A planet of radius `R=(1)/(10)xx(radius of Earth)` has the same mass density as Earth. Scientists dig a well of depth`(R )/(5)` on it and lower a wire of the same length and a linear mass density `10^(-3) 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 inplace is (take the radius of Earth`=6xx10^6m` and the acceleration due to gravity on Earth is `10ms^(-2)`

Options

A. (a)96N
B. (b)108N
C. (c)120N
D. (d)150N

Explanation

To solve the problem step by step, we need to find the force applied at the top of the wire by a person holding it in place. Here's how we can approach it:

### Step 1: Determine the radius of the planet
The radius of the planet \( R \) is given as:
\[ R = \frac{1}{10} \times R_E \]
where \( R_E \) is the radius of the Earth. Given \( R_E = 6 \times 10^6 \, \text{m} \):
\[ R = \frac{1}{10} \times 6 \times 10^6 = 6 \times 10^5 \, \text{m} \]

### Step 2: Calculate the depth of the well
The depth of the well is given as:
\[ \text{Depth} = \frac{R}{5} = \frac{6 \times 10^5}{5} = 1.2 \times 10^5 \, \text{m} \]

### Step 3: Calculate the mass density of the planet
Since the planet has the same mass density as Earth, we can use the density of Earth \( \rho_E \). The mass density \( \rho \) of the planet is:
\[ \rho = \rho_E \]

### Step 4: Calculate the gravitational acceleration at the depth of the well
The gravitational acceleration \( g' \) at a depth \( d \) inside a planet can be calculated using the formula:
\[ g' = g_E \left(1 - \frac{d}{R}\right) \]
where \( g_E = 10 \, \text{m/s}^2 \) is the gravitational acceleration at the surface of the Earth. Here \( d = 1.2 \times 10^5 \, \text{m} \) and \( R = 6 \times 10^5 \, \text{m} \):
\[ g' = 10 \left(1 - \frac{1.2 \times 10^5}{6 \times 10^5}\right) = 10 \left(1 - 0.2\right) = 10 \times 0.8 = 8 \, \text{m/s}^2 \]

### Step 5: Calculate the length of the wire
The length of the wire \( L \) is equal to the depth of the well:
\[ L = 1.2 \times 10^5 \, \text{m} \]

### Step 6: Calculate the mass of the wire
The linear mass density \( \mu \) of the wire is given as \( 10^{-3} \, \text{kg/m} \). Therefore, the mass \( m \) of the wire can be calculated as:
\[ m = \mu \times L = 10^{-3} \times 1.2 \times 10^5 = 120 \, \text{kg} \]

### Step 7: Calculate the force applied at the top of the wire
The force \( F \) applied at the top of the wire is given by:
\[ F = m \cdot g' = 120 \times 8 = 960 \, \text{N} \]

### Conclusion
The force applied at the top of the wire by a person holding it in place is \( 960 \, \text{N} \).