Question

Two particles each of mass m are placed at A and C as such AB = BC = L. The gravitational force on the third particle placed at D at a distance L metre on the perpendicular bisector of the line AC, is –

• A. $\frac { \mathrm { Gm } ^ { 2 } } { \sqrt { 2 } \mathrm {~L} ^ { 2 } }$ along BD
• B. $\frac { \mathrm { Gm } ^ { 2 } } { \sqrt { 2 } \mathrm {~L} ^ { 2 } }$along DB
• C. $\frac { \mathrm { Gm } ^ { 2 } } { \mathrm {~L} ^ { 2 } }$along AC
• D. None of these

Answer 1

Answer: (B)

$\frac { \mathrm { Gm } ^ { 2 } } { \sqrt { 2 } \mathrm {~L} ^ { 2 } }$along DB

Answer 2

Angle ÐADC = 90⁰

Þ distance AC = DC = $\sqrt { 2 }$l

Þ F_AB ^ F_CD