In given arrangement as shown in figure, the system is in equilibrium. Choose the correct option`(g = 10 m/s^2)`

Options

A. `T_1=100sqrt(5) N`
B. `theta =tan^-1 (2)`
C. `T_1=100sqrt(3) N`
D. Both (1) and (2)

Answer

To solve the problem step by step, we will analyze the forces acting on the system and apply the conditions for equilibrium.

### Step 1: Identify the Forces
In the given arrangement, we have two blocks with masses:
- Block 1: 20 grams (which is 0.02 kg)
- Block 2: 2 grams (which is 0.002 kg)

These blocks are connected by ropes and are in equilibrium. The forces acting on each block include the gravitational force and the tension in the ropes.

### Step 2: Calculate the Gravitational Forces
Using the formula for weight (W = mg), we can calculate the gravitational forces acting on each block:
- For Block 1 (20 grams):
\[ W_1 = m_1 \cdot g = 0.02 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.2 \, \text{N} \]
- For Block 2 (2 grams):
\[ W_2 = m_2 \cdot g = 0.002 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.02 \, \text{N} \]

### Step 3: Set Up the Equations for Tension
In equilibrium, the tension in the ropes must balance the weight of the blocks:
- Let \( T_2 \) be the tension in the rope connected to Block 1 (20 grams).
- Let \( T_3 \) be the tension in the rope connected to Block 2 (2 grams).

From the equilibrium conditions:
- For Block 1:
\[ T_2 = W_1 = 0.2 \, \text{N} \]
- For Block 2:
\[ T_3 = W_2 = 0.02 \, \text{N} \]

### Step 4: Relate Tensions Using Angles
Assuming the ropes make an angle \( \theta \) with the horizontal:
- The horizontal component of \( T_1 \) (the tension in the rope connected to Block 1) is:
\[ T_1 \cos \theta = T_3 \]
- The vertical component of \( T_1 \) is:
\[ T_1 \sin \theta = T_2 \]

### Step 5: Substitute the Tension Values
Substituting the values of \( T_2 \) and \( T_3 \):
- From \( T_1 \cos \theta = 0.02 \):
\[ T_1 \cos \theta = 0.02 \quad \text{(1)} \]
- From \( T_1 \sin \theta = 0.2 \):
\[ T_1 \sin \theta = 0.2 \quad \text{(2)} \]

### Step 6: Divide the Equations
Dividing equation (2) by equation (1):
\[ \frac{T_1 \sin \theta}{T_1 \cos \theta} = \frac{0.2}{0.02} \]
This simplifies to:
\[ \tan \theta = 10 \]

### Step 7: Calculate \( \theta \)
To find \( \theta \):
\[ \theta = \tan^{-1}(10) \]

### Step 8: Calculate \( T_1 \)
Using either equation (1) or (2) to find \( T_1 \):
From equation (1):
\[ T_1 = \frac{0.02}{\cos \theta} \]
Substituting \( \theta \):
\[ T_1 = \frac{0.02}{\cos(\tan^{-1}(10))} \]
Using the identity \( \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1+x^2}} \):
\[ T_1 = \frac{0.02}{\frac{1}{\sqrt{1+10^2}}} = 0.02 \sqrt{101} \]

### Conclusion
Thus, the values for \( T_1 \) and \( \theta \) are:
- \( T_1 = 0.02 \sqrt{101} \)
- \( \theta = \tan^{-1}(10) \)