JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 4)
A block 'A' takes 2 s 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 :
2.66 s
0.83 s
1.68 s
0.70 s
Explanation
$${\theta _1} = 30^\circ ,\,{\theta _2} = 45^\circ $$
$${a_1} = g\sin {\theta _1} = 5$$ m/s2, $${a_2} = g\sin {\theta _2} = 5\sqrt 2 $$ m/s2
$${{{t_1}} \over {{t_2}}} = {{\sqrt {{{2l} \over {{a_1}}}} } \over {\sqrt {{{2l} \over {{a_2}}}} }} = \sqrt {{{{a_2}} \over {{a_1}}}} $$
$${{{t_1}} \over {{t_2}}} = {(2)^{1/4}}$$
$${t_2} = {(2)^{3/4}}$$
$$ \approx 1.68$$ s
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