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A uniform rod of mass m and length l is ...

A uniform rod of mass `m` and length `l` is taken. Find the gravitational field intensity at point `P` at distance `d` which is on the perpendicular bisector of the rod as shown in Fig.

A

`(4Gm)/(d sqrt(4d^(2) + l^(2)))`

B

`(2Gm)/(d sqrt(l^(2)+ 4d^(2)))`

C

`(2sqrt(2)Gm)/(d sqrt(l^(2)+ 4d^(2)))`

D

none of the above

Text Solution

Verified by Experts

The correct Answer is:
B
Consider element of the rod of width `x` as shown in Fig.

Mass of element, `dm = (m)/(l) dx`
Gravitaitonal field intensity at `P` due to this element is
`dI = (Gdm)/(x^(2) + d^(2)) = (G)/((x^(2) + d^(2))) xx (m)/(l) dx` along `PC`
In `deltaOPC, x = d tan theta` and `dx = d sec^(2) theta d theta`
`:.x^(2) + d^(2) = d^(2) (tan^(2) theta + 1) = d^(2) sec^(2) theta`
`:. dI = (Gm)/(l) (d sec^(2) theta d theta)/(d^(2) sec^(2) theta) = (Gm)/(ld) d theta`
resolving `d vecI` into two rectangular components we have `dI cos theta` acts along `PO` and `dI sin theta` acts perpendicular to `PO`.
If we find the gravitaitonal field due to other elements of the rod at `P` and resolve them into two rectangular componets, we note that the components of gravitational field intensity perpendicular to `OP` will cancel out and components of gravitational intensity along `PO` will be added up. Therefore, total gravitational intensity at `P` due to entire rod is
`I = int_(-alpha)^(+alpha) dI cos theta = int_(-alpha)^(+alpha) (Gm)/(ld) cos theta d theta`
`= (Gm)/(ld) [2sin alpha] = (2Gm)/(ld) xx (l//2)/(sqrt((l//2)^(2) + d^(2)))`
`= (2Gm)/(dsqrt(l^(2) + 4d^(2)))`
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