Two masses 4g and 12g, are placed horizontally at a distance of 12cm from each other. Another mass of 0.5g 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 4g.
A
−6+6√3cm
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B
+6+6√3cm
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C
+6−6√3cm
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D
6√3cm
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Solution
The correct option is A−6+6√3cm Given:
Mass of the body 1, m1=4g
Mass of the body 2, m2=12g
Mass of the body 3, m3=0.5g
Let the third mass is placed at a distance xcm 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(12−x)2 3x2=x2−24x+144 2x2+24x−144=0 x2+12x−72=0 ⟹x=−6±6√3
But, x>0. So, x=−6+6√3cm