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A thin uniform disc (see figure) of mas...

A thin uniform disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass for point P on its axis to infinity is

A

`(2GM)/(7R)(4sqrt2-5)`

B

`-(2GM)/(7R)(4sqrt2-5)`

C

`(GM)/(4R)`

D

`(2GM)/(5R)(sqrt2-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Work required to take unit mass from infinity to point P =`V_P` br> `therefore` Work required to take it from P to infinity =`-V_P`
To calculate `V_P` , consider a ring element of radius r and width dr . Gravitational potential due to this ring at P is :
`dV_P =(-G(dm))/sqrt(r^2+(4R)^2)=(-G)/sqrt(r^2+16R^2) (M(2prdr))/((16pR^2-9pR^2)) =(-2GM)/(7R^2) (rdr)/sqrt(r^2+16R^2)`
`therefore V_P=intdV_P =(-2GM)/(7R^2) int_(3R)^(4R) (rdr)/sqrt(r^2+16R^2)`
Let `r^2+16R^2 =t^2 rArr 2r(dr)/(dt) = 2t rArr rdr = tdt`
`therefore int(rdr)/sqrt(r^2+16R^2) = int(tdt)/sqrt(t^2) =intdt=t =sqrt(r^2+16R^2)`
`therefore V_P=(-2GM)/(7R^2) [sqrt(r^2+16R^2)]_(3R)^(4R) =(-2GM)/(7R^2) [sqrt(32R^2) - sqrt(25R^2)]`
`=(-2GM)/(7R) (4sqrt2-5) therefore W_"reqd"=-V_P =(2GM)/(7R) (4sqrt2-5)`
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