A) 1m
B) \[0.8\] m
C) \[0.5\] m
D) \[1.25\]m
Correct Answer: B
Solution :
Given \[\frac{{{R}_{e}}}{{{R}_{P}}}\,=\frac{2}{3}\] and \[\frac{{{d}_{e}}}{{{d}_{P}}}=\frac{4}{5}\] As, \[Mg=gR_{e}^{2}\] and \[M={{d}_{e}}\times \frac{4}{3}\pi R_{e}^{3}\] \[{{d}_{e}}\times \,\frac{4}{3}\,\pi {{R}_{e}}\times G={{g}_{e}}\] ?(i) Similarly for planet, \[{{d}_{P}}\times \,\frac{4}{3}\,\pi {{R}_{P}}\,G={{g}_{p}}\] ?(ii) Dividing eq. (i) by eq. (ii) we get \[\frac{{{g}_{e}}}{{{g}_{P}}}=\frac{{{R}_{e}}}{{{R}_{P}}}\,\times \frac{{{d}_{e}}}{{{d}_{p}}}\] \[\Rightarrow \,\,\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{2}{3}\times \,\frac{4}{5}\,=\frac{8}{15}\] \[\because \,\,h=\frac{{{V}^{2}}}{2g}\,\,\,\Rightarrow \,\,\frac{{{h}_{p}}}{{{h}_{e}}}\,=\frac{{{g}_{e}}}{{{g}_{p}}}=\frac{8}{15}\] \[\Rightarrow \,\,{{h}_{p}}=\frac{8}{15}\times 1.5\,=\frac{8}{15}=0.8m\]
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