Answer: Option D. -> 20 m
:
D
Brandon experienced a free fall from the top of the tower and, thus, was under the acceleration due to gravity (g).
Distance travelled in 2 seconds is equal to the height of tower which can be found out by applying equation of motion.
From second equation of motion, we have,
$s=u\times t+\frac{1}{2}\times g{t}^2$
where s- height of tower, u- initial velocity, t-time taken and g- acceleration due to gravity.
Since he fell from rest, u = 0
$s=\frac{1}{2}\times g{t}^2=\frac{1}{2}\times 10\times 2^2=20m$
Therefore, height of tower is 20 m.

Answer: Option A. -> When density of the object is less than that of the liquid.
:
A
When an object is immersed in a liquid, the liquid exerts a force upwards on the object called 'Upthrust'. Also, the force due to gravitational attractionof the earth actson the object downwards.
Upthrust is equal to the weight of the liquid displaced. So if density of the object is less than that of the liquid then lesser liquid has to be displaced in order to balance the weight of the object and thus the object floats.

Answer: Option B. -> 10 N
:
B
Given,
Length of each side of the cube, $l=10cm=0.1m$
Density of water, $\rho =1000kg{m}^{-3}$
Acceleration due to gravity, $g=10m{s}^{-2}$
Volume of the cube, $V=l^3=(0.1{)}^3=0.001m^3$
Archimedes' principle states that the buoyant force acting on an object is equal to the weight of the fluid displaced by it.
In the given case, volume of cube is equal to the volume of water displaced.
Mass of waterdisplaced (m) = Density of water $\times$ Volume of water displaced
$\Rightarrow m=1000kg{m}^{-3}\times 0.001m^3$
$=1kg$
Weight of water displaced, W =mg
$\Rightarrow W=1\times 10$ =$10N$
Hence, the buoyant force is $10N$.

Answer: Option D. -> it supports increased water pressure at the depth
:
D
The pressure in the liquid increases with depth. Thus, as depth increases, more and more pressure is exerted by water on the wall of a dam. A thicker wall is required to withstand a greater pressure, therefore, the wall of the dam is made with the thickness increasing towards base.

Answer: Option A. -> Gravitational force of Earth
:
A
The atmosphere is held to the Earth due togravitational force ofEarth. The gases in the atmosphere have a mass and henceEarth attracts them, as a result of which the gases do not escape from Earth.

Answer: Option C. -> 6.7×10−11 Nm2kg−2
:
C
$G$ is universal gravitational constant. Its value remains constant everywhere in the universe and gravitational force between bodies $m$ and $M$ separated by distance $r$ is given by:
$F=\frac{GMm}{r^2}$
$\Rightarrow G=\frac{F{r}^2}{Mm}$
$\therefore$ Units of G are $N{m}^{2}k{g}^{-2}$
The value of Gravitational constant is $6.7\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer: Option B. -> F1=F2
:
B
Newton's third law of motion states that: For every action, there is an equal and opposite reaction.So, $F_1=F_2$. Both forces will remain the same as the value of $G$(Universal gravitational constant), the mass of the earth ($m_1$), the mass of the sun ($m_2$) and the distance between the two bodies$\left(r\right)$ do not change for both cases. However, the acceleration of the bodies will be inversely proportional to their masses.

Answer: Option D. -> It does not depend on any parameter
:
D
The value of Gravitational constant is a universal constant and does not depend on any factor. It remains constant anywhere in universe and its value is equal to $6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer: Option A. -> Pa and N
:
A

We know , Pressure $=\frac{Force}{Area}$

So, unit of pressure is $N{m}^{-2}$ which is called Pascal (Pa).
Since thrust is the perpendicular force acting on a body, its unit is Newton (N).

Answer: Option C. -> $F\propto \frac{1}{R^2}$
:
C

According to the Universal law of gravitation, the gravitational force between two bodies is directly proportional to product of their masses and inversely proportional to the square of the distance between them. Thus, in this case, the gravitational force between the point masses is inversely proportional to $R^2$.
Hence, $F\propto \frac{1}{R^2}$