Step 1: Understanding the Concept:
A body sliding down a smooth inclined plane moves with constant acceleration. The problem asks for the time taken to cover a fraction of the total distance. Since the body starts from rest, the distance covered is proportional to the square of the time.
Step 2: Key Formula or Approach:
For a body starting from rest ($u=0$) and moving with constant acceleration ($a$), the distance ($s$) covered in time ($t$) is given by:
\[ s = \frac{1}{2}at^2 \]
From this, we can see that $s \propto t^2$, which implies $t \propto \sqrt{s}$.
Step 3: Detailed Explanation:
Let the total length of the inclined plane be L.
Let the time taken to slide the full length L be T = 4 sec.
The acceleration of the body sliding down a smooth inclined plane of angle $\theta$ is $a = g \sin\theta$, which is constant.
Using the kinematic equation for the full journey:
\[ L = \frac{1}{2}aT^2 = \frac{1}{2}a(4^2) = \frac{1}{2}a(16) = 8a \quad \dots(1) \]
Now, let's consider the time ($t_1$) taken to slide a distance of $s_1 = L/4$.
Using the same kinematic equation:
\[ s_1 = \frac{1}{2}at_1^2 \]
\[ \frac{L}{4} = \frac{1}{2}at_1^2 \quad \dots(2) \]
We can solve this by setting up a ratio. Divide equation (2) by equation (1):
\[ \frac{L/4}{L} = \frac{\frac{1}{2}at_1^2}{\frac{1}{2}aT^2} \]
\[ \frac{1}{4} = \frac{t_1^2}{T^2} \]
Take the square root of both sides:
\[ \sqrt{\frac{1}{4}} = \frac{t_1}{T} \]
\[ \frac{1}{2} = \frac{t_1}{T} \]
So, $t_1 = \frac{T}{2}$.
Given that the total time T is 4 seconds:
\[ t_1 = \frac{4}{2} = 2 \text{ seconds} \]
Step 4: Final Answer:
The time taken to slide 1/4th of the length is 2 seconds. Therefore, option (B) is correct.