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Consider two configurations of a system ...

Consider two configurations of a system of three particles of masses `m`, `2m` and `3m`. The work done by gravity in changing the configuration of the system from figure (i) to figure (ii) is

A

zero

B

`(6 Gm^(2))/(a){1 + (1)/(sqrt(2))}`

C

`(6 Gm^(2))/(a){1 - (1)/(sqrt(2))}`

D

`(6 Gm^(2))/(a){2 - (1)/(sqrt(2))}`

Text Solution

Verified by Experts

The correct Answer is:
C

`W = - DeltaU = U_(i) - U_(f)`
`= - (Gmm)/(a) [{((1)(2))/(1) +((1) (3))/(1) + ((2)(3))/(sqrt(2))}]`
`-{((1 xx2))/(1) +((1xx3))/(1) + ((2xx3))/(1)}]`
`= (6 GM^(2))/(a)(1- (1)/(sqrt(2)))`
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Knowledge Check

  • The velocity (in m/s) of centre of mass of the system as shown in the figure is :

    A
    `(2 [1 + (4)/(sqrt2)] hat(i) + 8 [(1)/(sqrt2) + 2 ] hat(j))/(7)`
    B
    `(2 [1 + (4)/(sqrt2)] hat(i) + 8 [(1)/(sqrt2) - 2 ] hat(j))/(7)`
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    D
    `(8 [(1)/(sqrt2) - 2] hat(i) + 2 [1 + (4)/(sqrt2] hat(j)])/(7)`
  • Two masses m and M are attached to the strings as shown in the figure. If the system is in equilibrium, then

    A
    `tantheta=1+(2M)/(m)`
    B
    `tantheta=1+(2m)/(M)`
    C
    `cottheta=1+(2M)/(m)`
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  • Two point masses m and M are separated by a distance L. The distance of the centre of mass of the system from m is

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    C
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    D
    `L((m)/(m+M))`
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