An iron sphere of mass `100 kg` is suspended freely from a rigid support by means of a rope of length `2m`.The horizontal force required to displace it horizontally through `50cm` is nearly

Options

A. `980 N`
B. `490 N`
C. `245 N`
D. `112.5 N`

Explanation

To solve the problem of finding the horizontal force required to displace an iron sphere of mass 100 kg horizontally by 50 cm when suspended from a rope of length 2 m, we can follow these steps:

### Step 1: Understand the Setup
The iron sphere is suspended from a rigid support by a rope. When a horizontal force \( F \) is applied, the sphere is displaced horizontally by 50 cm. The rope forms an angle \( \theta \) with the vertical.

### Step 2: Draw a Diagram
Draw a diagram showing the sphere, the rope, and the horizontal displacement. Label the following:
- Length of the rope \( L = 2 \, \text{m} \)
- Vertical displacement \( y = L - \sqrt{L^2 - (0.5)^2} \)
- Horizontal displacement \( x = 0.5 \, \text{m} \)

### Step 3: Calculate the Angle \( \theta \)
Using the sine function, we can find the angle \( \theta \):
\[ \sin \theta = \frac{x}{L} = \frac{0.5}{2} = \frac{1}{4} \]
Thus, \( \theta = \sin^{-1}\left(\frac{1}{4}\right) \).

### Step 4: Resolve Forces
At equilibrium, the forces acting on the sphere can be resolved into two components:
- Vertical component: \( T \cos \theta = mg \)
- Horizontal component: \( F = T \sin \theta \)

Where:
- \( T \) is the tension in the rope,
- \( m = 100 \, \text{kg} \) (mass of the sphere),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 5: Calculate the Weight of the Sphere
Calculate the weight \( mg \):
\[ mg = 100 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 981 \, \text{N} \]

### Step 6: Relate the Forces
From the vertical component:
\[ T \cos \theta = 981 \, \text{N} \quad \text{(1)} \]
From the horizontal component:
\[ F = T \sin \theta \quad \text{(2)} \]

### Step 7: Divide the Two Equations
Dividing equation (2) by equation (1):
\[ \frac{F}{T \cos \theta} = \frac{T \sin \theta}{T \cos \theta} \]
This simplifies to:
\[ F = mg \tan \theta \]

### Step 8: Calculate \( \tan \theta \)
Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ \sin \theta = \frac{1}{4} \]
Using Pythagorean theorem:
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \]
Thus:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}} \]

### Step 9: Substitute Back to Find \( F \)
Now substitute \( \tan \theta \) back into the equation for \( F \):
\[ F = mg \tan \theta = 981 \times \frac{1}{\sqrt{15}} \]

### Step 10: Calculate the Final Value
Calculating \( F \):
\[ F \approx 981 \times 0.2582 \approx 253.5 \, \text{N} \]

### Conclusion
The horizontal force required to displace the sphere horizontally through 50 cm is approximately **250 N**.