Sphere Geometry

$\displaystyle \frac{V'}{V} = \frac{172.8}{100} = \frac{\displaystyle \frac{4}{3} \pi R^{.3}}{\displaystyle \frac{4}{3} \pi R^3}$
$\displaystyle \frac{R'}{R} = 1.2$
Now, ratio of surface area $= \displaystyle \frac{S'}{S} = \frac{4 \pi R^{.2}}{4 \pi R^3}$
$= \displaystyle \frac{S'}{S} = 1.44$
Hence surface area increased by 44%

Curved surface area of sphere$=4\pi r^2$
diametre after decreases by $25\%$
$A _D=\pi d^2$
$d _1=\displaystyle\frac{75}{100}d=\frac{3d}{4}$
$A _1=\pi \left(\displaystyle\frac{3d}{4}\right)^2=\frac{9}{16}\pi d^2$
$\%$ decrease=$\displaystyle\frac{\displaystyle\frac{9}{16}\pi d^2-\pi d^2}{\pi d^2}\times 100=-43.75\%$

The significant figures are the number of digits in a value, that are significant. In $1.06m$, the number of significant figures is 3. In the answer also we must maintain 3 significant figures only. Hence option B is correct.

Let n number of grains are in the vessel.

So, volume of cuboid vessel is equal to n times volume of spheres. i.e.

$ 1\times 0.6\times 0.5=n\left( \dfrac{4}{3}\times 3.14\times {{\left( \dfrac{3}{2000} \right)}^{3}} \right) $

$ 0.3=n\dfrac{4}{3}\times 3.14\times 3.375\times {{10}^{-9}} $

$ 0.3=n\times 14.13\times {{10}^{-9}} $

$ n=\dfrac{0.3}{14.13\times {{10}^{-9}}} $

$ n=0.021\times {{10}^{9}}\, $

$ n=2\times {{10}^{7}}$

The size of a nanoparticle is  about 50000 times smaller than the diameter of a single strand of human hair and 1/100th of a typical bacterium.

Thermal expansion affects the entire material uniformly, including the boundaries of any cavities. As the metal expands outward, the cavity effectively expands as if it were made of the same material.

Volume of wire (cyl)=Volume of sphere 
$\displaystyle \Rightarrow \pi r^{2}\times 72=\dfrac{4}{3}\pi \times 3^{3}$
$\displaystyle \Rightarrow r=\dfrac{1}{\sqrt{2}}cm$
Thus diameter$\displaystyle \dfrac{2}{\sqrt{2}}cm=\sqrt{2}cm=1.4cm$

The average size of the colloids is of the order of $10^{-7} :m$ to $10^{-9}:m$.

The particle size of the colloidal particle in dispersed phase typically ranges from $1 \ nanometer$ to $ 1 \ micrometer$ or. ${10}^{-6}$ to ${10}^{-9}m$

Planets are like spheres. Volume of a sphere is given by, 
V = $\displaystyle{\frac{4}{3}\pi r^3}$ 
Let the r$ _1$, r$ _2$ be radii of the Jupiter and the Earth respectively. 
r$ _1$ $= 11r _2$ _(1)
Volume of Jupiter, V$ _1$ = $\displaystyle{\frac{4}{3}\pi r^3 _1}$ _(2) 
Volume of Earth, V$ _2$ = $\displaystyle{\frac{4}{3}\pi r^3 _2}$ _(3) 
$\displaystyle{\frac{(2)}{(3)} \Rightarrow \frac{V _1}{V _2} = \frac{\frac{4}{3}\pi r _1^3}{\frac{4}{3} \pi r _2^3} = \frac{r _1^3}{r _2^3}}$ 
$\displaystyle{\frac{V _1}{V _2} = \frac{r _1^3}{r _2^3} = \frac{(11r _2)^3}{r _2^3} = (11)^3}$ = 1331 
Jupiter can accommodate 1331 number of Earths within it.

Given the external  radius of hollow sphere is 4 cm and internal radius 3 cm 

The Volume of a sphere$=\dfrac{4}{3}\pi \ \text{R}^{3}$
Here Radius $R=7.5 \ cm$
On solving We get Volume$=1767.15\ {cm}^{3}$ 

Radius $=\dfrac{1}{2}\times\sqrt{\dfrac{\text{surface area}}{\pi}}$
on solving we get radius $=7$ cm