1.

(3) Fdistance d on a line. A third point massbe placed at a point on the line such that to igravitational force on it is zero.9. Two point masses m and 4m are separatat thenass(The GraGravity13. Thenum4mt isThe distance of that point from the m mass is(2) 42(3(3) 3(4) 5

Answer»

let the mass z be keptata disance of y units from m,

hence the distance from 4m will be (d-z)units

now since the attractive force from both the masses is equal and opposite so

Gmz / (y2)= G4mz / (d-y)2

or, 1 / y2 = 4 / (d-y)2

or, 1/ y = 2 / (d-y)

or, 2y = d-y

or, 3y = d

or y = d /3

hence the mass should be kept at distance of d/3 units form m particle.

how did 4m became 2m in the next step

please specify the step where u have problem

1/y = 4 /(d-y)2y = d- y it is 4 in the first step nd then how did 2 came in next step


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