Physics - NLM Question with Solution | TestHub

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,

$a=g\sin \theta$

Using equation of motion for constant acceleration,

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

$\Rightarrow l=\frac{1}{2}g\sin 30°{\left(2\right)}^2 ...\left(1\right)$

When the inclination is changed to $45°$,

$l=\frac{1}{2}g\sin 45°t^2 ...\left(2\right)$

Equating the above equations,

$\frac{1}{2}g\sin 30°{\left(2\right)}^2=\frac{1}{2}g\sin 45°{\left(t\right)}^2$

$$\begin{gathered} \Rightarrow \left(\frac{1}{2}\right)\left(4\right)=\frac{1}{\sqrt{2}}{t}^2 \\ \Rightarrow t=\sqrt{2\sqrt{2}}=1.68 \mathrm{s} \end{gathered}$$