A block released from rest from the tope of a smooth inclined plane of angle of inclination `theta_(1) = 30^(@)` reaches the bottom in time `t_(1)`. The same block, released from rest from the top of another smooth inclined plane of angle of inclination `theta_(2) = 45^(@)` reaches the bottom in time `t_(2)`. Time `t_(1)` and `t_(2)` are related as ( Assume that the initial heights of blocks in the two cases are equal ).

Options

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

Answer

To solve the problem, we will analyze the motion of a block sliding down two different inclined planes with angles of inclination \( \theta_1 = 30^\circ \) and \( \theta_2 = 45^\circ \). We will derive the relationship between the times \( t_1 \) and \( t_2 \) taken by the block to slide down each incline.

### Step 1: Understand the setup
We have two inclined planes with angles \( \theta_1 \) and \( \theta_2 \). The block is released from rest from the same height \( h \) in both cases. We need to find the relationship between the times \( t_1 \) and \( t_2 \).

### Step 2: Determine the length of the incline
The length \( L \) of the incline can be related to the height \( h \) and the angle \( \theta \) using the sine function:
\[ L = \frac{h}{\sin(\theta)} \]
Thus, for the two cases:
\[ L_1 = \frac{h}{\sin(\theta_1)} \quad \text{and} \quad L_2 = \frac{h}{\sin(\theta_2)} \]

### Step 3: Calculate the acceleration along the incline
The acceleration \( a \) of the block along the incline is given by:
\[ a = g \sin(\theta) \]
where \( g \) is the acceleration due to gravity. Therefore, for the two inclines:
\[ a_1 = g \sin(\theta_1) \quad \text{and} \quad a_2 = g \sin(\theta_2) \]

### Step 4: Use the kinematic equation to find time
Using the kinematic equation \( s = ut + \frac{1}{2} a t^2 \) where \( u = 0 \) (initial velocity), we have:
\[ L = \frac{1}{2} a t^2 \]
Rearranging for time \( t \):
\[ t = \sqrt{\frac{2L}{a}} \]
Substituting for \( L \) and \( a \):
\[ t_1 = \sqrt{\frac{2L_1}{g \sin(\theta_1)}} \quad \text{and} \quad t_2 = \sqrt{\frac{2L_2}{g \sin(\theta_2)}} \]

### Step 5: Substitute \( L_1 \) and \( L_2 \)
Substituting \( L_1 \) and \( L_2 \):
\[ t_1 = \sqrt{\frac{2 \left( \frac{h}{\sin(\theta_1)} \right)}{g \sin(\theta_1)}} = \sqrt{\frac{2h}{g \sin^2(\theta_1)}} \]
\[ t_2 = \sqrt{\frac{2 \left( \frac{h}{\sin(\theta_2)} \right)}{g \sin(\theta_2)}} = \sqrt{\frac{2h}{g \sin^2(\theta_2)}} \]

### Step 6: Find the ratio \( \frac{t_2}{t_1} \)
Now, we can find the ratio of the times:
\[ \frac{t_2}{t_1} = \frac{\sqrt{\frac{2h}{g \sin^2(\theta_2)}}}{\sqrt{\frac{2h}{g \sin^2(\theta_1)}}} = \frac{\sin(\theta_1)}{\sin(\theta_2)} \]

### Step 7: Substitute the angles
Substituting \( \theta_1 = 30^\circ \) and \( \theta_2 = 45^\circ \):
\[ \sin(30^\circ) = \frac{1}{2} \quad \text{and} \quad \sin(45^\circ) = \frac{1}{\sqrt{2}} \]
Thus,
\[ \frac{t_2}{t_1} = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{1}{2} \cdot \sqrt{2} = \frac{\sqrt{2}}{2} \]

### Final Result
The relationship between the times \( t_1 \) and \( t_2 \) is:
\[ \frac{t_2}{t_1} = \frac{\sqrt{2}}{2} \]