A block released from rest from the top of a smooth inclined plane of angle `theta_1` reaches the bottom in time `t_1`. The same block released from rest from the top of another smooth inclined plane of angle `theta_2` reaches the bottom in time `t_2` If the two inclined planes have the same height, the relation between `t_1 and t_2` is

Options

A. `t_2/t_1=((sin theta_1)/(sin theta_2))^(1//2)`
B. `t_2/t_1=1`
C. `t_2/t_1=(sin theta_1)/(sin theta_2)`
D. `t_2/t_1 =(sin^2 theta_1)/(sin^2 theta_2)`

Answer

To solve the problem, we need to analyze the motion of a block sliding down two different inclined planes with the same height but different angles. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We have two inclined planes with angles θ₁ and θ₂, both having the same height (H). A block is released from rest at the top of each incline, and we want to find the relationship between the times taken to slide down each incline, t₁ and t₂.

### Step 2: Identify the Forces
When the block is on the incline, the gravitational force acting on it can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline (causing the block to slide down): \( mg \sin \theta \)

### Step 3: Determine the Acceleration
The acceleration of the block down each incline can be expressed as:
- For incline 1 (angle θ₁):
\[ a_1 = g \sin \theta_1 \]
- For incline 2 (angle θ₂):
\[ a_2 = g \sin \theta_2 \]

### Step 4: Relate Height to Length of Incline
Using trigonometry, the length of the incline can be related to the height:
- For incline 1:
\[ L_1 = \frac{H}{\sin \theta_1} \]
- For incline 2:
\[ L_2 = \frac{H}{\sin \theta_2} \]

### Step 5: Use Kinematic Equations
Using the kinematic equation for motion under constant acceleration, where initial velocity \( u = 0 \):
\[ s = ut + \frac{1}{2} a t^2 \]
We can write for both inclines:
- For incline 1:
\[ L_1 = \frac{1}{2} g \sin \theta_1 t_1^2 \]
- For incline 2:
\[ L_2 = \frac{1}{2} g \sin \theta_2 t_2^2 \]

### Step 6: Substitute Lengths
Substituting the expressions for \( L_1 \) and \( L_2 \):
- For incline 1:
\[ \frac{H}{\sin \theta_1} = \frac{1}{2} g \sin \theta_1 t_1^2 \]
- For incline 2:
\[ \frac{H}{\sin \theta_2} = \frac{1}{2} g \sin \theta_2 t_2^2 \]

### Step 7: Set Up Ratios
Now, we can set up the ratios of the two equations:
\[ \frac{H/\sin \theta_1}{H/\sin \theta_2} = \frac{\frac{1}{2} g \sin \theta_1 t_1^2}{\frac{1}{2} g \sin \theta_2 t_2^2} \]
This simplifies to:
\[ \frac{\sin \theta_2}{\sin \theta_1} = \frac{\sin \theta_1 t_1^2}{\sin \theta_2 t_2^2} \]

### Step 8: Solve for Time Ratio
Cross-multiplying gives us:
\[ \sin^2 \theta_2 t_1^2 = \sin^2 \theta_1 t_2^2 \]
Taking the square root of both sides, we find:
\[ \frac{t_1}{t_2} = \frac{\sin \theta_2}{\sin \theta_1} \]
Thus, we can express the relationship as:
\[ \frac{t_2}{t_1} = \frac{\sin \theta_1}{\sin \theta_2} \]

### Final Result
The relation between \( t_1 \) and \( t_2 \) is:
\[ \frac{t_2}{t_1} = \frac{\sin \theta_1}{\sin \theta_2} \]