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Question Number 173558 by MadhumitaSamanta last updated on 13/Jul/22

A metal sphere has radius R and mass m. A spherical hollow of diameter R is made in this sphere such that its surface passes through the centre of the metal sphere and touches the outside surface of the metal sphere. A unit mass is placed at a distance from the centre of the metal sphere. The gravitational field at that point is (a) ((GM)/R^2 ) (1−(1/(8(1−((2R)/a))^2 ))) (b) ((GM)/a^2 ) (1−(1/(8(1−(R/(2a)))^2 ))) (c) ((GM)/((R+a)^2 )) (1−(1/(8(1−(R/(2a)))^2 ))) (d) ((GM)/((R−a)^2 )) (1−(1/(8(1−((2a)/R))^2 )))

$$\mathrm{A}\:\mathrm{metal}\:\mathrm{sphere}\:\mathrm{has}\:\mathrm{radius}\:{R}\:\mathrm{and}\:\mathrm{mass}\:{m}.\:\mathrm{A}\:\mathrm{spherical} \\ $$$$\mathrm{hollow}\:\mathrm{of}\:\mathrm{diameter}\:{R}\:\mathrm{is}\:\mathrm{made}\:\mathrm{in}\:\mathrm{this}\:\mathrm{sphere}\:\mathrm{such}\:\mathrm{that}\:\mathrm{its} \\ $$$$\mathrm{surface}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{metal}\:\mathrm{sphere} \\ $$$$\mathrm{and}\:\mathrm{touches}\:\mathrm{the}\:\mathrm{outside}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{metal}\:\mathrm{sphere}. \\ $$$$\mathrm{A}\:\mathrm{unit}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{metal} \\ $$$$\mathrm{sphere}.\:\mathrm{The}\:\mathrm{gravitational}\:\mathrm{field}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}\:\mathrm{is} \\ $$$$\left(\mathrm{a}\right)\:\frac{{GM}}{{R}^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{2}{R}}{{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{b}\right)\:\frac{{GM}}{{a}^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{c}\right)\:\frac{{GM}}{\left({R}+{a}\right)^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{d}\right)\:\frac{{GM}}{\left({R}−{a}\right)^{\mathrm{2}} }\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{2}{a}}{{R}}\right)^{\mathrm{2}} }\right) \\ $$

Commented by MadhumitaSamanta last updated on 13/Jul/22

Answered by aleks041103 last updated on 13/Jul/22

We can say that the metal element is a superposition of a sphere of radius R and mass m and a sphere of diameter R and mass −(m/8) ⇒F=((Gm)/a^2 )+((G(−(m/8)))/((a−R/2)^2 ))=((Gm)/a^2 )(1−(1/(8(1−(R/(2a)))^2 ))) ⇒Ans. (b)

$${We}\:{can}\:{say}\:{that}\:{the}\:{metal}\:{element}\:{is} \\ $$$${a}\:{superposition}\:{of}\:{a}\:{sphere}\:{of}\:{radius}\:{R}\: \\ $$$${and}\:{mass}\:{m}\:{and}\:{a}\:{sphere}\:{of}\:{diameter}\:{R} \\ $$$${and}\:{mass}\:−\frac{{m}}{\mathrm{8}} \\ $$$$\Rightarrow{F}=\frac{{Gm}}{{a}^{\mathrm{2}} }+\frac{{G}\left(−\frac{{m}}{\mathrm{8}}\right)}{\left({a}−{R}/\mathrm{2}\right)^{\mathrm{2}} }=\frac{{Gm}}{{a}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\frac{{R}}{\mathrm{2}{a}}\right)^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{Ans}.\:\left({b}\right) \\ $$

Commented by Tawa11 last updated on 13/Jul/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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