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QJEEM22Ph264b: JEE Main 2022 Physics - Class 11 JEEM22Ph264b Solution: Two inclined planes are placed as shown in figure. A block is…

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(2+1)t(\sqrt{2}+1) s. The value of t is ___________.

(use  g=10 m/s2\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2} )

JEE Main 2022 (Online) 27th July Evening Shift Physics - Laws of Motion Question 44 English

Step-by-Step Verified Solution

AB=102AB = 10\sqrt 2 m

vA=2×10×10=102{v_A} = \sqrt {2 \times 10 \times 10} = 10\sqrt 2 m/s

vC=102{v_C} = 10\sqrt 2 m/s

aBC=gsin⁡(30∘)=5{a_{BC}} = g\sin (30^\circ ) = 5 m/s2

tBC=22 s (vcaBC){t_{BC}} = 2\sqrt 2 \,s\,\left( {{{{v_c}} \over {{a_{BC}}}}} \right)

tAB=vA52=2 s{t_{AB}} = {{{v_A}} \over {5\sqrt 2 }} = 2\,s

tAB+tBC=2(2+1){t_{AB}} + {t_{BC}} = 2(\sqrt 2 + 1)

⇒t=2\Rightarrow t = 2

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