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A block $' \mathrm{A}^{\prime}$ takes $2 \mathrm{~s}$ to slide down a frictionless incline of $30^{\circ}$ and length $\mathrm{ ' l^{\prime}}$ , kept inside a lift going up with uniform velocity $\mathrm{' v '. }$ If the incline is changed to $45^{\circ}$ , the time taken by the block, to slide down the incline, will be approximately :
Option: 1
$2.66\: \mathrm{s}$

Option: 2
$0.83\: \mathrm{s}$

Option: 3
$1.68\: \mathrm{s}$

Option: 4
$0.70\: \mathrm{s}$

Answers (1)

best_answer

For a lift moving with constant velocity there is no pseudo force acting on the block placed inside it due to lifts motion.

from the equation of motion ,

$\mathrm{S=ut+\frac{1}{2}at^{2}}$

$\mathrm{-\ell=0+\frac{1}{2}\left ( -g\sin \Theta \right )t^{2}}$

$\mathrm{t=\sqrt{\frac{2\ell}{g\sin \Theta }}}$

The above equation gives the time taken by block to slide down a frictionless incline of angle $\mathrm{\Theta }$ with the horizontal and length $\mathrm{\ell}$

Case (1) for $\mathrm{\Theta =30\degree,t=2s}$

$\mathrm{t_{1}=2=\sqrt{\frac{2\ell}{g\sin 30\degree}}}$

Case (2)

$\mathrm{t_{2}=\sqrt{\frac{2\ell}{g\sin 45\degree}}}$

$\mathrm{\frac{t_{1}}{t_{2}}=\sqrt{\frac{\sin 45\degree}{\sin 30\degree}}=\sqrt{\frac{1/\sqrt{2}}{1/2}}}$

$\mathrm{\frac{2}{t_{2}}=2^{1/4}}$

$\mathrm{t_{2}=2^{3/4}s}$

$\mathrm{\therefore t_{2}=1.68\: s}$

Hence 3 is correct option

Posted by

vishal kumar

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