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Gravitation

Section: Physics

Question : 16 of 21

Marks: +1, -0

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

[JEE Adv 2010 P1]

$\frac{2GM}{7R} (4\sqrt{2}-5)$
- $\frac{2GM}{7R} (4\sqrt{2}-5)$
$\frac{GM}{4R}$
$\frac{2GM}{5R} (\sqrt{2}-1)$

Validate

Solution:

The sine component of gravitational force from a mass element da at a distance 3R + dr and its counterpart at opposite end cancel each other. Thus, the total gravitational force at point p is the mass density σ, which is given as

σ =

\frac{M}{2\pi R^{2}(16-9)}

\frac{M}{14\pi R^{2}}

Now, V = -

\int\limits_{3R}^{4R}\frac{G\sigma 2\pi r dr}{\sqrt{r^{2}+16R^{2}}}

-G 2\pi \int\limits_{3R}^{4R}\frac{r dr}{\sqrt{r^{2}+16R^{2}}}

V =

G\sigma 2\pi \left| \sqrt{r^{2}+16R^{2}} \right|_{3R}^{4R}

G\sigma 2\pi R (4\sqrt{2}-5)

Substituting the value of σ, we get

V =

\frac{2GM}{7R} (4\sqrt{2}-5)

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