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Physics

Mechanics

Easy

Question

Four particles of masses m, 2 m, 3 m and 4 m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be

$\frac{24 m^{2} G}{a^{2}}$
$\frac{6 m^{2} G}{a^{2}}$
$\frac{4 \sqrt{2} G m^{2}}{a^{2}}$
zero

The correct answer is: $fraction numerator 4 square root of 2 G m squared over denominator a squared end fraction$

If two particles of mass m are placed x distance apart then force of attraction $fraction numerator G m m over denominator x squared end fraction equals F$ (Let)
$F subscript P A end subscript equals$ force at point P due to particle $A equals fraction numerator G m m over denominator x squared end fraction equals F$
Similarly, $F subscript P B end subscript equals fraction numerator G 2 m m over denominator x squared end fraction equals 2 F comma F subscript P C end subscript equals fraction numerator G 3 m m over denominator x squared end fraction equals 3 F$ and $F subscript P D end subscript equals fraction numerator G 4 m m over denominator x squared end fraction equals 4 F$
$therefore F with rightwards arrow on top subscript n e t end subscript equals 2 square root of 2 fraction numerator G m m over denominator x squared end fraction equals 2 square root of 2 fraction numerator G m squared over denominator open parentheses fraction numerator a over denominator square root of 2 end fraction close parentheses squared end fraction equals fraction numerator 4 square root of 2 G m squared end root over denominator a squared end fraction$

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