JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Hard
- MCQ
A rectangles solid box of length \(0.3\, m\) is held horizontally, with one of its sides on the edge of a platform of height \(5\,m\). When released, it slips off the table in a very short time \(= 0.01\, s\), remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(0.5\)
Step-by-step Solution
Detailed explanation
\(Angular \,impulse = change\, in\, angular\, momentum.\)
\(\tau \,\Delta t = \Delta L\)
\(mg\frac{\ell }{2} \times 0.01 = \frac{{m{\ell ^2}}}{3}\omega \)
\(\omega = \frac{{3g \times 0.01}}{{2\ell }} = \frac{{3 \times 10 \times 0.01}}{{2 \times 0.3}} = \frac{1}{2} = 0.5\,rad/s\)
Time taken by rod to hit the ground
\(t\sqrt {\frac{{2h}}{g}} = \sqrt {\frac{{2 \times 5}}{{10}}} = 1\,\sec .\)
In this time angule rotate by rod
\(\theta = \omega t = 0.5 \times 1 = 0.5\,radian\)
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