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

Step 1: Define the surface mass density of the annular disc.

The inner radius is $b = 3R$ and the outer radius is $a = 4R$. The total mass of the disc is $M$.

$$\sigma = \frac{M}{\pi(a^2 - b^2)}$$ $$\sigma = \frac{M}{\pi((4R)^2 - (3R)^2)}$$ $$\sigma = \frac{M}{7\pi R^2}$$

Step 2: Set up the integral for the gravitational potential at point $P$.

Point $P$ is on the axis of the disc at a distance $z = 4R$ from the center. Consider an elemental ring of radius $r$ and width $dr$. Its mass $dm$ is:

$$dm = \sigma(2\pi r \, dr)$$

The gravitational potential $dV$ at point $P$ due to this elemental ring is:

$$dV = -\frac{G \, dm}{\sqrt{r^2 + z^2}}$$ $$dV = -\frac{2\pi G \sigma r \, dr}{\sqrt{r^2 + z^2}}$$

Step 3: Integrate to find the total potential $V(P)$.

Integrate from the inner radius $b$ to the outer radius $a$.

$$V(P) = -2\pi G \sigma \int_{b}^{a} \frac{r \, dr}{\sqrt{r^2 + z^2}}$$ $$V(P) = -2\pi G \sigma \left[ \sqrt{r^2 + z^2} \right]_{b}^{a}$$ $$V(P) = -2\pi G \sigma \left( \sqrt{a^2 + z^2} - \sqrt{b^2 + z^2} \right)$$

Step 4: Substitute the given values.

Substitute $a = 4R$, $b = 3R$, and $z = 4R$ into the square root terms.

$$\sqrt{a^2 + z^2} = \sqrt{(4R)^2 + (4R)^2} = 4\sqrt{2}R$$ $$\sqrt{b^2 + z^2} = \sqrt{(3R)^2 + (4R)^2} = 5R$$

Now substitute these values and $\sigma$ into the expression for $V(P)$.

$$V(P) = -2\pi G \left( \frac{M}{7\pi R^2} \right) \left( 4\sqrt{2}R - 5R \right)$$ $$V(P) = -\frac{2GM}{7R} \left( 4\sqrt{2} - 5 \right)$$

Step 5: Calculate the work required to move a unit mass to infinity.

The work required $W$ to move a unit mass from point $P$ to infinity is given by the potential difference.

$$W = V(\infty) - V(P)$$

Since $V(\infty) = 0$:

$$W = -V(P)$$ $$W = \frac{2GM}{7R} \left( 4\sqrt{2} - 5 \right)$$

Answer: The work required is $\dfrac{2GM}{7R} \left( 4\sqrt{2} - 5 \right)$.