Question 24 of 210

A block $A$ takes $2s$ to slide down a frictionless incline of $30^{\circ}$ and length $l$, kept inside a lift going up with uniform velocity $v$. If the incline is changed to $45^{\circ}$, the time taken by the block, to slide down the incline, will be approximately:

A$2.66s$B$0.83s$C$1.68s$D$0.70s$

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Answer: (C) $1.68s$

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,
$s=ut+\dfrac{1}{2}a{t}^{2}$
$\Rightarrow l=\dfrac{1}{2}g\sin 30^{\circ}{\left(2\right)}^{2}...\left(1\right)$
When the inclination is changed to $45^{\circ}$,
$l=\dfrac{1}{2}g\sin 45^{\circ}{t}^{2}...\left(2\right)$
Equating the above equations,
$\dfrac{1}{2}g\sin 30^{\circ}{\left(2\right)}^{2}=\dfrac{1}{2}g\sin 45^{\circ}{\left(t\right)}^{2}$
$\Rightarrow \left(\dfrac{1}{2}\right)\left(4\right)=\dfrac{1}{\sqrt{2}}{t}^{2} \Rightarrow t=\sqrt{2\sqrt{2}}=1.68s$