(1)

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Let gravitational field at point $P$ is zero which is situated at a distance $x$ from mass $m$.

$\therefore$    $\frac{Gm}{x^2}=\frac{G\left(9m\right)}{{(R-x)}^2}$

or,  ${(R-x)}^2=9x^2$

or,  $\frac{R-x}{x}=3 \text{or} x=\frac{R}{4}$                                ....(i)

Gravitational potential at point $P,$ $V_P=-\frac{Gm}{x}-\frac{G\left(9m\right)}{R-x}$

or  $V_P=-\frac{Gm}{R/4}-\frac{G\left(9m\right)}{3R/4} \text{(using (i))}$

or   $V_P=-\frac{Gm}{R}(4+12)=-16\frac{Gm}{R}$

(2)

Here, $F=G\frac{M_1{M}_2}{r^2}$

Gravitational constant,  $G=\frac{F\times r^2}{M_1\times M_2}=\left[M^{-1}{L}^3{T}^{-2}\right]$

Gravitational potential energy, $U=\frac{-G{M}_1{M}_2}{r}=\left[M^1{L}^2{T}^{-2}\right]$

Gravitational potential, $V=-\frac{GM}{r}=\left[M^0{L}^2{T}^{-2}\right]$

Gravitational intensity,  $I=\frac{GM}{r^2}=\left[M^0{L}^1{T}^{-2}\right]$

So, (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

(4)

Work done = Change in potential energy

   $=u_f-u_i=\frac{{\displaystyle -GMm}}{{\displaystyle (R+h)}}-\left(\frac{{\displaystyle -GMm}}{{\displaystyle R}}\right)$

where $M$ is the mass of Earth and $R$ is the radius of Earth.

$\therefore$   $W=GMm[\frac{1}{R}-\frac{1}{(R+h)}]$

Now, $h=R$

$\therefore$    $W=GMm[\frac{1}{R}-\frac{1}{2R}]=\frac{GMm}{2R}$

$\Rightarrow W=\frac{mgR}{2}$   $[\because g=\frac{GM}{R^2}]$

(3)

Gravitation potential at a height $h$ from the surface of earth, $V_h=-5.4\times {10}^7$ $\text{J}$ ${\mathrm{kg}}^{-1}$

At the same point, acceleration due to gravity, $g_h=6$ $\text{m}$$s^{-2}$

$R=6400\text{km}=6.4\times {10}^6\text{m}$

We know, $V_h=-\frac{GM}{(R+h)},$

       $g_h=\frac{GM}{{{\displaystyle (}{\displaystyle R}{\displaystyle +}{\displaystyle h}{\displaystyle )}}^2}=-\frac{V_h}{R+h}$

$\Rightarrow R+h=-\frac{V_h}{g_h}$

$\therefore$   $h=-\frac{V_h}{g_h}-R=-\frac{(-5.4\times {10}^7)}{6}-6.4\times {10}^6$

       $=9\times {10}^6-6.4\times {10}^6=2600$$\text{km}$