JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 6)
Three identical particles $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of mass $$100 \mathrm{~kg}$$ each are placed in a straight line with $$\mathrm{AB}=\mathrm{BC}=13 \mathrm{~m}$$. The gravitational force on a fourth particle $$\mathrm{P}$$ of the same mass is $$\mathrm{F}$$, when placed at a distance $$13 \mathrm{~m}$$ from the particle $$\mathrm{B}$$ on the perpendicular bisector of the line $$\mathrm{AC}$$. The value of $$\mathrm{F}$$ will be approximately :
21 G
100 G
59 G
42 G
Explanation
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m = 100 kg
$${F_{AP}} = {{G{m^2}} \over {{{\left( {13\sqrt 2 } \right)}^2}}}$$
$${F_{BP}} = {{G{m^2}} \over {{{13}^2}}}$$
$${F_{CP}} = {{G{m^2}} \over {{{\left( {13\sqrt 2 } \right)}^2}}}$$
$${F_{net}} = {F_{BP}} + {F_{AP}}\cos 45^\circ + {F_{CP}}\cos 45^\circ $$
$$ = {{G{m^2}} \over {{{13}^2}}}\left( {1 + {1 \over {\sqrt 2 }}} \right)$$
$$ = {{G{{100}^2}} \over {169}}(1 + 0.707)$$
$$ \simeq 100\,G$$
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