JEE Main 2022 · 27 Jul · Shift 2 · Physics Q60

← JEE Main 2022 · 27 Jul · Shift 2

PhysicsInteger+4 / -1Question 60 of 90

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t\left(\sqrt2+1\right)$ s. The value of t is ______.

(use g = 10 m/s $^2$ )

Two smooth inclines meeting at B, 10 m high: AB at 45° on the left and BC at 30° on the right.

Your answer

Answer and worked solution

Official answer: 2

Up AB ( $10\sqrt2$ m, deceleration $5\sqrt2$ ): reverse of starting from rest, $t_1=\sqrt{\frac{2\cdot10\sqrt2}{5\sqrt2}}=2$ s.

Down BC (20 m, acceleration 5): $t_2=\sqrt8=2\sqrt2$ s.

Total $2(\sqrt2+1)$ s.

Answer: 2

Topics:Newton's laws Motion in a straight line

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JEE Main 2022 · 27 Jul · Shift 2, Physics question 60