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A straight of length L extends from x=a ...

A straight of length L extends from x=a to x=L+a. the gravitational force it exerts on a point mass 'm' at x=0, if the mass per unit length of the rod is `A+Bx^(2), ` is given by :

A

`Gm [A((1)/(a+L)-(1)/(A))-BL]`

B

`Gm [A((1)/(a)-(1)/(a+L))+BL]`

C

`Gm [A((1)/(a+L)-(1)/(a))+BL]`

D

`Gm [A((1)/(a)-(1)/(a+L))-BL]`

Text Solution

Verified by Experts

The correct Answer is:
B
dm=rdx
`=(A+Bx^2)dx`
`F=int_(a)^(a+L)(Gmd M)/(x^2)=Gm int_(a)^(a+L)((A+Bx^2dx)dx)/(x^2)`
`=Gm[int_(a)^(a+L) (A)/(x^2)dx+a int_(a)^(a+L)Bdx]`
`F=4m[A((1)/(a)-(1)/(a+L))]=BL`
`=GM[A((L)/(a(a+L)))+Bl]`
.
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