A.

$\frac{1}{\mathrm{R}}$

Kinetic energy of the satellite

$$\begin{gathered} \mathrm{KE} = \frac{1}{2}{\mathrm{mv}}_{\mathrm{o}}^2 ...... 1 \\ \mathrm{Where} {\mathrm{v}}_{\mathrm{o}} =\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \\ \mathrm{Now} \mathrm{putting} \mathrm{the} \mathrm{value} \mathrm{of} {\mathrm{v}}_{\mathrm{o}} \mathrm{is} {\mathrm{eq}}^{\mathrm{n}}\left(1\right) \\ \mathrm{KE} =\frac{1}{2}\mathrm{m}{\left(\sqrt{\frac{\mathrm{GM}}{{\mathrm{R}}^2}}\right)}^2 \\ =\frac{1}{2}\frac{\mathrm{mGM}}{\mathrm{R}} \\ \therefore \mathrm{KE} \propto \frac{1}{\mathrm{R}} \end{gathered}$$

So the kinetic energy of the satellite is inversely proportional 

to R. Therefore R increases then KE decreases.

B.

4 s

When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by

$$\begin{gathered} \nu \ge \sqrt{\mathrm{g} \mathrm{r}} \\ also \nu = r\omega =r\frac{2\mathrm{\pi }}{T} \\ Hence, \frac{2\mathrm{\pi r}}{T}\ge \sqrt{r g} \\ T\le \frac{2\mathrm{\pi r}}{\sqrt{r g}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{r}}{\mathrm{g}}} \\ =2\mathrm{\pi }\sqrt{\frac{4}{9.8}} \\ =2\times 3.14\times 0.6389 \\ =4.0 \sec \\ \mathrm{so} {\mathrm{T}}_{\max } =4 \sec \end{gathered}$$

C.

400 m/s

B.

2 × 10² cm³

Using relation for volume

Given:- length of a block  =  12 cm

          breadth of a block  =  6 cm

        thickness of a block  =  2.45cm

V =length  × breadth  × thickness

  =12  × 6  × 2.45  = 176.4

 =1.764 ×10² cm³

The minimum number of significant figures is 1 in thickness, hence the volume will contain only one significant figure.

Therefore V= 2 ×10² cm³ 

B.

0.1 kg m²

The moment of inertia of given system that contains 5 particles each of mass = 2 kg on the rim of circular disc of radius 0.1 m and of negligible mass is given by :

MI of disc + MI of particle

Since the mass of the disc is negligible

∴ MI of the system = MI of particle

= 5 × 2 × (0.1)²

= 0.1 Kg m²