Question

Question:-
(Refer to the attachment).
________...
ANTHE 2018 IXth question paper (Physics). ​

Answer 1

Mass is both a property of a physical body and a measure of its resistance to acceleration when a net force is applied. An object's mass also determines the strength of its gravitational attraction to other bodies. The basic SI unit of mass is the kilogram.

Answer 2

Question

Two point masses M and 3M are placed at a distance L apart. Another point mass m is placed in between on the line joining them so that the net gravitational force acting on it due to masses M and 3M is zero. The magnitude of gravitational force acting due to mass M on mass m will be

Answer :

According to Newton's law of gravitation,

$\bigstar$The froce if attraction between two particles is proportional to the product of their masses and inversely proportional to the square of distance between them. .i.e.,

${\boxed{\sf{F = \dfrac{GmM}{r^2}}}}$

where,

F denotes force

G denotes gravity

r denotes distance between m and M

froce on m is zero because net froce acting on any mass at the distance between 2 masses is zero.

According to question,

$:\implies{\sf{F_{mM} = F_{m3M}}}$

$:\implies{\sf{ \dfrac{GmM}{r^2} = \dfrac{Gm3M}{(L - r)^{2} }}}$

$:\implies{\sf{ \dfrac{( L- r)^2}{r^2} = \dfrac{Gm3M}{GmM }}}$

$:\implies{ \sf{{(L - r)}^{2} = 3 {r}^{2} }}$

$:\implies{ \sf{ L- r = \sqrt{3}r }}$

$:\implies{ \sf{ L= \sqrt{3} r + r}}$

$:\implies{ \sf{r = \dfrac{L}{ \sqrt{3} + 1} }}$

we know that,

$:\implies{\sf{F = \dfrac{GmM}{r^2}}}$

$:\implies{\sf{F = \dfrac{GmM}{ \bigg(\dfrac{L}{ \sqrt{3} + 1 } \bigg)^{2} }}}$

$:\implies \underline{\boxed{\purple{\sf F = \dfrac{GmM( \sqrt{3} + 1)^{2} }{ {L}^{2} }}}}$

Hence, option A is correct answer.