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Question

Two masses 4 g and 12 g, are placed horizontally at a distance of 12 cm from each other. Another mass of 0.5 g is placed in between them such that there is no net force acting on it. Find the distance (in cm) of the third mass from the mass of 4 g.

A
6+63 cm
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B
+6+63 cm
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C
+663 cm
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D
63 cm
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Solution

The correct option is A 6+63 cm
Given:
Mass of the body 1, m1=4 g
Mass of the body 2, m2=12 g
Mass of the body 3, m3=0.5 g
Let the third mass is placed at a distance x cm from m1 such that there is no net force acting on it.
If r13 is the distance between m1 and m3, and r23 is the distance between m2 and m3, then for the net force on m3 to be zero:
F13=F23
Gm1m3r213=Gm2m3r223
m1r213=m2r223
m1m2=r213r223
412=x2(12x)2
3x2=x224x+144
2x2+24x144=0
x2+12x72=0
x=6±63
But, x>0. So,
x=6+63 cm

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