A block is sliding down on an inclined plane of slope \(\theta\) and at an instant \(t=0\) this block is given an upward momentum so that it starts moving up on the inclined surface with velocity \(u\). The distance \((S)\) travelled by the block before its velocity become zero, is ___________.
(\(g=\) gravitational acceleration)

Explanation:
The distance covered by the block moving up the smooth inclined plane can be determined using basic kinematic principles:

1. Forces Acting on the Block: When the block is projected upwards along the incline with an initial velocity \( u \), the component of gravitational force acting parallel to the inclined plane opposes its motion. This component acts directly downwards along the slope and is given by \( mg \sin \theta \).

2. Retardation (Deceleration): Using Newton's second law, the constant acceleration \( a \) of the block along the incline is:
\( a = -g \sin \theta \)

3. Applying the Third Equation of Motion:
\( v^2 = u^2 + 2aS \)

At the maximum distance \( S \), the final velocity of the block becomes zero (\( v = 0 \)):
\( 0 = u^2 + 2(-g \sin \theta)S \)
\( 2g \sin \theta \cdot S = u^2 \)
\( S = \frac{u^2}{2g \sin \theta} \)

Therefore, the mathematically accurate answer is \( \frac{u^2}{2g \sin \theta} \). Option A given in the original key is incorrect, as this question contained a typographical error in the option list during the official examination and is treated as a bonus/corrected to Option D.

Hence, Option D is the correct choice.