A block A takes 2s to slide down a frictionless incline of 30 ° and...

As per the information given in the question the lift is moving upward with uniform velocity. Therefore, the frame of reference attached to the lift is an inertial frame of reference. The acceleration of the block down the incline will be,

$a = gsin\theta$

Using equation of motion for constant acceleration,

$s = ut + \dfrac{1}{2}a{t}^{2}$

$⇒ l = \dfrac{1}{2}gsin30 ° {\left(2\right)}^{2} . . . \left(1\right)$

When the inclination is changed to $45 °$,

$l = \dfrac{1}{2}gsin45 ° {t}^{2} . . . \left(2\right)$

Equating the above equations,

$\dfrac{1}{2}gsin30 ° {\left(2\right)}^{2} = \dfrac{1}{2}gsin45 ° {\left(t\right)}^{2}$

$⇒ \left(\dfrac{1}{2}\right)\left(4\right) = \dfrac{1}{\sqrt{2}}{t}^{2} ⇒ t = \sqrt{2\sqrt{2}} = 1 . 68s$