AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(d) The given situation is shown in the following figure,
The acceleration given to wedge to the left is a. The block has a pseudo acceleration to the right, pressing against the wedge because of which the block does not move. In this case,
∴‌mg‌sin‌θ=ma‌cos‌θ
⇒‌a=‌

g‌sin‌θ

cos‌θ

‌‌......(i)
Force exerted by the wedge on the block is equal to normal reaction on the block,
i.e. R=mg‌cos‌θ+ma‌sin‌θ
=mg‌cos‌θ+m⋅g‌

sin‌θ

cos‌θ

⋅sin‌θ‌‌ [From Eq. (i)]
‌‌=‌

mg

cos‌θ

[cos2θ+sin2θ]
⇒‌‌R‌‌=‌

mg

cos‌θ

=mg‌sec‌θ