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Q.

In the arrangement shown, the wedge W is moving to the right with constant acceleration ‘a’ and the 2kg block remains stationary relative to the wedge. All surfaces are frictionless. Then the normal reaction of the wedge on the block is g=10m/s2

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a

40 N

b

403 N

c

203 N

d

103 N

answer is A.

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Detailed Solution

(1)Vertical component of acceleration of the block is zero. So net vertical force acting on the block must be zero.∴   N  cos600−mg=0⇒   N=2 ×  10cos 600N=40N.
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