A planet of radius $R = \frac{1}{10} \times (\text{radius of Earth})$ has the same mass density as Earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and a linear mass density $10^{-3}\,\text{kg m}^{-1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $= 6 \times 10^{6}\,\text{m}$ and the acceleration due to gravity on Earth is $10\,\text{ms}^{-2}$)

Step 1: Find the acceleration due to gravity at the surface of the planet.

The acceleration due to gravity at the surface of a planet with uniform mass density $\rho$ is given by:

$$g_s = \frac{4}{3}\pi G\rho R$$

Since the planet has the same mass density as Earth, the surface gravity is directly proportional to its radius:

$$g_s = g_E \left(\frac{R}{R_E}\right)$$

Substitute the given values $g_E = 10\text{ m/s}^2$ and $R = \frac{R_E}{10}$:

$$g_s = 10 \times \frac{1}{10} = 1\text{ m/s}^2$$

Step 2: Express the acceleration due to gravity as a function of depth.

At a distance $r$ from the center of the planet ($r \le R$), the acceleration due to gravity is:

$$g(r) = g_s \frac{r}{R}$$

Let $x$ be the depth below the surface, so that $r = R - x$:

$$g(x) = g_s \left(1 - \frac{x}{R}\right)$$

Step 3: Set up the integral for the total weight of the wire.

The tension $T$ at the top of the wire is equal to the total gravitational force acting on the wire. For an infinitesimal element of length $dx$ at depth $x$, the mass is $dm = \lambda \, dx$.

$$dT = g(x) \, dm = \lambda g_s \left(1 - \frac{x}{R}\right) dx$$

Integrate from the surface $x = 0$ to the bottom of the well $x = \frac{R}{5}$:

$$T = \int_{0}^{R/5} \lambda g_s \left(1 - \frac{x}{R}\right) dx$$ $$T = \lambda g_s \left[ x - \frac{x^2}{2R} \right]_{0}^{R/5}$$ $$T = \lambda g_s \left(\frac{R}{5} - \frac{R}{50}\right)$$ $$T = \lambda g_s \frac{9R}{50}$$

Step 4: Calculate the numerical value of the force.

The radius of the planet is:

$$R = \frac{R_E}{10} = \frac{6 \times 10^6}{10} = 6 \times 10^5\text{ m}$$

Substitute $\lambda = 10^{-3}\text{ kg/m}$, $g_s = 1\text{ m/s}^2$, and $R = 6 \times 10^5\text{ m}$ into the expression for $T$:

$$T = 10^{-3} \times 1 \times \frac{9 \times 6 \times 10^5}{50}$$ $$T = \frac{5400}{50}$$ $$T = 108\text{ N}$$

Answer: The force applied at the top of the wire to hold it in place is $108\text{ N}$.