(1)

Given, $m=100\text{kg}, g=10{\text{m/s}}^2$

$W_A=m{g}_A=mg{\left(\frac{R_E}{R_E+H_B}\right)}^2=mg{\left(\frac{R_E}{R_E+2R_E}\right)}^2=\frac{mg}{9}$

$W_B=m{g}_B=mg{\left(\frac{R_E}{R_E+H_B}\right)}^2=mg{\left(\frac{R_E}{R_E+\frac{3R_E}{2}}\right)}^2=\frac{4mg}{25}$

Change in weight $=W_B-W_A$

$=\frac{4mg}{25}-\frac{mg}{9}=\frac{36mg-25mg}{225}$

$=\frac{11mg}{225}=\frac{11}{225}\times 100\times 10=49$ $\text{N}$

(4)

Given, $M_P=\frac{M_E}{10}$ and $D_P=\frac{D_E}{2}$ or $R_P=\frac{R_E}{2}$

As,  $g_E=\frac{G{M}_E}{R_E^2}$                                ...(i)

        $g_P=\frac{G{M}_P}{R_P^2}$                                ...(ii)

From equation (i) and (ii)

$\frac{g_P}{g_E}=\frac{M_P}{R_P^2}\times \frac{R_E^2}{M_E}=\left(\frac{{\displaystyle M_E}}{{\displaystyle 10}}\right)\times \left(\frac{{\displaystyle 4}}{{\displaystyle R_E^2}}\right)\times \frac{R_E^2}{M_E}$

$\therefore$    $g_P=\frac{2}{5}{g}_E\Rightarrow g_P=\frac{2}{5}\times 9.8=3.92$ ${\text{ms}}^{-2}$

(3)

The relation between acceleration due to gravity and radius of earth is $g=\frac{GM}{R^2}$

We know $M=\rho V$

Then, $g=\frac{4}{3}\pi GR\rho \text{or} \rho =\frac{3g}{4\pi GR}$

(4)

If universal gravitational constant becomes ten times, then $G\text{'}=10G$.

So, acceleration due to gravity increases. i.e., (d) is the wrong option.