ExamBro
ExamBro
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.
Verified Solution

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

\(\theta  = \omega t = 0.5 \times 1 = 0.5\,radian\)

Same subject
Explore more questions on app