A body has a weight 90, kg on earth's surface, mass of moon is 1/9 that of earth's mass and

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7.Gravitation

medium

A body has a weight $90\, kg$ on the earth's surface, the mass of the moon is $1/9$ that of the earth's mass and its radius is $1/2$ that of the earth's radius. On the moon the weight of the body is .......... $kg$

$45$

$202.5$

$90$

$40$

Solution

(d) $\frac{{{g_m}}}{{{g_e}}} = \frac{{{M_m}}}{{{M_e}}} \times {\left( {\frac{{{R_e}}}{{{R_m}}}} \right)^2} = \left( {\frac{1}{9}} \right)\;{\left( {\frac{2}{1}} \right)^2} = \frac{4}{9}$ $⇒$ ${g_m} = \frac{4}{9}{g_e}$

${W_{\rm{m}}} = \frac{4}{9} \times {W_e} = \frac{4}{9} \times 90 = 40\;kg$

Standard 11

Physics

The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$)

medium

In the reported figure of earth, the value of acceleration due to gravity is same at point $A$ and $C$ but it is smaller than that of its value at point $B$ (surface of the earth). The value of OA $: AB$ will be $x : y .$ The value of $x$ is $\ldots \ldots \ldots$

hard

(JEE MAIN-2021)

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easy

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normal

The weight of a body at the surface of earth is $18\,N$. The weight of the body at an altitude of $3200\,km$ above the earth's surface is $……..\,N$ (given, radius of earth $R _{ e }=6400\,km$ )