h} = \dfrac{r}{{(\dfrac{{P} _{a} +  \rho gh}{{P} _{a}})}^{1/3}}$
${r} _{h} = \dfrac{r}{{(1 + 0.0981h)}^{1/3}}$              eq(2)

(a) at depth 30m
h = 30m
put h=30 in eq(2)
${r} _{h} = \dfrac{r}{1.5}$
false

(b) at depth 70m
h = 70m
put h=70 in eq(2)
${r} _{h} = \dfrac{r}{2}$
true

(c) at depth 140m
h = 140m
put h=140 in eq(2)
${r} _{h} = \dfrac{r}{2.4}$
false

(d) at depth 260m
h = 260m
put h=260 in eq(2)
${r} _{h} = \frac{r}{3}$
true

Answer is D.

Let the new pressure inside be $P$ after air introduced.

Evaporating dish has a shallow bottom to increase the surface area of a liquid.
Evaporating dish is used to evaporate volatile solvent from a solution.

Thin film interference, which causes the colors seen on oil slicks, requires the thickness of the film to be comparable to the wavelength of visible light. Visible light wavelengths are in the range of hundreds of nanometers, or 10^-7 to 10^-6 meters.

Density determines buoyancy. A liquid with lower density (lighter) will float on top of a liquid with higher density (heavier).

Liquids exert equal pressure on the wall of the container at equal depth. This is because the density of the liquid at the same depth will be same.

A standard rain gauge has a funnel with a diameter of 20.3 cm while the measuring cylinder is of 12.5 cm in height. If rain intensity is beyond 12.5 cm, the classification for rainfall is “Very Heavy Rainfall”. Any rain water in excess of 12.5 cm overflows into an outer metal cylinder. This prevents loss of rain water which may otherwise occur through splashing out of the funnel or spilling. The height of 12.5 cm was determined as per the commonly recorded pattern of rainfall in 24 hours. 

The relative permittivity of water id defined as:

Using the equilibrium condition for suspended charges in a medium: K = rho / (rho - sigma).

Saturn is the least dense planet. In fact, its density is less than water and it can hence float on water.