The maximum vertical distance through which a fully dressed astro… - Vidyadip

MCQ

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is $0.5\, m$. If mean density of the moon is two-thirds that of the earth and radius is one quarter that of the earth, the maximum vertical distance through which he can jump on the moon and the ratio of time of duration of jump on the moon to that on the earth are

✓
$3\, m, 6 : 1$
B
$6\, m, 3 : 1$
C
$3\, m, 1 : 6$
D
$6\, m, 1 : 6$

✓

Answer

Correct option: A.

$3\, m, 6 : 1$

a
$(I)$ $g^{\prime}=$ acceleration due to gravity on the moon

$=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{\prime} \rho^{\prime}=\mathrm{G} \cdot \frac{4}{3} \pi \times \frac{\mathrm{R}}{4} \times \frac{2}{3} \rho=\frac{\mathrm{g}}{6}$

$\text { Now, } \quad \mathrm{h}_{\max .}=\frac{\mathrm{u}^{2}}{2 \mathrm{g}} \quad \text { or } \quad \mathrm{h}_{\mathrm{m}} \propto \frac{1}{\mathrm{g}}$

${\text { Hence, }} {\mathrm{h}_{\mathrm{m}}^{\prime} \mathrm{g}^{\prime}=\mathrm{h}_{\mathrm{m}} \mathrm{g}}$

${\mathrm{h}_{\mathrm{m}}^{\prime}=6 \mathrm{h}_{\mathrm{m}}=6 \times 0.5=3 \mathrm{m}}$

$(II)$ $t=\frac{2 u}{g} \quad$ or $\quad t \propto \frac{1}{g}$

$\mathrm{t'g}^{\prime}=\mathrm{tg} \quad$ or $\quad \frac{\mathrm{t}^{\prime}}{\mathrm{t}}=\frac{\mathrm{g}}{\mathrm{g}^{\prime}}=\frac{6}{1}$

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