- 1(current)
Q. No.Two inclined planes are placed as shown in the figure. A block is projected from the Point \(A\) of the inclined plane \(AB\) along its surface with a velocity just sufficient to carry it to the top point \(B\) at a height of \(10~\text m.\) After reaching the point \(B\) the block slides down on the inclined plane \(BC.\) The time it takes to reach the point \(C\) from the point \(A\) is \(t( \sqrt 2 + 1 )~\text{s}.\) The value of \(t\) is:
(Take \(g=10~\text{m/s}^2\) )
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
(Take \(g=10~\text{m/s}^2\) )
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)
Subtopic: Uniformly Accelerated Motion |
To view explanation, please take trial in the course below.
NEET 2026 - Target Batch - Vital
To view explanation, please take trial in the course below.
NEET 2026 - Target Batch - Vital
Please attempt this question first.
Launched MCQ Practice Books
Prefer Books for Question Practice? Get NEETprep's Unique MCQ Books with Online Audio/Video/Text Solutions via Telegram Bot
NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology- 1(current)
Q. No.
© 2025 GoodEd Technologies Pvt. Ltd.