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JEE Main 2023 · Physics · Gravitation · hard

A uniform solid sphere of radius R contains a spherical cavity of radius R/2 whose centre lies at a distance R/2 from the centre of the sphere. Let…

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A uniform solid sphere of radius RR contains a spherical cavity of radius R/2R/2 whose centre lies at a distance R/2R/2 from the centre of the sphere. Let g1g_1 be the gravitational field at the surface point nearest to the cavity and g2g_2 the field at the diametrically opposite surface point, both on the line through the two centres. The ratio g1/g2g_1/g_2 is
  1. A. 12\dfrac{1}{2}
  2. B. 1718\dfrac{17}{18}
  3. C. 917\dfrac{9}{17}
  4. D. 817\dfrac{8}{17}
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Answer: (C) 917\dfrac{9}{17}

Use superposition: cavity == (solid sphere of mass MM) ++ (sphere of mass −M/8-M/8 in the cavity's place), since the cavity volume is 18\tfrac18 of the whole. Let g0=GM/R2g_0=GM/R^2. Nearest point: distance R/2R/2 from the cavity centre, so g1=g0−G(M/8)(R/2)2=g0−g02=g02g_1=g_0-\dfrac{G(M/8)}{(R/2)^2}=g_0-\dfrac{g_0}{2}=\dfrac{g_0}{2}. Farthest point: distance R+R/2=3R2R+R/2=\tfrac{3R}{2} from the cavity centre, so g2=g0−G(M/8)(3R/2)2=g0(1−118)=17g018g_2=g_0-\dfrac{G(M/8)}{(3R/2)^2}=g_0\left(1-\dfrac1{18}\right)=\dfrac{17g_0}{18}. Ratio =1/217/18=917=\dfrac{1/2}{17/18}=\dfrac9{17}.
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