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In the figure if block A and wedge B will move with same acceleration, then the magnitude of normal reaction between the block and the wedge will be (There is no friction between block and the wedge and the wedge moves on horizontal surface as shown.)

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2 mg/cos $θ$

2 mg cos $θ$

mg cos $θ$

none of these

answer is A.

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Detailed Solution

FBD of block A

$N \cos θ = 2 mg - - - - - - - - - - - (i)$

[For accelerations of both the block A and wedge B be same, they will move along the horizontal]

$N \sin θ = ma - - - - - - (ii)$

$\cos θ = \frac{2 g}{a} - - - - - - (iii)$

a = 2g tan $θ$

From (ii), we get

$N = \frac{m}{\sin θ} . 2 \tan θ \Rightarrow N = \frac{2 mg}{\cos θ}$

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