Three blocks A, B and C of masses 5 kg, 5 kg and 10 kg respectively are connected with two light strings, one of them is passing over an ideal pulley The system is released from rest on smooth inclined plane of inclination 30" as shown in figure. The value of tension `T_1 /T_2` is

Options

A. ` 1/2`
B. 2
C. `sqrt (3) / 2`
D. `2 /sqrt (3)`

Answer

To find the ratio of tensions \( \frac{T_1}{T_2} \) in the given system of blocks, we will analyze the forces acting on each block and apply Newton's second law.

### Step-by-Step Solution:

1. **Identify the Forces on Each Block:**
- Block A (mass = 5 kg) and Block B (mass = 5 kg) are on the inclined plane.
- Block C (mass = 10 kg) is hanging vertically.
- The angle of inclination \( \theta = 30^\circ \).

2. **Calculate the Gravitational Force Components:**
- For Block A and Block B, the gravitational force acting down the incline is given by:
\[ F_{\text{gravity, incline}} = m \cdot g \cdot \sin(\theta) \]
- For both Block A and Block B:
\[ F_{\text{gravity, incline}} = 5 \cdot 9.8 \cdot \sin(30^\circ) = 5 \cdot 9.8 \cdot 0.5 = 24.5 \, \text{N} \]

3. **Set Up the Equations of Motion:**
- For Block A:
\[ T_2 - m_A \cdot g \cdot \sin(30^\circ) = m_A \cdot a \]
\[ T_2 - 24.5 = 5a \quad \text{(1)} \]

- For Block B (similar to Block A):
\[ T_2 - m_B \cdot g \cdot \sin(30^\circ) = m_B \cdot a \]
\[ T_2 - 24.5 = 5a \quad \text{(2)} \]

- For Block C:
\[ T_1 - m_C \cdot g = -m_C \cdot a \]
\[ T_1 - 10 \cdot 9.8 = -10a \]
\[ T_1 = 98 - 10a \quad \text{(3)} \]

4. **Equate the Tension Equations:**
- From equations (1) and (2), we see that both give the same expression for \( T_2 \):
\[ T_2 = 24.5 + 5a \]
- Substitute \( T_2 \) into equation (3):
\[ T_1 = 98 - 10a \]

5. **Find the Acceleration \( a \):**
- Since the system is connected, we can express the net force:
\[ T_1 - 2T_2 = 0 \quad \text{(since both A and B are pulling against C)} \]
- Substitute \( T_1 \) and \( T_2 \):
\[ (98 - 10a) - 2(24.5 + 5a) = 0 \]
\[ 98 - 10a - 49 - 10a = 0 \]
\[ 49 = 20a \]
\[ a = \frac{49}{20} = 2.45 \, \text{m/s}^2 \]

6. **Calculate \( T_1 \) and \( T_2 \):**
- Substitute \( a \) back into \( T_2 \):
\[ T_2 = 24.5 + 5(2.45) = 24.5 + 12.25 = 36.75 \, \text{N} \]
- Substitute \( a \) back into \( T_1 \):
\[ T_1 = 98 - 10(2.45) = 98 - 24.5 = 73.5 \, \text{N} \]

7. **Find the Ratio \( \frac{T_1}{T_2} \):**
\[ \frac{T_1}{T_2} = \frac{73.5}{36.75} = 2 \]

### Final Answer:
The value of \( \frac{T_1}{T_2} \) is \( 2 \).

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