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A cavity of radius R/2 is made inside a solid sphere of radius R. The centre of the cavity is located at a distance R/2 from the centre of the sphere. The gravitational force on a particle of mass ' m ' at a distance R/2 from the centre of the sphere on the line joining both the centres of sphere and cavity is (opposite to the centre of cavity).

[Here  g=\frac{GM}{R^{2}}, where M is the mass of the sphere ] 

Option: 1

\frac{m\: g}{2}


Option: 2

\frac{3\: g\: m}{8}


Option: 3

\frac{m\: g}{16}


Option: 4

 none of these


Answers (1)

best_answer

Gravitation field at mass ‘m’ due to full solid sphere

\vec{E}_{1}=\frac{\rho \vec{r}}{3\: \varepsilon _{0}}=\frac{\rho R}{6\; \varepsilon _{0}}....\left [ \varepsilon _{0}=\frac{1}{4\pi G} \right ]

Gravitational field at mass ‘m’ due to cavity \left ( -\rho \right )

       

\vec{E}=\frac{\left ( -\rho \right )\left ( R/2 \right )^{3}}{3\: \varepsilon _{0}R^{2}}

=\frac{\left ( -\rho \right )R^{3}}{24\varepsilon _{0}R^{2}}=\frac{-\rho R}{24\varepsilon _{0}}

Net gravitational field \vec{E}=\vec{E}_{1}+\vec{E}_{2}=\frac{\rho R}{6\varepsilon _{0}}-\frac{\rho R}{24\varepsilon _{0}}=\frac{\rho R}{8\varepsilon _{0}}

Net force on  'm'\rightarrow F=\frac{m\rho R}{8\varepsilon _{0}}

Here \rho =\frac{M}{4/3\pi R^{3}}   &   \varepsilon _{0}=\frac{1}{4\pi G},   then   F=\frac{3mg}{8}

 

 

Posted by

Ritika Jonwal

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