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A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is,
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A)2GM7R(425)B)2GM7R(425)C)GM4RD)2GM5R(21)

Answer
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Hint: We must know that to bring a unit mass from point P to infinity, work has to be done against the gravitational field of the disc. So, we will find the gravitational potential of a small element of thickness dr. Then by finding the potential at P due to this small element and integrating it from the limit of inner radius to outer, we can find the work required to take a unit mass from point P to infinity.

Complete step-by-step Solution:
Let us consider that the work needed to take a unit mass from point P to is Vp where, Vp is the gravitational potential due to the disc. For finding Vp, we will consider a small element of thickness dr from the disc .

Now, let us take such an element at distance r from the center of the disc as shown in the figure.
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Now, the mass of the small element is given by,
dm=M(2πrdr)π(4R)2π(3R)2=2Mrdr7R2

So, the gravitational potential due to the entire disc is given by integrating the potential due to the number of these small elements within the limit 3R to 4R. That is,
Vp=3R4RGdmr2+16R2

Here, the distance of a small element from P is found using Pythagoras theorem.
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Now, substituting dm in the above equation,
Vp=3R4R2MGrdr7R2r2+16R2=2MG7R23R4Rrdrr2+16R2
Let us assume r2+16R2 as x2. That is,
r2+16R2=x2

Differentiating both sides, we will get
2rdr=2xdxrdr=xdx
Now, when r=3R, x=9R2+16R2=5R
Now, when r=4R, x=16R2+16R2=42R

So, our limits will be changing according to this.
Vp=2MG7R25R42Rxdxr2+16R2Vp=2MG7R25R42Rdx=2MG7R2[x]5R42RVp=2MG7R2(42R5R)
Or we can write this as, Vp=2MG7R(425)

Therefore the work required to take a unit mass from point P to infinity is found to be Vp=2MG7R(425).

So, the option B is correct.

Note:
The key element of this type of questions is understanding that the work needed to be done will be equal to the potential of the system. The negative sign will be there as work needed to be done against the potential. Also, the tricky part is changing the limits of the integration to our desired variable. So, we must be very careful while doing it.
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