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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 )))

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

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fdx(√(25))

$${fdx}\sqrt{\mathrm{25}} \\ $$

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A rocket vertically from the surface of the earth with an initil velocity(v_o ) show that its velocity v at height h is given by v_o ^2 −v^2 =((2gh)/(1+(h/R))) where R is radius of earth and g is the acceleration due to gravity at the earth surface

$${A}\:{rocket}\:{vertically}\:{from} \\ $$$${the}\:{surface}\:{of}\:{the}\:{earth} \\ $$$${with}\:{an}\:{initil}\:{velocity}\left({v}_{{o}} \right) \\ $$$${show}\:{that}\:{its}\:{velocity}\:{v} \\ $$$${at}\:{height}\:{h}\:{is}\:{given}\:{by} \\ $$$${v}_{{o}} ^{\mathrm{2}} −{v}^{\mathrm{2}} =\frac{\mathrm{2}{gh}}{\mathrm{1}+\frac{{h}}{{R}}} \\ $$$${where}\:{R}\:{is}\:{radius}\:{of}\:{earth} \\ $$$${and}\:\:{g}\:{is}\:{the}\:{acceleration} \\ $$$${due}\:{to}\:{gravity}\:{at}\:{the}\:{earth} \\ $$$${surface} \\ $$

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with reference to book i am posting some basic question for students. 1)two charged coducting sphere of radius r_(1 ) and r_2 are positvely charged separated at a distance d from each other. force of interaction is F which is correct answer a)F=(1/(4πε))((q_1 q_2 )/d^2 ) b)F>(1/(4πε))((q_1 q_2 )/d^2 ) c)F<(1/(4πε))((q_1 q_2 )/d^2 ) d)none of these 2)same question as 1 but one is +ve q_1 and another is −ve q_2 then which option is correct a)F=F_(columb) b)F>F_(columb) c)F<F_(columb) d)none of these pls give answer with explanatiin..

$${with}\:{reference}\:{to}\:{book}\:\:{i}\:{am}\:{posting}\:{some}\:{basic} \\ $$$${question}\:{for}\:{students}. \\ $$$$\left.\mathrm{1}\right){two}\:{charged}\:{coducting}\:{sphere}\:{of}\:{radius} \\ $$$${r}_{\mathrm{1}\:} {and}\:{r}_{\mathrm{2}} \:{are}\:{positvely}\:{charged}\:{separated} \\ $$$${at}\:{a}\:{distance}\:{d}\:{from}\:{each}\:{other}. \\ $$$${force}\:{of}\:{interaction}\:{is}\:{F} \\ $$$${which}\:{is}\:{correct}\:{answer}\: \\ $$$$\left.{a}\right){F}=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon}\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\left.{b}\right){F}>\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon}\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\left.{c}\right){F}<\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon}\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\left.{d}\right){none}\:{of}\:{these} \\ $$$$\left.\mathrm{2}\right){same}\:{question}\:{as}\:\mathrm{1}\:{but}\:{one}\:{is}\:+{ve}\:{q}_{\mathrm{1}} \:{and} \\ $$$${another}\:{is}\:−{ve}\:{q}_{\mathrm{2}} \:{then}\:{which}\:{option}\:{is}\:{correct} \\ $$$$\left.{a}\right){F}={F}_{{columb}} \\ $$$$\left.{b}\right){F}>{F}_{{columb}} \\ $$$$\left.{c}\right){F}<{F}_{{columb}} \\ $$$$\left.{d}\right){none}\:{of}\:{these} \\ $$$${pls}\:{give}\:{answer}\:{with}\:{explanatiin}.. \\ $$

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two sphere with 10 cm radious and 1 kg mass and distence between this two is 1 m .after what time they will touch each other?

$$\mathrm{two}\:\mathrm{sphere}\:\mathrm{with}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{radious}\:\mathrm{and} \\ $$$$\mathrm{1}\:\mathrm{kg}\:\mathrm{mass}\:\mathrm{and}\:\mathrm{distence}\:\mathrm{between} \\ $$$$\mathrm{this}\:\mathrm{two}\:\mathrm{is}\:\mathrm{1}\:\mathrm{m}\:.\mathrm{after}\:\mathrm{what}\:\mathrm{time}\:\mathrm{they} \\ $$$$\mathrm{will}\:\mathrm{touch}\:\mathrm{each}\:\mathrm{other}? \\ $$

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an elevator of mass 250kg is carrying 3 person whose masses 60kg, 80kg, 100kg and the force exacted by the motion is 5000N. a) With what acceleration will the elevator ascend b) Starting from rest,how far will it go in 5s. (acceleration to gravity G=9.8ms^(−2) ).

$${an}\:{elevator}\:{of}\:{mass}\:\mathrm{250}{kg}\:{is}\:{carrying}\:\mathrm{3}\:{person}\:{whose}\:{masses}\:\mathrm{60}{kg},\:\mathrm{80}{kg},\:\mathrm{100}{kg}\:{and}\:{the}\:{force}\:{exacted}\:{by}\:{the}\:{motion}\:{is}\:\mathrm{5000}{N}. \\ $$$$\left.{a}\right)\:{With}\:{what}\:{acceleration}\:{will}\:{the}\:{elevator}\:{ascend} \\ $$$$\left.{b}\right)\:{Starting}\:{from}\:{rest},{how}\:{far}\:{will}\:{it}\:{go}\:{in}\:\mathrm{5}{s}. \\ $$$$\left({acceleration}\:{to}\:{gravity}\:{G}=\mathrm{9}.\mathrm{8}{ms}^{−\mathrm{2}} \right). \\ $$

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If a planet is suddenly stopped in its orbit, supposed to be circular, then it would fall into the sun in a time (T/(4(√2))), where T is the time period of revolution. Prove this.

$${If}\:{a}\:{planet}\:{is}\:{suddenly}\:{stopped}\:{in}\:{its} \\ $$$${orbit},\:{supposed}\:{to}\:{be}\:{circular},\:{then}\:{it} \\ $$$${would}\:{fall}\:{into}\:{the}\:{sun}\:{in}\:{a}\:{time}\:\frac{{T}}{\mathrm{4}\sqrt{\mathrm{2}}}, \\ $$$${where}\:{T}\:{is}\:{the}\:{time}\:{period}\:{of} \\ $$$${revolution}.\:{Prove}\:{this}. \\ $$

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