Menu Close

Two-balls-each-of-radius-R-equal-mass-and-density-are-placed-in-contact-then-the-force-of-gravitation-between-them-is-proportional-to-1-F-1-R-2-2-F-R-3-F-R-4-4-F-1-R-

Tinku Tara 0 Comments
Pin




Question Number 26959 by Tinkutara last updated on 31/Dec/17
Two balls, each of radius R, equal mass and density are placed in contact, then the force of gravitation between them is proportional to (1) F ∝ (1/R^2 ) (2) F ∝ R (3) F ∝ R^4 (4) F ∝ (1/R)
$$\mathrm{Two}\:\mathrm{balls},\:\mathrm{each}\:\mathrm{of}\:\mathrm{radius}\:{R},\:\mathrm{equal}\:\mathrm{mass} \\ $$$$\mathrm{and}\:\mathrm{density}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{contact},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{force}\:\mathrm{of}\:\mathrm{gravitation}\:\mathrm{between}\:\mathrm{them} \\ $$$$\mathrm{is}\:\mathrm{proportional}\:\mathrm{to} \\ $$$$\left(\mathrm{1}\right)\:{F}\:\propto\:\frac{\mathrm{1}}{{R}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:{F}\:\propto\:{R} \\ $$$$\left(\mathrm{3}\right)\:{F}\:\propto\:{R}^{\mathrm{4}} \\ $$$$\left(\mathrm{4}\right)\:{F}\:\propto\:\frac{\mathrm{1}}{{R}} \\ $$
Answered by prakash jain last updated on 31/Dec/17
mass of each ball=(4/3)ρπR^3 F=((Gm_1 m_2 )/R^2 )∝R^4
$$\mathrm{mass}\:\mathrm{of}\:\mathrm{each}\:\mathrm{ball}=\frac{\mathrm{4}}{\mathrm{3}}\rho\pi\mathrm{R}^{\mathrm{3}} \\ $$$$\mathrm{F}=\frac{{Gm}_{\mathrm{1}} {m}_{\mathrm{2}} }{{R}^{\mathrm{2}} }\propto{R}^{\mathrm{4}} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 31/Dec/17
Yes this answer is correct according to one book but I have another book which says:
Commented by Tinkutara last updated on 31/Dec/17
Commented by prakash jain last updated on 31/Dec/17
Maybe ajfour/mrW1 can comment. In my view if R is increased mass will increase and distance will also increase so force will be proportinal to R^4 . In the question density is contant and not mass.
$$\mathrm{Maybe}\:{ajfour}/\mathrm{mrW1}\:\mathrm{can}\:\mathrm{comment}. \\ $$$$\mathrm{In}\:\mathrm{my}\:\mathrm{view}\:\mathrm{if}\:\mathrm{R}\:\mathrm{is}\:\mathrm{increased} \\ $$$$\mathrm{mass}\:\mathrm{will}\:\mathrm{increase}\:\mathrm{and}\:\mathrm{distance} \\ $$$$\mathrm{will}\:\mathrm{also}\:\mathrm{increase}\:\mathrm{so}\:\mathrm{force}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{proportinal}\:\mathrm{to}\:\mathrm{R}^{\mathrm{4}} . \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{question}\:\mathrm{density}\:\mathrm{is}\:\mathrm{contant} \\ $$$$\mathrm{and}\:\mathrm{not}\:\mathrm{mass}. \\ $$
Commented by Tinkutara last updated on 31/Dec/17
But to me both answers seem correct. But what to do if this question really comes in exam?
Commented by Femmy last updated on 31/Dec/17
it is an inverse square law remember?
$$\mathrm{it}\:\mathrm{is}\:\mathrm{an}\:\mathrm{inverse}\:\mathrm{square}\:\mathrm{law}\:\mathrm{remember}? \\ $$
Commented by Tinkutara last updated on 31/Dec/17
But the book from which I posted this question says (3). What is the mistake if we go by other method?
Commented by prakash jain last updated on 31/Dec/17
R^4 is correct for this question since mass also depends on R. F∝((m_1 m_2 )/r^2 ) Given ρ is constant FαR^4
$$\mathrm{R}^{\mathrm{4}} \:\mathrm{is}\:\mathrm{correct}\:\mathrm{for}\:\mathrm{this}\:\mathrm{question}\:\mathrm{since} \\ $$$$\mathrm{mass}\:\mathrm{also}\:\mathrm{depends}\:\mathrm{on}\:{R}. \\ $$$${F}\propto\frac{{m}_{\mathrm{1}} {m}_{\mathrm{2}} }{{r}^{\mathrm{2}} } \\ $$$$\mathrm{Given}\:\rho\:\mathrm{is}\:\mathrm{constant}\:{F}\alpha{R}^{\mathrm{4}} \\ $$
Commented by Tinkutara last updated on 31/Dec/17
So when the book from which I posted image is correct?
Commented by mrW1 last updated on 31/Dec/17
I think the answer should be (1). The question asks the force between the two balls. They are fix units, concerning mass and size.
$${I}\:{think}\:{the}\:{answer}\:{should}\:{be}\:\left(\mathrm{1}\right). \\ $$$${The}\:{question}\:{asks}\:{the}\:{force}\:{between} \\ $$$${the}\:{two}\:{balls}.\:{They}\:{are}\:{fix}\:{units},\: \\ $$$${concerning}\:{mass}\:{and}\:{size}. \\ $$
Commented by prakash jain last updated on 31/Dec/17
Hi mrW1, The force in this case is proportion to R^4 since mass is dependent on R. Say another sumi lar question is when a ball is melted to form a wire then resistance is proportinal to l^? . Force between two bodies of a given mass is propotional to (1/R^2 ). But when the question says constanst density (and not given mass) force will be propotional to R^4 . What is wrong with my argument?
$$\mathrm{Hi}\:\mathrm{mrW1}, \\ $$$$\mathrm{The}\:\mathrm{force}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{is}\:\mathrm{proportion} \\ $$$$\mathrm{to}\:\mathrm{R}^{\mathrm{4}} \:\mathrm{since}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{dependent}\:\mathrm{on}\:\mathrm{R}. \\ $$$$\mathrm{Say}\:\mathrm{another}\:\mathrm{sumi}\:\mathrm{lar}\:\mathrm{question}\:\mathrm{is}\:\mathrm{when} \\ $$$$\mathrm{a}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{melted}\:\mathrm{to}\:\mathrm{form}\:\mathrm{a}\:\mathrm{wire}\:\mathrm{then} \\ $$$$\mathrm{resistance}\:\mathrm{is}\:\mathrm{proportinal}\:\mathrm{to}\:{l}^{?} . \\ $$$$\mathrm{Force}\:\mathrm{between}\:\mathrm{two}\:\mathrm{bodies}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{given}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{propotional}\:\mathrm{to}\:\frac{\mathrm{1}}{{R}^{\mathrm{2}} }. \\ $$$$\mathrm{But}\:\mathrm{when}\:\mathrm{the}\:\mathrm{question}\:\mathrm{says}\:\mathrm{constanst} \\ $$$$\mathrm{density}\:\left(\mathrm{and}\:\mathrm{not}\:\mathrm{given}\:\mathrm{mass}\right)\:\mathrm{force} \\ $$$$\mathrm{will}\:\mathrm{be}\:\mathrm{propotional}\:\mathrm{to}\:{R}^{\mathrm{4}} . \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{wrong}\:\mathrm{with}\:\mathrm{my}\:\mathrm{argument}? \\ $$
Commented by mrW1 last updated on 31/Dec/17
you are also right sir. the question is just not clear enough. it can be understood differently, depending on if the mass of the balls is seen as constant or not.
$${you}\:{are}\:{also}\:{right}\:{sir}.\:{the}\:{question}\:{is} \\ $$$${just}\:{not}\:{clear}\:{enough}.\:{it}\:{can}\:{be} \\ $$$${understood}\:{differently},\:{depending} \\ $$$${on}\:{if}\:{the}\:{mass}\:{of}\:{the}\:{balls}\:{is}\:{seen} \\ $$$${as}\:{constant}\:{or}\:{not}. \\ $$
Answered by Femmy last updated on 31/Dec/17
=> 1
$$=> \\ $$$$\mathrm{1} \\ $$
Commented by Tinkutara last updated on 31/Dec/17
I want to know which option is really correct.
Answered by ajfour last updated on 31/Dec/17
mass remains constant or density stays constant must be mentioned. Both expeiments are possible. we can choose to change balls for new radii keeping mass the same or density the same.
$${mass}\:{remains}\:{constant}\:{or} \\ $$$${density}\:{stays}\:{constant}\:{must}\:{be} \\ $$$${mentioned}. \\ $$$${Both}\:{expeiments}\:{are}\:{possible}. \\ $$$${we}\:{can}\:{choose}\:{to}\:{change}\:{balls} \\ $$$${for}\:{new}\:{radii}\:{keeping}\:{mass}\:{the} \\ $$$${same}\:{or}\:{density}\:{the}\:{same}. \\ $$
Commented by prakash jain last updated on 31/Dec/17
Question says copper balls so density is constant.
$$\mathrm{Question}\:\mathrm{says}\:\mathrm{copper}\:\mathrm{balls}\:\mathrm{so} \\ $$$$\mathrm{density}\:\mathrm{is}\:\mathrm{constant}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *