- 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: Â Uniformly Accelerated Motion |
 60%
From NCERT
JEE
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NEET 2026 - Target Batch - Vital
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NEET 2026 - Target Batch - Vital
- 1(current)
Q. No.
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