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 the block on 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 reach the bottom of the two planes.

### Step 1: Determine the distance traveled down the inclined planes

Given that both blocks are released from the same height \( h \):

1. For the first inclined plane (angle \( \theta_1 = 30^\circ \)):
\[ L_1 = \frac{h}{\sin(30^\circ)} = 2h \]

2. For the second inclined plane (angle \( \theta_2 = 45^\circ \)):
\[ L_2 = \frac{h}{\sin(45^\circ)} = h \sqrt{2} \]

### Step 2: Calculate the acceleration of the blocks

The acceleration of the block down the inclined plane is given by:
\[ a = g \sin(\theta) \]

1. For the first inclined plane:
\[ a_1 = g \sin(30^\circ) = g \cdot \frac{1}{2} = \frac{g}{2} \]

2. For the second inclined plane:
\[ a_2 = g \sin(45^\circ) = g \cdot \frac{1}{\sqrt{2}} = \frac{g}{\sqrt{2}} \]

### Step 3: Apply the equation of motion to find time taken

Using the equation of motion \( s = ut + \frac{1}{2} a t^2 \) where \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \]
Thus, we can rearrange to find time \( t \):
\[ t = \sqrt{\frac{2s}{a}} \]

1. For the first inclined plane:
\[ t_1 = \sqrt{\frac{2L_1}{a_1}} = \sqrt{\frac{2 \cdot 2h}{\frac{g}{2}}} = \sqrt{\frac{4h}{\frac{g}{2}}} = \sqrt{\frac{8h}{g}} \]

2. For the second inclined plane:
\[ t_2 = \sqrt{\frac{2L_2}{a_2}} = \sqrt{\frac{2 \cdot h\sqrt{2}}{\frac{g}{\sqrt{2}}}} = \sqrt{\frac{2h\sqrt{2}}{\frac{g}{\sqrt{2}}}} = \sqrt{\frac{4h}{g}} = \sqrt{\frac{4h}{g}} \]

### Step 4: Find the ratio \( \frac{t_1}{t_2} \)

Now we can find the ratio of the times:
\[ \frac{t_1}{t_2} = \frac{\sqrt{\frac{8h}{g}}}{\sqrt{\frac{4h}{g}}} = \sqrt{\frac{8h}{g} \cdot \frac{g}{4h}} = \sqrt{2} = \frac{t_1}{t_2} \]

Thus, we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = \frac{t_1}{\sqrt{2}} \]

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