Question

please answer these questions

Answer 1

Here,
By using FBD of the block of mass √2m and by equating forces in vertical direction we get.

$mg \cos( \alpha ) + mg \cos( \alpha ) = ( \sqrt{2} m)g$

So,
$2mg \cos( \alpha ) = mg \sqrt{2}$

$\cos( \alpha ) = \frac{1}{ \sqrt{2} }$

so
$\alpha = {45}^{0}$

Answer 2

here 120 , 45 and option d of que 3 is the answer by tricky methods