A) 1 m
B) 0.8 m
C) 0.5 m
D) 125 m
Correct Answer: B
Solution :
[b] to K.E. of the person = P.E. at the maximum height \[\frac{1}{2}m{{v}^{2}}=m\frac{GM}{{{R}^{2}}}h.\] \[\therefore m\frac{GM}{{{R}^{2}}}.h=mG\frac{M'}{R{{'}^{2}}}h'\] Given \[\rho =\frac{4}{5}p',\frac{R}{R'}=\frac{2}{3}\] \[\therefore mG.\frac{4}{3}\frac{\pi {{R}^{3}}\rho }{{{R}^{2}}}.h=mG.\frac{4}{3}\frac{\pi R{{'}^{3}}\rho '}{R{{'}^{2}}}.h'\] \[\Rightarrow \frac{2}{3}R.\frac{4}{5}\rho '1.5=R'\rho 'h'\therefore h'=0.8m.\]
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