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A mass M at rest is broken into two piec...

A mass M at rest is broken into two pieces having masses m and (M-m). The two masses are then separated by a distance r. The gravitational force between them will be maximum when the ratio of the masses [m : (M-m) of the two parts is

A

`1:1`

B

`1:2`

C

`1:3`

D

`1:4`

Text Solution

Verified by Experts

The correct Answer is:
A
Using. `F=(Gm_(1)m_(2))/(r^(2))`(Universal law of gravitation)
Here, `m_(1)=m` and `m_(2)=(M-m) :. F=G(m(M-m))/(r^(2))=0`
or `M-2m=0`
`rArr m=(M)/(2) :. (m)/((M-m))=(M//2)/((M-M//2))=(1)/(1)`
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Knowledge Check

  • A mass M splits into two parts m and (M - m) , which are then separated by a certain distance. Wha ratio (m/M) maximize the gravitational force between the path ?

    A
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    B
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    C
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    D
    `(1)/(3)`

  • The reduce mass of two particles having masses m and 2 m is

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    2 m
    B
    3 m
    C
    2 m/3
    D
    m/2

  • The gravitational force between two particles of masses m_(1) and m_(2) separeted by the same distance, then the gravitational force between them will be

    A
    greater than `F`
    B
    less than `F`
    C
    `F`
    D
    zero

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