# Answer this question

Soln (a):

$h=ut+\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}Forfreefall:initialvelocity,u=0\phantom{\rule{0ex}{0ex}}h=\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}g=\frac{2h}{{t}^{2}}=\frac{2\times 5}{5\times 5}=0.4m/{s}^{2}$

Soln b:

Gravitational force on a body of mass m due to planet A: $G\frac{{M}_{A}m}{{{R}_{A}}^{2}}=mg$

Gravitational force on a body of mass m due to planet B: $G\frac{{M}_{B}m}{{{R}_{B}}^{2}}=mg\text{'}$

Given, ${R}_{A}=\frac{{R}_{B}}{2}$

For $g\text{'}=\frac{g}{2}\phantom{\rule{0ex}{0ex}}\frac{G{M}_{B}}{{{R}_{B}}^{2}}=\frac{1}{2}\frac{G{M}_{A}}{{{R}_{A}}^{2}}=\frac{4}{2}\frac{G{M}_{A}}{{{R}_{B}}^{2}}\phantom{\rule{0ex}{0ex}}\therefore {M}_{B}=2{M}_{A}$

Soln (c):

Mass on earth, m = 5 kg

Weight on earth, w = mg = 49N

$\therefore g=9.8m/{s}^{2}$

${g}_{moon}=\frac{1}{6}\times g=\frac{1}{6}\times 9.8=1.63m/{s}^{2}$

mass on moon will remain same i.e. 5 kg

weight on moon = $\frac{1}{6}\times 49=8.17N$

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