Two particle start together from a point `O` and slide down along straight smooth wire inclined at `30^(@)` and `60^(@)` to the vertical plane and on the same slide of vertical through `O` . The relative acceleration of second with respect to first will be of magnitude

Options

A. `(a)(g)/(2)`
B. `(b)(sqrt3 g)/(2)`
C. `(c)(g)/(sqrt3)`
D. `(d)g`

Explanation

To solve the problem of finding the relative acceleration of the second particle with respect to the first, we will follow these steps:

### Step 1: Identify the angles and the gravitational acceleration components
- We have two particles sliding down two different inclined wires. The first particle is on a wire inclined at \(30^\circ\) to the vertical, and the second particle is on a wire inclined at \(60^\circ\) to the vertical.
- The gravitational acceleration acting on both particles is \(g\).

### Step 2: Calculate the acceleration of each particle
- For the first particle (Particle 1) inclined at \(30^\circ\):
\[ A_1 = g \cdot \cos(30^\circ) = g \cdot \frac{\sqrt{3}}{2} \]

- For the second particle (Particle 2) inclined at \(60^\circ\):
\[ A_2 = g \cdot \cos(60^\circ) = g \cdot \frac{1}{2} \]

### Step 3: Use the formula for relative acceleration
- The relative acceleration of Particle 2 with respect to Particle 1 is given by:
\[ A_{21} = A_2 - A_1 \]

### Step 4: Substitute the values of \(A_1\) and \(A_2\)
- Now substituting the values we calculated:
\[ A_{21} = \left( g \cdot \frac{1}{2} \right) - \left( g \cdot \frac{\sqrt{3}}{2} \right) \]
- This simplifies to:
\[ A_{21} = g \left( \frac{1}{2} - \frac{\sqrt{3}}{2} \right) = g \cdot \frac{1 - \sqrt{3}}{2} \]

### Step 5: Calculate the magnitude of the relative acceleration
- The magnitude of the relative acceleration can be calculated using the Pythagorean theorem since the two accelerations are acting at an angle:
\[ |A_{21}| = \sqrt{A_1^2 + A_2^2 - 2A_1A_2 \cos(\theta)} \]
- Here, \(\theta\) is the angle between the two acceleration vectors, which is \(30^\circ\) (the difference between \(60^\circ\) and \(30^\circ\)).

### Step 6: Substitute the values into the formula
- Substituting the values:
\[ |A_{21}| = \sqrt{\left(g \cdot \frac{\sqrt{3}}{2}\right)^2 + \left(g \cdot \frac{1}{2}\right)^2 - 2 \left(g \cdot \frac{\sqrt{3}}{2}\right) \left(g \cdot \frac{1}{2}\right) \cos(30^\circ)} \]
- This becomes:
\[ |A_{21}| = \sqrt{\frac{3g^2}{4} + \frac{g^2}{4} - 2 \cdot \frac{\sqrt{3}g^2}{4} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}} \]
- Simplifying further:
\[ |A_{21}| = \sqrt{\frac{3g^2}{4} + \frac{g^2}{4} - \frac{3g^2}{8}} \]
- Combine the terms:
\[ |A_{21}| = \sqrt{\frac{4g^2}{4} - \frac{3g^2}{8}} = \sqrt{g^2 - \frac{3g^2}{8}} = \sqrt{\frac{8g^2}{8} - \frac{3g^2}{8}} = \sqrt{\frac{5g^2}{8}} = g \sqrt{\frac{5}{8}} \]

### Final Result
The magnitude of the relative acceleration of the second particle with respect to the first is:
\[ |A_{21}| = \frac{g \sqrt{5}}{2\sqrt{2}} = \frac{g \sqrt{10}}{4} \]