JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 15)
A rectangular solid box of length 0.3 m is held
horizontally, with one of its sides on the edge
of a platform of height 5m. When released, it
slips off the table in a very short time t = 0.01s,
remaining essentially horizontal. The angle by
which it would rotate when it hits the ground
will be (in radians) close to :-
_8th_April_Evening_Slot_en_15_1.png)
0.28
0.02
0.3
0.5
Explanation
Angular impulse = change in angular momentum.
$$mg{l \over 2} \times 0.01 = {{m{l^2}} \over 3}\omega $$
$$\omega = {{3g \times 0.01} \over {2l}} = {{3 \times 10 \times 0.01} \over {2 \times 0.3}}$$
$$ = $$ $${1 \over 2} = 0.5\,rad/s$$
Time taken by rod to hit the ground
$$t = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 5} \over {10}}} = 1\,\sec .$$
In this time angle rotate by rod
$$\theta $$ = $$\omega $$t = 0.5 × 1 = 0.5 radian
$$mg{l \over 2} \times 0.01 = {{m{l^2}} \over 3}\omega $$
$$\omega = {{3g \times 0.01} \over {2l}} = {{3 \times 10 \times 0.01} \over {2 \times 0.3}}$$
$$ = $$ $${1 \over 2} = 0.5\,rad/s$$
Time taken by rod to hit the ground
$$t = \sqrt {{{2h} \over g}} = \sqrt {{{2 \times 5} \over {10}}} = 1\,\sec .$$
In this time angle rotate by rod
$$\theta $$ = $$\omega $$t = 0.5 × 1 = 0.5 radian
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