The ratio of respective acceleration due to gravity on the surface of two planets having masses in the ratio x and density in the ratio y is

\frac{x^{2}}{y^{2}}

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\frac{2}{3}

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x^{2} y

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(x y^{2})^{1 / 3}

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Solution

The correct option is D $(x y^{2})^{1 / 3}$
Acceleration of due to gravity is given by
$\begin{matrix} m g = \frac{G m n}{R^{2}} [o n s u r f a c e o f p l a n e w i t h r a d i i R a n d m a s s M] I f ρ b e t h e d e n s i t y T h e n, ρ = \frac{3 M}{4 π R^{3}} = {(\frac{3 M}{4 π ρ})}^{\frac{1}{3}} g = \frac{G M}{{(\frac{3 M}{4 π ρ})}^{\frac{2}{3}}} = {(\frac{4 π}{3})}^{\frac{2}{3}} G M^{\frac{1}{3}} ρ^{\frac{1}{3}} G i v e n \frac{ρ_{1}}{ρ_{2}} = y \frac{M_{1}}{M_{2}} = k H e n c e, \frac{g_{1}}{g_{2}} = {(k y^{2})}^{\frac{1}{3}} H e n c e, o p t i o n D i s t h e r e q u i r e d a n s w e r . \end{matrix}$

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The ratio of respective acceleration due to gravity on the surface of two planets having masses in the ratio x and density in the ratio y is

The correct option is D (xy2)1/3Acceleration of due to gravity is given by mg=GmnR2[onsurfaceofplanewithradiiRandmassM]IfρbethedensityThen,ρ=3M4πR3=(3M4πρ)13g=GM(3M4πρ)23=(4π3)23GM13ρ13Givenρ1ρ2=yM1M2=kHence,g1g2=(ky2)13Hence,optionDistherequiredanswer.